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G12 ELITE STREAM  |  AP Physics C: E&M

End of Term 3 — Question Bank

239 MCQs + 95 FRQs  |  25 Objectives  |  Free Access

Q1 [1 mark]

What is the SI unit of electromotive force (EMF)?

AAmpere (A)
BVolt (V)
COhm (Ω)
DWeber (Wb)
Answer
B
EMF is measured in Volts (V). 1 V = 1 J C⁻¹.
Q2 [1 mark]

Which of the following correctly expresses 1 Volt in base SI units?

Akg·m·s⁻²·A⁻¹
Bkg·m²·s⁻³·A⁻¹
Ckg·m²·s⁻²·A⁻¹
Dkg·m·s⁻³·A⁻²
Answer
B
1 V = 1 J C⁻¹ = 1 kg m² s⁻³ A⁻¹.
Q3 [1 mark]

A battery maintains a potential difference of 12 V across its terminals. This means it supplies how many joules per coulomb?

A1.2 J/C
B12 J/C
C120 J/C
D0.12 J/C
Answer
B
Voltage = energy per unit charge. 12 V = 12 J C⁻¹.
Q4 [1 mark]

Which quantity has the same SI unit as EMF?

AElectric current
BElectric charge
CElectric potential difference
DElectric resistance
Answer
C
Both EMF and electric potential difference are measured in Volts (V).
Q5 [1 mark]

The SI unit of EMF can also be written as:

AW/A
BJ/A
CC/s
DN/C
Answer
A
Power P = εI, so ε = P/I, giving units W A⁻¹ = V.
Q6 [1 mark]

Which of the following is NOT equivalent to 1 Volt?

A1 J·C⁻¹
B1 W·A⁻¹
C1 N·m·C⁻¹
D1 kg·m·s⁻²·A⁻¹
Answer
D
1 V = 1 J C⁻¹ = 1 W A⁻¹ = 1 N m C⁻¹. Option D has wrong dimensions.
Q7 [1 mark]

A source of EMF does 48 J of work moving 4 C of charge around a circuit. What is the EMF?

A4 V
B8 V
C12 V
D48 V
Answer
C
ε = W/Q = 48/4 = 12 V.
Q8 [1 mark]

The EMF of a source is defined as:

AThe current it drives through a circuit
BThe work done per unit charge by the source
CThe resistance of the source
DThe power dissipated in the external circuit
Answer
B
EMF = work done per unit charge = W/Q, measured in Volts.
Q9 [1 mark]

If a source moves 0.5 C of charge and does 3 J of work, its EMF is:

A1.5 V
B3.0 V
C6.0 V
D0.17 V
Answer
C
ε = W/Q = 3/0.5 = 6.0 V.
Q10 [1 mark]

Which of the following devices is the primary source of EMF in a simple DC circuit?

AResistor
BCapacitor
CBattery
DInductor
Answer
C
A battery converts chemical energy to electrical energy, acting as a source of EMF.
Q11 [4 marks]

What is the SI unit of voltage (EMF)?

Mark Scheme
The SI unit of EMF (electromotive force) is the **Volt (V)**. 1 V = 1 J C⁻¹ = 1 kg m² s⁻³ A⁻¹ (in base SI units).
Q12 [4 marks]

A source of EMF does 60 J of work in moving 5 C of charge around a circuit. (a) Calculate the EMF of the source. (b) State the SI unit of your answer.

Mark Scheme
(a) ε = W/Q = 60/5 = **12 V** (b) The SI unit is the **Volt (V)**.
Q13 [4 marks]

Explain the physical meaning of EMF and distinguish it from terminal voltage.

Mark Scheme
EMF (electromotive force) is the work done per unit charge by the source in moving charge around the complete circuit: ε = W/Q. Terminal voltage is the potential difference across the terminals when current flows; it is less than EMF due to internal resistance r: V_terminal = ε − Ir.
Q14 [4 marks]

State two equivalent expressions for the unit Volt (V) in terms of other SI units.

Mark Scheme
1 V = 1 J C⁻¹ (energy per unit charge) 1 V = 1 W A⁻¹ (power per unit current)
Q1 [1 mark]

Magnetic flux through a loop is defined as:

AΦ = B/A
BΦ = BA cos θ
CΦ = BA sin θ
DΦ = B²A
Answer
B
Φ_B = BA cos θ, where θ is the angle between B and the area normal vector.
Q2 [1 mark]

A flat circular loop of area 0.04 m² is in a uniform field B = 2.0 T perpendicular to the plane of the loop. What is the magnetic flux?

A0.02 Wb
B0.04 Wb
C0.08 Wb
D0.16 Wb
Answer
C
B perpendicular to plane → parallel to normal → θ = 0°. Φ = BA = 2.0 × 0.04 = 0.08 Wb.
Q3 [1 mark]

A loop is placed in a magnetic field B such that the field is parallel to the plane of the loop. The flux through the loop is:

ABA
BBA/2
C0
D2BA
Answer
C
B parallel to plane → perpendicular to normal (θ = 90°). Φ = BA cos 90° = 0.
Q4 [1 mark]

A circular loop of radius 0.1 m is in a uniform field B = 0.5 T. The field makes an angle of 30° with the normal to the loop. What is the flux?

A0.01360 Wb
B0.00785 Wb
C0.00681 Wb
D0.01571 Wb
Answer
A
A = π(0.1)² = 0.03142 m². Φ = BA cosθ = 0.5 × 0.03142 × cos 30° = 0.5 × 0.03142 × 0.866 = 0.01360 Wb.
Q5 [1 mark]

The SI unit of magnetic flux is:

ATesla (T)
BHenry (H)
CWeber (Wb)
DAmpere (A)
Answer
C
Magnetic flux is measured in Webers (Wb). 1 Wb = 1 T·m².
Q6 [1 mark]

A square loop of side 0.2 m is placed in a field B = 0.3 T at 60° to the plane of the loop. What is the flux?

A0.0052 Wb
B0.0104 Wb
C0.0120 Wb
D0.0060 Wb
Answer
B
Angle to plane = 60° → angle to normal = 30°. A = 0.04 m². Φ = 0.3 × 0.04 × cos 30° = 0.0104 Wb.
Q7 [1 mark]

Which of the following increases the magnetic flux through a loop?

ARotating the loop so B is parallel to the plane
BIncreasing the area of the loop
CDecreasing the magnetic field strength
DTilting the loop to 90° from the field
Answer
B
Φ = BA cos θ. Increasing area A directly increases flux.
Q8 [1 mark]

A rectangular loop (0.3 m × 0.4 m) is in a field B = 1.5 T perpendicular to the loop. What is the flux?

A0.12 Wb
B0.18 Wb
C0.24 Wb
D0.36 Wb
Answer
B
A = 0.3 × 0.4 = 0.12 m². Φ = 1.5 × 0.12 × cos 0° = 0.18 Wb.
Q9 [1 mark]

If the magnetic flux through a loop is zero, which of the following must be true?

AB = 0
BA = 0
CB is parallel to the plane of the loop
DThe loop is circular
Answer
C
Φ = 0 when cos θ = 0 → θ = 90° between B and normal → B is parallel to the plane.
Q10 [1 mark]

A circular loop of radius r is in a field B at angle θ to the plane of the loop. The flux is:

ABπr² cos θ
BBπr² sin θ
CBπr² tan θ
DBπr²
Answer
B
θ from plane → angle from normal = 90°−θ → cos(90°−θ) = sin θ. Φ = Bπr² sin θ.
Q11 [4 marks]

A circular loop of radius r is placed in a magnetic field B at angle θ to the plane of the loop. What is the magnetic flux through the loop?

Mark Scheme
Φ_B = B·A·cos α, where α is the angle between B and the area normal. Since θ is measured from the **plane**, α = 90° − θ. Area: A = πr² ∴ Φ_B = Bπr² cos(90° − θ) = **Bπr² sin θ**
Q12 [4 marks]

A rhombus-shaped loop of side L is placed in a magnetic field B at angle θ to the normal of the loop. What is the magnetic flux?

Mark Scheme
Here θ is measured from the **normal**, so use cos θ directly. Area of rhombus with equal sides L (square rhombus): A = L² ∴ Φ_B = **BL² cos θ**
Q13 [4 marks]

A rectangular loop (length 0.5 m, width 0.3 m) is in a uniform field B = 0.8 T. (a) Find the flux when B is perpendicular to the plane of the loop. (b) Find the flux when B makes an angle of 40° with the normal to the loop.

Mark Scheme
(a) θ = 0°: Φ = BA = 0.8 × 0.15 = **0.12 Wb** (b) θ = 40°: Φ = 0.12 × cos 40° = 0.12 × 0.766 = **0.092 Wb**
Q14 [4 marks]

Explain why magnetic flux is maximum when B is perpendicular to the plane of the loop and zero when B is parallel to the plane.

Mark Scheme
Φ_B = BA cos θ, where θ is the angle between B and the area normal. • B ⊥ plane → B ∥ normal → θ = 0° → cos 0° = 1 → Φ = BA (maximum) • B ∥ plane → B ⊥ normal → θ = 90° → cos 90° = 0 → Φ = 0 (minimum)
Q1 [1 mark]

Faraday's Law states that the induced EMF in a loop is:

Aε = dΦ/dt
Bε = −dΦ/dt
Cε = BLv
Dε = LdI/dt
Answer
B
Faraday's Law: ε = −dΦ_B/dt. The negative sign indicates Lenz's Law.
Q2 [1 mark]

A loop has magnetic flux Φ_B = 3t² − 2t (Wb). What is the induced EMF at t = 2 s?

A−8 V
B−10 V
C8 V
D10 V
Answer
B
dΦ/dt = 6t − 2. At t = 2: dΦ/dt = 10. ε = −10 V.
Q3 [1 mark]

The magnetic flux through a 50-turn coil changes from 0.02 Wb to 0.08 Wb in 0.1 s. What is the magnitude of the induced EMF?

A0.6 V
B30 V
C3 V
D60 V
Answer
B
|ε| = N|ΔΦ/Δt| = 50 × (0.08−0.02)/0.1 = 50 × 0.6 = 30 V.
Q4 [1 mark]

A loop has flux Φ_B = 5 sin(2t) Wb. What is the induced EMF at t = π/4 s?

A10 V
B−10 V
C0 V
D5 V
Answer
C
dΦ/dt = 10 cos(2t). At t = π/4: cos(π/2) = 0. ε = 0 V.
Q5 [1 mark]

Which of the following will NOT increase the induced EMF in a coil?

AIncreasing the number of turns
BIncreasing the rate of change of flux
CUsing a stronger magnet
DIncreasing the resistance of the coil
Answer
D
Resistance affects current, not EMF. EMF = −N dΦ/dt.
Q6 [1 mark]

A single-turn loop has flux Φ_B = 4t² + 2 Wb. What is the induced EMF at t = 3 s?

A−24 V
B24 V
C−38 V
D38 V
Answer
A
dΦ/dt = 8t. At t = 3: dΦ/dt = 24. ε = −24 V.
Q7 [1 mark]

The flux through a loop changes at a rate of 0.5 Wb/s. The loop has 40 turns. What is the magnitude of the induced EMF?

A0.5 V
B20 V
C40 V
D80 V
Answer
B
|ε| = N|dΦ/dt| = 40 × 0.5 = 20 V.
Q8 [1 mark]

A loop has flux Φ_B = 4t² − 4t (Wb). At what time is the induced EMF zero?

At = 0 s
Bt = 0.5 s
Ct = 1.0 s
Dt = 2.0 s
Answer
B
dΦ/dt = 8t − 4 = 0 → t = 0.5 s.
Q9 [1 mark]

Faraday's Law applies to:

AOnly circular loops
BOnly rectangular loops
CAny closed conducting loop
DOnly loops in uniform fields
Answer
C
Faraday's Law applies to any closed conducting loop regardless of shape.
Q10 [1 mark]

A 200-turn coil (area = 0.01 m²) is in a field B = 0.5 sin(100t) T. What is the maximum induced EMF?

A1 V
B100 V
C50 V
D200 V
Answer
B
ε_max = NBAω = 200 × 0.5 × 0.01 × 100 = 100 V.
Q11 [1 mark]

A loop has flux Φ_B = 4t² − 4t (Wb). What is the induced EMF at t = 3 s?

A−20 V
B20 V
C−24 V
D24 V
Answer
A
dΦ/dt = 8t − 4. At t = 3: 24 − 4 = 20. ε = −20 V.
Q12 [4 marks]

State and explain Faraday's Law of Induction.

Mark Scheme
**Faraday's Law:** ε = −dΦ_B/dt The induced EMF in a closed loop equals the negative rate of change of magnetic flux through the loop. The **negative sign** (Lenz's Law) means the induced EMF drives a current whose magnetic field opposes the change in flux — a consequence of conservation of energy.
Q13 [4 marks]

A loop has magnetic flux Φ_B = 4t² − 4t (Wb). Calculate the induced EMF at t = 3 s.

Mark Scheme
dΦ_B/dt = 8t − 4 ε = −dΦ_B/dt = −(8t − 4) At t = 3 s: ε = −(24 − 4) = **−20 V** Magnitude of induced EMF = 20 V
Q14 [4 marks]

A loop has magnetic flux Φ_B = 4t² + 2 (Wb). Calculate the induced EMF at t = 3 s.

Mark Scheme
dΦ_B/dt = 8t ε = −8t At t = 3 s: ε = −8(3) = **−24 V** Magnitude = 24 V
Q15 [4 marks]

A 100-turn rectangular coil (area = 0.05 m²) is in a field that changes at 0.4 T/s. (a) Calculate the induced EMF. (b) If the coil has resistance 20 Ω, find the induced current.

Mark Scheme
(a) dΦ/dt = A × dB/dt = 0.05 × 0.4 = 0.02 Wb/s ε = N × dΦ/dt = 100 × 0.02 = **2 V** (b) I = ε/R = 2/20 = **0.1 A**
Q1 [1 mark]

Lenz's Law states that the induced current flows in a direction that:

AMaximises the flux change
BOpposes the change in magnetic flux
CIs always clockwise
DIs always counter-clockwise
Answer
B
Lenz's Law: the induced current opposes the change in magnetic flux that produced it.
Q2 [1 mark]

Lenz's Law is a consequence of which conservation principle?

AConservation of momentum
BConservation of charge
CConservation of energy
DConservation of angular momentum
Answer
C
If the induced current aided the change, it would create a perpetual increase of energy — violating conservation of energy.
Q3 [1 mark]

A bar magnet is pushed toward a conducting loop. The induced current in the loop will:

ACreate a magnetic field that attracts the magnet
BCreate a magnetic field that repels the magnet
CNot create any magnetic field
DCreate a field perpendicular to the magnet
Answer
B
By Lenz's Law, the induced current opposes the increase in flux, so it repels the approaching magnet.
Q4 [1 mark]

The negative sign in Faraday's Law ε = −dΦ/dt represents:

AThe direction of the magnetic field
BLenz's Law — the opposition to flux change
CEnergy loss in the circuit
DThe decreasing current
Answer
B
The negative sign encodes Lenz's Law — the induced EMF opposes the change in flux.
Q5 [1 mark]

A magnet is pulled away from a coil. The induced current will:

ARepel the magnet
BAttract the magnet
CHave no effect on the magnet
DReverse the polarity of the magnet
Answer
B
When flux decreases (magnet moving away), the induced current creates a field to maintain the flux, attracting the magnet.
Q6 [1 mark]

A conducting ring is dropped through a region of uniform magnetic field. As it enters the field, the induced current:

AAccelerates the ring downward
BHas no effect on the ring's motion
CDecelerates the ring
DReverses the direction of the field
Answer
C
The induced current creates a force opposing the motion (Lenz's Law), decelerating the ring.
Q7 [1 mark]

In a generator, the mechanical work done to rotate the coil against the opposing magnetic torque is converted into:

AHeat only
BElectrical energy
CKinetic energy
DPotential energy
Answer
B
Lenz's Law means work is done against the opposing torque; this work is converted to electrical energy.
Q8 [1 mark]

A loop is in a decreasing magnetic field directed out of the page. By Lenz's Law, the induced current will flow:

AClockwise to oppose the decrease
BCounter-clockwise to oppose the decrease
CClockwise to aid the decrease
DCounter-clockwise to aid the decrease
Answer
B
Flux out of page decreasing → induced current creates flux out of page → counter-clockwise.
Q9 [1 mark]

Which of the following is an application of Lenz's Law?

AElectromagnetic braking in trains
BOhm's Law in resistors
CCharging a capacitor
DResonance in LC circuits
Answer
A
Electromagnetic braking uses Lenz's Law — induced currents create opposing forces that slow the train.
Q10 [1 mark]

A magnet is held stationary inside a coil. The induced EMF is:

AMaximum
BZero
CEqual to the battery EMF
DProportional to B
Answer
B
No change in flux → no induced EMF. EMF requires dΦ/dt ≠ 0.
Q11 [4 marks]

State Faraday's Law of electromagnetic induction.

Mark Scheme
The induced EMF in any closed circuit equals the negative of the time rate of change of the magnetic flux through the circuit: **ε = −dΦ_B/dt**
Q12 [4 marks]

State Lenz's Law.

Mark Scheme
The direction of the induced current is such that the magnetic field it creates **opposes** the change in magnetic flux that produced it. In other words, the induced current always acts to maintain the original flux.
Q13 [4 marks]

Lenz's Law is a consequence of which conservation law? Explain why.

Mark Scheme
Lenz's Law is a consequence of **Conservation of Energy**. If the induced current aided the change in flux, it would create more flux, inducing more current — an impossible perpetual increase of energy with no external work done. By opposing the change, the induced current ensures that external work must be done to change the flux, consistent with energy conservation.
Q14 [4 marks]

A conducting ring falls toward a horizontal bar magnet with its N pole pointing up. (a) Describe the direction of the induced current (viewed from above) as the ring approaches. (b) Describe the force on the ring. (c) Explain how this is consistent with energy conservation.

Mark Scheme
(a) Upward flux increases → induced current creates downward flux → clockwise (viewed from above). (b) The induced current creates a S pole facing the N pole of the magnet → repulsion → force is upward, opposing downward motion. (c) Work must be done by gravity against the opposing force. This work is converted to electrical energy (heat) in the ring — consistent with energy conservation.
Q1 [1 mark]

A circular loop in a uniform magnetic field directed INTO the page has its area DECREASING. The induced current flows:

ACounter-clockwise
BClockwise
CNo current flows
DAlternating
Answer
B
Flux into page decreasing → induced B must be into page → clockwise.
Q2 [1 mark]

A circular loop in a uniform magnetic field directed INTO the page has its area INCREASING. The induced current flows:

ACounter-clockwise
BClockwise
CNo current flows
DAlternating
Answer
A
Flux into page increasing → induced B must be out of page → counter-clockwise.
Q3 [1 mark]

A rectangular loop in a uniform magnetic field directed OUT OF the page has its area DECREASING. The induced current flows:

ACounter-clockwise
BClockwise
CNo current flows
DAlternating
Answer
A
Flux out of page decreasing → induced B must be out of page → counter-clockwise.
Q4 [1 mark]

A rectangular loop in a uniform magnetic field directed OUT OF the page has its area INCREASING. The induced current flows:

ACounter-clockwise
BClockwise
CNo current flows
DAlternating
Answer
B
Flux out of page increasing → induced B must be into page → clockwise.
Q5 [1 mark]

The rate of change of area of a loop in a field B = 0.5 T is 0.02 m²/s. What is the induced EMF?

A0.01 V
B0.025 V
C0.05 V
D0.1 V
Answer
A
ε = B × dA/dt = 0.5 × 0.02 = 0.01 V.
Q6 [1 mark]

A loop shrinks in a field B = 2 T at a rate of 0.05 m²/s. The induced EMF magnitude is:

A0.025 V
B0.1 V
C0.4 V
D2.5 V
Answer
B
ε = B × |dA/dt| = 2 × 0.05 = 0.1 V.
Q7 [1 mark]

A conducting loop is squeezed so its area decreases from 0.04 m² to 0.01 m² in 0.1 s in a field B = 0.5 T. The induced EMF is:

A0.075 V
B0.15 V
C0.3 V
D0.5 V
Answer
B
ΔΦ = 0.5 × (0.04 − 0.01) = 0.015 Wb. ε = 0.015/0.1 = 0.15 V.
Q8 [1 mark]

A loop in a field B into the page is being stretched (area increasing). The induced current creates a magnetic field inside the loop that is:

AInto the page
BOut of the page
CParallel to the plane of the loop
DZero
Answer
A
Flux into page increases → induced B into page (oppose) → clockwise → B inside is into page.
Q9 [1 mark]

Which of the following changes to a loop in a constant magnetic field would induce an EMF?

AMoving the loop parallel to B
BRotating the loop so its projected area onto B changes
CMoving the loop perpendicular to B without changing orientation
DKeeping the loop stationary
Answer
B
Rotating the loop changes the effective area (and thus flux), inducing an EMF.
Q10 [4 marks]

A circular loop is in a uniform magnetic field directed into the page (×). The area of the loop decreases at a constant rate. Determine the direction of the induced current and explain using Lenz's Law.

Mark Scheme
Flux into the page is **decreasing** (Φ = BA, A is decreasing). By Lenz's Law, the induced current must oppose this decrease by creating B into the page. By right-hand rule (thumb pointing into page): current flows **Clockwise**.
Q11 [4 marks]

A circular loop is in a uniform magnetic field directed into the page (×). The area of the loop increases at a constant rate. Determine the direction of the induced current.

Mark Scheme
Flux into the page is **increasing**. By Lenz's Law, induced current must create B out of the page (oppose increase). By right-hand rule: current flows **Counter-clockwise**.
Q12 [4 marks]

A rectangular loop is in a uniform magnetic field directed out of the page (•). The area decreases at a constant rate. Determine the direction of the induced current.

Mark Scheme
Flux out of the page is **decreasing**. Induced current must create B out of the page (oppose decrease). By right-hand rule: current flows **Counter-clockwise**.
Q13 [4 marks]

A rectangular loop is in a uniform magnetic field directed out of the page (•). The area increases at a constant rate. Determine the direction of the induced current.

Mark Scheme
Flux out of the page is **increasing**. Induced current must create B into the page (oppose increase). By right-hand rule: current flows **Clockwise**.
Q1 [1 mark]

A bar magnet (N pole facing down) is moved toward a horizontal coil from above. Viewed from above, the induced current flows:

AClockwise
BCounter-clockwise
CNo current
DAlternating
Answer
B
N pole approaching from above → upward flux increases → coil creates N pole facing up → counter-clockwise.
Q2 [1 mark]

A bar magnet (N pole facing down) is moved away from a horizontal coil upward. Viewed from above, the induced current flows:

AClockwise
BCounter-clockwise
CNo current
DAlternating
Answer
A
N pole moves away → downward flux decreases → coil creates S pole facing up (attract) → clockwise.
Q3 [1 mark]

A bar magnet (S pole facing up) is moved toward a horizontal coil from below. Viewed from above, the induced current flows:

AClockwise
BCounter-clockwise
CNo current
DAlternating
Answer
B
S pole approaching from below → downward flux increases → coil creates S pole facing down (repel) → counter-clockwise.
Q4 [1 mark]

A bar magnet (S pole facing up) is moved away from a horizontal coil downward. Viewed from above, the induced current flows:

AClockwise
BCounter-clockwise
CNo current
DAlternating
Answer
A
S pole moves away → downward flux decreases → coil creates N pole facing down (attract) → clockwise.
Q5 [1 mark]

A magnet is held stationary inside a coil. The induced EMF is:

AMaximum
BZero
CEqual to battery EMF
DProportional to B
Answer
B
No change in flux → no induced EMF.
Q6 [1 mark]

A bar magnet is moved toward a coil. The coil repels the magnet. This is because:

AThe coil becomes a S pole facing the approaching N pole
BThe coil becomes a N pole facing the approaching N pole
CThe induced current creates an attractive force
DThe coil has no magnetic effect
Answer
B
By Lenz's Law, the coil creates a N pole facing the approaching N pole, causing repulsion.
Q7 [1 mark]

A magnet is dropped through a vertical copper tube. Compared to free fall, the magnet:

AFalls faster
BFalls at the same speed
CFalls slower
DStops immediately
Answer
C
Lenz's Law: induced currents create opposing forces, slowing the magnet below free fall speed.
Q8 [1 mark]

The induced EMF in a coil is greatest when the magnet is:

AStationary inside the coil
BMoving fastest through the coil
CFar from the coil
DParallel to the coil axis
Answer
B
ε = −dΦ/dt. Greatest rate of flux change occurs at maximum speed.
Q9 [1 mark]

A coil has 500 turns. A magnet causes the flux through each turn to change at 0.002 Wb/s. The induced EMF is:

A0.002 V
B0.4 V
C1.0 V
D250 V
Answer
C
ε = N × dΦ/dt = 500 × 0.002 = 1.0 V.
Q10 [4 marks]

A bar magnet (S pole on top, N pole on bottom) is held above a horizontal coil and moved DOWN toward the coil. Determine the direction of the induced current viewed from above.

Mark Scheme
N pole approaches from above → upward flux through coil **increases**. Coil must create a N pole facing up to oppose (repel approaching N pole). By right-hand rule (thumb up): current flows **Counter-clockwise** when viewed from above.
Q11 [4 marks]

A bar magnet (S pole on top, N pole on bottom) is held above a horizontal coil and moved UP away from the coil. Determine the direction of the induced current viewed from above.

Mark Scheme
N pole moves away → downward flux **decreases**. Coil must create a S pole facing up to maintain flux (attract magnet). Thumb down → current flows **Clockwise** when viewed from above.
Q12 [4 marks]

A bar magnet (S pole on top, N pole on bottom) is held below a horizontal coil and moved UP toward the coil. Determine the direction of the induced current viewed from above.

Mark Scheme
S pole approaches from below → downward flux through coil **increases**. Coil must create a S pole facing down to repel. Thumb up → current flows **Counter-clockwise** when viewed from above.
Q13 [4 marks]

A bar magnet (S pole on top, N pole on bottom) is held below a horizontal coil and moved DOWN away from the coil. Determine the direction of the induced current viewed from above.

Mark Scheme
S pole moves away → downward flux **decreases**. Coil must create a N pole facing down to attract. Thumb down → current flows **Clockwise** when viewed from above.
Q14 [4 marks]

A conducting ring falls toward a bar magnet with its N pole pointing upward. (a) As the ring approaches, describe the direction of the induced current (viewed from above). (b) What is the direction of the force on the ring? (c) As the ring moves away from the magnet, what is the direction of the induced current?

Mark Scheme
(a) Upward flux increases → induced current creates downward flux → **clockwise** (viewed from above). (b) The induced current creates a S pole facing the N pole → **repulsion** → force is upward (opposing downward motion). (c) Upward flux decreases → induced current creates upward flux → **counter-clockwise** (viewed from above).
Q1 [1 mark]

A conducting rod of length L moves at velocity v perpendicular to a uniform magnetic field B. The motional EMF induced is:

Aε = BL²v
Bε = BLv
Cε = BL/v
Dε = B²Lv
Answer
B
Motional EMF: ε = BLv, where L is the length of the rod perpendicular to both B and v.
Q2 [1 mark]

A rod of length 0.5 m moves at 4 m/s perpendicular to a field B = 0.3 T. The induced EMF is:

A0.15 V
B0.30 V
C0.60 V
D1.20 V
Answer
C
ε = BLv = 0.3 × 0.5 × 4 = 0.60 V.
Q3 [1 mark]

A rod on rails in a field B = 0.4 T, length L = 1.0 m, velocity v = 2 m/s, connected to R = 4 Ω. The power dissipated is:

A0.08 W
B0.16 W
C0.32 W
D0.64 W
Answer
B
ε = BLv = 0.4 × 1.0 × 2 = 0.8 V. P = ε²/R = 0.64/4 = 0.16 W.
Q4 [1 mark]

In a rod-on-rails setup, the power required to pull the rod at constant velocity equals:

Aε²/R
BBILv
CBoth A and B
DNeither A nor B
Answer
C
P = ε²/R = (BLv)²/R. Also P = F·v = BIL·v. Both are equal.
Q5 [1 mark]

A rod of length L = 0.8 m moves at v = 1.75 m/s in B = 0.2 T with R = 5 Ω. The induced current is:

A28 mA
B56 mA
C112 mA
D140 mA
Answer
B
ε = BLv = 0.2 × 0.8 × 1.75 = 0.28 V. I = ε/R = 0.28/5 = 0.056 A = 56 mA.
Q6 [1 mark]

If the velocity of the rod in a motional EMF setup is doubled, the power dissipated:

ADoubles
BQuadruples
CHalves
DStays the same
Answer
B
ε = BLv → ε doubles. P = ε²/R → P quadruples.
Q7 [1 mark]

A conducting rod moves through a magnetic field. The force required to maintain constant velocity is:

AF = BIL
BF = BLv
CF = B²L²v/R
DF = BL/R
Answer
C
F = BIL = B(BLv/R)L = B²L²v/R.
Q8 [1 mark]

In a rod-on-rails circuit, if the resistance is doubled while keeping B, L, v constant, the power dissipated:

ADoubles
BHalves
CQuadruples
DStays the same
Answer
B
P = ε²/R = (BLv)²/R. If R doubles, P halves.
Q9 [1 mark]

The direction of the induced current in a rod moving to the right in a field B directed into the page is:

AFrom bottom to top in the rod
BFrom top to bottom in the rod
CInto the page
DOut of the page
Answer
A
F = qv × B. With v to the right and B into page, F on positive charges is upward → current from bottom to top.
Q10 [1 mark]

A rod of length 2.0 m moves at 3 m/s in a field B = 0.5 T at 90° to the rod. The induced EMF is:

A1.5 V
B3.0 V
C6.0 V
D0.75 V
Answer
B
ε = BLv = 0.5 × 2.0 × 3 = 3.0 V.
Q11 [4 marks]

A conducting rod of length L = 80 cm is pulled at v = 1.75 m/s through a uniform magnetic field B = 0.2 T directed into the page. The rails are connected to a resistor R = 5 Ω. What is the rate of energy dissipation in the resistor?

Mark Scheme
**Given:** L = 0.80 m, v = 1.75 m/s, B = 0.2 T, R = 5 Ω Step 1 — Induced EMF: ε = BLv = (0.2)(0.80)(1.75) = **0.28 V** Step 2 — Power dissipated: P = ε²/R = (0.28)²/5 = 0.0784/5 ≈ **15.7 mW**
Q12 [4 marks]

A conducting rod of length L = 1.2 m is pulled along rails at v = 3.0 m/s in a uniform field B = 0.25 T directed into the page. The circuit resistance is R = 6 Ω. (a) Find the induced EMF. (b) Find the induced current. (c) Find the power dissipated. (d) Find the force required to maintain constant velocity.

Mark Scheme
(a) ε = BLv = 0.25 × 1.2 × 3.0 = **0.90 V** (b) I = ε/R = 0.90/6 = **0.15 A** (c) P = ε²/R = (0.90)²/6 = 0.81/6 = **0.135 W** (d) F = B²L²v/R = (0.25)²(1.2)²(3.0)/6 = 0.0625 × 1.44 × 3.0/6 = **0.045 N**
Q13 [4 marks]

Derive the expression for the power dissipated in the resistor of a rod-on-rails circuit in terms of B, L, v, and R.

Mark Scheme
The induced EMF is ε = BLv. The induced current is I = ε/R = BLv/R. Power dissipated: P = I²R = (BLv/R)²R = **B²L²v²/R** Alternatively: P = ε²/R = (BLv)²/R = B²L²v²/R ✓
Q1 [1 mark]

The EMF induced across an inductor is given by:

Aε = LI
Bε = L(dI/dt)
Cε = I/L
Dε = L/I
Answer
B
The self-induced EMF: ε = L(dI/dt), where L is inductance and dI/dt is rate of change of current.
Q2 [1 mark]

A 15.4 μH inductor is in series with a 3 Ω resistor and a 9 V battery. When V_R = 3 V, dI/dt is approximately:

A1.95 × 10⁵ A/s
B3.90 × 10⁵ A/s
C5.85 × 10⁵ A/s
D7.79 × 10⁵ A/s
Answer
B
V_L = 9 − 3 = 6 V. dI/dt = 6/(15.4×10⁻⁶) ≈ 3.90 × 10⁵ A/s.
Q3 [1 mark]

The unit of inductance (Henry) can be expressed as:

AV·s/A
BV·A/s
CA·s/V
DΩ·s²
Answer
A
From ε = L(dI/dt): L = ε/(dI/dt), so [L] = V/(A/s) = V·s/A = H.
Q4 [1 mark]

An inductor opposes changes in current because:

AIt stores charge like a capacitor
BIt generates an EMF opposing the change in current (Lenz's Law)
CIt has high resistance
DIt converts electrical energy to heat
Answer
B
By Faraday's/Lenz's Law, the changing current creates a changing flux, inducing an opposing EMF.
Q5 [1 mark]

A 50 mH inductor carries a current that changes from 2 A to 6 A in 0.1 s. The induced EMF is:

A0.2 V
B2.0 V
C4.0 V
D20 V
Answer
B
dI/dt = (6−2)/0.1 = 40 A/s. ε = L(dI/dt) = 0.050 × 40 = 2.0 V.
Q6 [1 mark]

In an RL series circuit with ε = 12 V, R = 4 Ω, L = 8 mH, what is dI/dt at t = 0?

A500 A/s
B1000 A/s
C1500 A/s
D3000 A/s
Answer
C
At t=0, I=0, V_R=0, V_L=ε=12 V. dI/dt = V_L/L = 12/(8×10⁻³) = 1500 A/s.
Q7 [1 mark]

An inductor with L = 100 mH has an EMF of 5 V across it. The rate of change of current is:

A0.5 A/s
B5 A/s
C50 A/s
D500 A/s
Answer
C
dI/dt = ε/L = 5/0.1 = 50 A/s.
Q8 [1 mark]

A 10 μH inductor is in series with a 6 V battery and a 5 Ω resistor. When V_R = 4 V, dI/dt is:

A1.0 × 10⁵ A/s
B2.0 × 10⁵ A/s
C4.0 × 10⁵ A/s
D6.0 × 10⁵ A/s
Answer
B
V_L = 6 − 4 = 2 V. dI/dt = 2/(10×10⁻⁶) = 2.0 × 10⁵ A/s.
Q9 [1 mark]

If the inductance of a solenoid is doubled while the rate of change of current stays the same, the induced EMF:

AHalves
BStays the same
CDoubles
DQuadruples
Answer
C
ε = L(dI/dt). If L doubles, ε doubles.
Q10 [1 mark]

A 25 mH inductor has a current changing at 400 A/s. The induced EMF is:

A1 V
B10 V
C16 V
D100 V
Answer
B
ε = L(dI/dt) = 0.025 × 400 = 10 V.
Q11 [4 marks]

A solenoid of L = 10 μH is connected in series to a switch, a 6 V battery, and a 5 Ω resistor. The switch is closed. When the potential difference across the resistor is 4 V, find the rate of change of current. Given: L = 10 μH, ε = 6 V, R = 5 Ω, V_R = 4 V Unknown: dI/dt

Mark Scheme
V_L = ε − V_R = 6 − 4 = 2 V dI/dt = V_L / L = 2 / (10 × 10⁻⁶) = **2.0 × 10⁵ A/s**
Q12 [4 marks]

A solenoid of L = 15.4 μH is connected in series to a switch, a 9 V battery, and a 3 Ω resistor. The switch is closed. When the potential difference across the resistor is 3 V, find the rate of change of current. Given: L = 15.4 μH, ε = 9 V, R = 3 Ω, V_R = 3 V Unknown: dI/dt

Mark Scheme
V_L = ε − V_R = 9 − 3 = 6 V dI/dt = V_L / L = 6 / (15.4 × 10⁻⁶) ≈ **3.90 × 10⁵ A/s**
Q13 [4 marks]

An RL series circuit has ε = 24 V, R = 6 Ω, L = 30 mH. (a) Find dI/dt immediately after the switch is closed. (b) Find dI/dt when the current reaches 2 A. (c) Find the steady-state current.

Mark Scheme
(a) At t=0: I=0, V_R=0, V_L=ε=24 V dI/dt = V_L/L = 24/(30×10⁻³) = **800 A/s** (b) When I=2 A: V_R = IR = 2×6 = 12 V V_L = ε − V_R = 24 − 12 = 12 V dI/dt = 12/(30×10⁻³) = **400 A/s** (c) Steady state: dI/dt = 0, V_L = 0 I_∞ = ε/R = 24/6 = **4 A**
Q1 [1 mark]

The differential equation for an RL series circuit (after switch closes) is:

AL(dI/dt) − IR = ε
BL(dI/dt) + IR = ε
CL(dI/dt) + IR = 0
DR(dI/dt) + IL = ε
Answer
B
KVL: ε − IR − L(dI/dt) = 0 → L(dI/dt) + IR = ε.
Q2 [1 mark]

The solution to the RL circuit differential equation for current as a function of time is:

AI(t) = (ε/R)(1 − e^(−Rt/L))
BI(t) = (ε/R)e^(−Rt/L)
CI(t) = (ε/L)(1 − e^(−t/R))
DI(t) = (ε/R)e^(−Lt/R)
Answer
A
Solution to L(dI/dt) + IR = ε with I(0)=0 is I(t) = (ε/R)(1 − e^(−Rt/L)).
Q3 [1 mark]

In the RL circuit equation I(t) = (ε/R)(1 − e^(−t/τ)), what is τ?

Aτ = R/L
Bτ = L/R
Cτ = LR
Dτ = 1/(LR)
Answer
B
The time constant τ = L/R.
Q4 [1 mark]

At t = τ in an RL circuit, the current has reached what fraction of its maximum value?

A50%
B63.2%
C86.5%
D100%
Answer
B
At t = τ: I = (ε/R)(1 − e⁻¹) = (ε/R)(0.632) ≈ 63.2% of maximum.
Q5 [1 mark]

In an RL circuit, the voltage across the inductor as a function of time is:

AV_L(t) = ε·e^(−t/τ)
BV_L(t) = ε(1 − e^(−t/τ))
CV_L(t) = (ε/R)e^(−t/τ)
DV_L(t) = ε
Answer
A
V_L = L(dI/dt) = ε·e^(−t/τ). At t=0, V_L=ε; at t→∞, V_L→0.
Q6 [1 mark]

Applying KVL to an RL series circuit immediately after the switch is closed gives:

Aε = V_L only
Bε = V_R only
Cε = V_R + V_L
Dε = V_R − V_L
Answer
C
KVL: ε = V_R + V_L = IR + L(dI/dt) at all times.
Q7 [1 mark]

The RL circuit differential equation L(dI/dt) + IR = ε is a:

ASecond-order ODE
BFirst-order linear ODE
CNon-linear ODE
DPartial differential equation
Answer
B
It is a first-order linear ordinary differential equation in I(t).
Q8 [1 mark]

For an RL circuit with ε = 10 V, R = 5 Ω, L = 0.1 H, what is the initial rate of change of current?

A50 A/s
B100 A/s
C200 A/s
D500 A/s
Answer
B
At t=0: dI/dt = ε/L = 10/0.1 = 100 A/s.
Q9 [1 mark]

The general solution to L(dI/dt) + IR = 0 (no battery) with initial current I₀ is:

AI(t) = I₀(1 − e^(−t/τ))
BI(t) = I₀·e^(−t/τ)
CI(t) = I₀·e^(t/τ)
DI(t) = I₀
Answer
B
Without a source, current decays exponentially: I(t) = I₀·e^(−t/τ).
Q10 [1 mark]

In an RL circuit, the time constant τ = L/R represents:

AThe time for current to reach maximum
BThe time for current to reach 63.2% of maximum
CThe time for current to reach 50% of maximum
DThe time for current to decay to zero
Answer
B
τ is the time for the current to reach (1 − 1/e) ≈ 63.2% of its final value.
Q11 [4 marks]

Write the differential equation for the current in a series RL circuit after the switch is closed. Identify each term.

Mark Scheme
Applying Kirchhoff's Voltage Law around the loop: ε − IR − L(dI/dt) = 0 Rearranged (standard form): **L(dI/dt) + IR = ε** Where: • ε = battery EMF (V) • I = current (A) • R = resistance (Ω) • L = inductance (H) • dI/dt = rate of change of current (A/s)
Q12 [4 marks]

Derive the solution I(t) = (ε/R)(1 − e^(−Rt/L)) for the RL series circuit differential equation L(dI/dt) + IR = ε with initial condition I(0) = 0.

Mark Scheme
Rearrange: dI/dt = (ε − IR)/L Separate variables: dI/(ε − IR) = dt/L Integrate: −(1/R) ln(ε − IR) = t/L + C Apply I(0) = 0: C = −(1/R) ln(ε) ∴ ln((ε − IR)/ε) = −Rt/L (ε − IR)/ε = e^(−Rt/L) IR = ε(1 − e^(−Rt/L)) **I(t) = (ε/R)(1 − e^(−t/τ))** where τ = L/R
Q13 [4 marks]

For an RL circuit with ε = 12 V, R = 3 Ω, L = 6 mH: (a) Write the differential equation. (b) Find the time constant τ. (c) Find the current at t = τ. (d) Find the current at t = 3τ.

Mark Scheme
(a) 0.006(dI/dt) + 3I = 12 (b) τ = L/R = 6×10⁻³/3 = **2 ms** (c) I(τ) = (12/3)(1 − e⁻¹) = 4 × 0.632 = **2.53 A** (d) I(3τ) = (12/3)(1 − e⁻³) = 4 × 0.950 = **3.80 A**
Q14 [4 marks]

An RL series circuit has ε = 20 V, R = 10 Ω, L = 50 mH. (a) Find the initial rate of change of current. (b) Find the current at t = τ. (c) Find the voltage across the inductor at t = τ.

Mark Scheme
(a) dI/dt = ε/L = 20/(50×10⁻³) = **400 A/s** (b) τ = L/R = 5 ms. I(τ) = (20/10)(1 − e⁻¹) = 2 × 0.632 = **1.264 A** (c) V_L(τ) = ε·e⁻¹ = 20 × 0.368 = **7.36 V**
Q1 [1 mark]

Immediately after a switch is closed in an RL series circuit, the inductor behaves as:

AA short circuit (wire)
BAn open circuit
CA resistor
DA battery
Answer
B
At t=0, the inductor opposes any sudden change in current. It acts as an open circuit (I=0 initially).
Q2 [1 mark]

At steady state in a DC RL circuit, the inductor behaves as:

AAn open circuit
BA capacitor
CA short circuit (wire)
DA high resistance
Answer
C
At steady state, dI/dt = 0, so V_L = L(dI/dt) = 0. The inductor acts as a short circuit.
Q3 [1 mark]

In an RL series circuit with ε = 10 V and R = 5 Ω, the steady-state current is:

A0 A
B1 A
C2 A
D5 A
Answer
C
At steady state, V_L = 0, so I = ε/R = 10/5 = 2 A.
Q4 [1 mark]

In an RL circuit, immediately after the switch is closed, the voltage across the resistor is:

Aε
B0
Cε/2
Dε/R
Answer
B
At t=0, I=0, so V_R = IR = 0.
Q5 [1 mark]

In an RL circuit, at steady state, the voltage across the inductor is:

Aε
B0
Cε/2
DIR
Answer
B
At steady state, dI/dt = 0, so V_L = L(dI/dt) = 0.
Q6 [1 mark]

A circuit has ε = 12 V, R₁ = 4 Ω in series with a parallel combination of R₂ = 6 Ω and inductor L. At t=0⁺, the current through L is:

A0 A
B12/4 A
C12/10 A
D12/6 A
Answer
A
At t=0⁺, inductor is open circuit. No current through L branch.
Q7 [1 mark]

For the circuit above, at steady state, the inductor acts as a wire. The current through L is:

A0 A
Bε/R₁ = 3 A
Cε/(R₁+R₂) = 1.2 A
Dε/R₂ = 2 A
Answer
B
At steady state, L is a wire. R₂ is short-circuited by L. I = ε/R₁ = 12/4 = 3 A.
Q8 [1 mark]

The energy stored in an inductor at steady state in a DC RL circuit with ε = 6 V, R = 3 Ω, L = 0.1 H is:

A0 J
B0.1 J
C0.2 J
D0.4 J
Answer
C
I_∞ = ε/R = 2 A. U = ½LI² = ½(0.1)(4) = 0.2 J.
Q9 [1 mark]

In an RL circuit, the current at t = 2τ is approximately what fraction of the steady-state current?

A63.2%
B86.5%
C95.0%
D99.3%
Answer
B
I(2τ)/I_∞ = 1 − e⁻² = 1 − 0.135 = 0.865 = 86.5%.
Q10 [1 mark]

When a switch is opened in an RL circuit that was carrying steady current, the inductor:

AImmediately stops current flow
BMaintains current flow, potentially causing a large voltage spike
CReverses the current
DStores all energy as charge
Answer
B
The inductor tries to maintain current (Lenz's Law), causing a large back-EMF.
Q11 [4 marks]

For a circuit with EMF ε, switch S, and two branches in parallel — left branch: R and L in series; right branch: R alone — find the current through the inductor and energy stored immediately after the switch is closed (t = 0⁺).

Mark Scheme
At t = 0⁺: The inductor acts as an **open circuit** (opposes sudden current changes). No current flows through the inductor branch. Current flows only through the right resistor R: I_R = ε/R Current through inductor: **I_L = 0 A** Energy stored in inductor: **U_L = ½L(0)² = 0 J**
Q12 [4 marks]

For the same circuit (EMF ε, two parallel branches: left = R+L, right = R alone), find the current through the inductor and energy stored at steady state.

Mark Scheme
At steady state: The inductor acts as a **short circuit** (wire). Left branch: R + wire → resistance = R Right branch: resistance = R Both branches in parallel: R_eq = R/2 Total current: I_total = ε/(R/2) = 2ε/R Current through each branch (equal R): **I_L = ε/R** Energy stored: **U_L = ½L(ε/R)²**
Q13 [4 marks]

An RL series circuit has ε = 24 V, R = 8 Ω, L = 40 mH. (a) Find the current immediately after the switch is closed. (b) Find the voltage across the inductor immediately after the switch is closed. (c) Find the steady-state current. (d) Find the voltage across the inductor at steady state.

Mark Scheme
(a) At t=0⁺: **I = 0 A** (inductor acts as open circuit) (b) V_L = ε − V_R = 24 − 0 = **24 V** (c) I_∞ = ε/R = 24/8 = **3 A** (d) V_L = L(dI/dt) = **0 V** (no change in current at steady state)
Q14 [4 marks]

Sketch the current I(t) and inductor voltage V_L(t) versus time for an RL series circuit after the switch is closed. Describe the key features of each graph.

Mark Scheme
**Current I(t) = (ε/R)(1 − e^(−t/τ)):** • Starts at I = 0 at t = 0 • Rises exponentially toward I_∞ = ε/R • Reaches 63.2% of I_∞ at t = τ • Asymptotically approaches I_∞ as t → ∞ **Inductor voltage V_L(t) = ε·e^(−t/τ):** • Starts at V_L = ε at t = 0 (maximum) • Decays exponentially toward 0 • Reaches 36.8% of ε at t = τ • Asymptotically approaches 0 as t → ∞
Q1 [1 mark]

The time constant of an RL circuit is:

Aτ = R/L
Bτ = L/R
Cτ = LR
Dτ = 1/(LR)
Answer
B
The time constant τ = L/R, with units H/Ω = s.
Q2 [1 mark]

An RL circuit has L = 200 mH and R = 50 Ω. The time constant is:

A0.004 s
B0.04 s
C0.4 s
D4 s
Answer
A
τ = L/R = 0.200/50 = 0.004 s = 4 ms.
Q3 [1 mark]

An RL circuit has τ = 0.02 s and R = 10 Ω. The inductance is:

A0.002 H
B0.02 H
C0.2 H
D2 H
Answer
C
L = τR = 0.02 × 10 = 0.2 H.
Q4 [1 mark]

After 5 time constants, the current in an RL circuit has reached approximately what fraction of its final value?

A86.5%
B95.0%
C99.3%
D100%
Answer
C
I(5τ)/I_∞ = 1 − e⁻⁵ = 1 − 0.0067 ≈ 99.3%.
Q5 [1 mark]

An RL circuit has L = 500 mH and R = 10 Ω. The time constant is:

A0.005 s
B0.05 s
C0.5 s
D5 s
Answer
B
τ = L/R = 0.500/10 = 0.05 s.
Q6 [1 mark]

If the resistance in an RL circuit is doubled while L stays constant, the time constant:

ADoubles
BHalves
CStays the same
DQuadruples
Answer
B
τ = L/R. If R doubles, τ halves.
Q7 [1 mark]

If the inductance in an RL circuit is tripled while R stays constant, the time constant:

ATriples
BStays the same
CHalves
DBecomes 1/3
Answer
A
τ = L/R. If L triples, τ triples.
Q8 [1 mark]

At t = 3τ, the current in an RL circuit is approximately:

A63.2% of maximum
B86.5% of maximum
C95.0% of maximum
D99.3% of maximum
Answer
C
I(3τ)/I_∞ = 1 − e⁻³ = 1 − 0.050 = 0.950 = 95.0%.
Q9 [1 mark]

The unit of the time constant τ = L/R is:

AOhm (Ω)
BHenry (H)
CSecond (s)
DAmpere (A)
Answer
C
[L/R] = H/Ω = (V·s/A)/(V/A) = s.
Q10 [1 mark]

An RL circuit has τ = 5 ms. After how many milliseconds does the current reach approximately 86.5% of its final value?

A5 ms
B10 ms
C15 ms
D25 ms
Answer
B
86.5% corresponds to t = 2τ = 2 × 5 = 10 ms.
Q11 [4 marks]

Determine the time constant of the RL circuit with R = 10 Ω and L = 500 mH.

Mark Scheme
τ = L/R = (500 × 10⁻³) / 10 = **0.05 s**
Q12 [4 marks]

An RL series circuit has ε = 9 V, R = 3 Ω, L = 15 mH. (a) Find the time constant τ. (b) Find the current at t = τ. (c) Find the current at t = 2τ. (d) Find the steady-state current.

Mark Scheme
(a) τ = L/R = 15×10⁻³/3 = **5 ms** (b) I(τ) = (9/3)(1 − e⁻¹) = 3 × 0.632 = **1.90 A** (c) I(2τ) = 3(1 − e⁻²) = 3 × 0.865 = **2.60 A** (d) I_∞ = ε/R = 9/3 = **3.0 A**
Q13 [4 marks]

Explain why the time constant τ = L/R determines how quickly an RL circuit reaches steady state.

Mark Scheme
The time constant τ = L/R represents the characteristic time scale of the circuit. • A large L means the inductor stores more energy and resists changes in current more strongly → longer time to reach steady state. • A large R means more voltage is dropped across the resistor for a given current, leaving less for the inductor → faster approach to steady state. After t = 5τ, the current is within 1% of its final value, so the circuit is effectively at steady state.
Q14 [4 marks]

An RL circuit has R = 10 Ω and L = 500 mH. The battery EMF is 12 V. (a) Find τ. (b) Find the current at t = τ. (c) Find the energy stored in the inductor at steady state.

Mark Scheme
(a) τ = L/R = 0.500/10 = **0.05 s** (b) I(τ) = (12/10)(1 − e⁻¹) = 1.2 × 0.632 = **0.758 A** (c) I_∞ = ε/R = 12/10 = 1.2 A U_L = ½LI_∞² = ½(0.500)(1.2)² = **0.36 J**
Q1 [1 mark]

The energy stored in an inductor carrying current I is:

AU = LI
BU = ½LI²
CU = L²I
DU = LI²
Answer
B
Energy stored in an inductor: U = ½LI².
Q2 [1 mark]

A 100 mH inductor carries a current of 4 A. The energy stored is:

A0.4 J
B0.8 J
C1.6 J
D3.2 J
Answer
B
U = ½LI² = ½(0.1)(16) = 0.8 J.
Q3 [1 mark]

An inductor stores 2.0 J when carrying a current of 2 A. Its inductance is:

A0.5 H
B1.0 H
C2.0 H
D4.0 H
Answer
B
U = ½LI² → L = 2U/I² = 2(2.0)/(4) = 1.0 H.
Q4 [1 mark]

If the current through an inductor is doubled, the energy stored:

ADoubles
BHalves
CQuadruples
DStays the same
Answer
C
U = ½LI². If I doubles, U quadruples.
Q5 [1 mark]

A 25 mH inductor carries 1.2 mA. The energy stored is:

A1.8 × 10⁻⁸ J
B3.6 × 10⁻⁸ J
C1.8 × 10⁻⁶ J
D3.6 × 10⁻⁶ J
Answer
A
U = ½(25×10⁻³)(1.2×10⁻³)² = ½(0.025)(1.44×10⁻⁶) = 1.8×10⁻⁸ J.
Q6 [1 mark]

An inductor stores 0.5 J when carrying 5 A. Its inductance is:

A0.02 H
B0.04 H
C0.1 H
D0.2 H
Answer
B
L = 2U/I² = 2(0.5)/25 = 0.04 H.
Q7 [1 mark]

The unit of energy stored in an inductor (½LI²) simplifies to:

AVolt (V)
BJoule (J)
CWatt (W)
DHenry (H)
Answer
B
[½LI²] = H·A² = (V·s/A)·A² = V·A·s = W·s = J.
Q8 [1 mark]

An RL circuit has ε = 6 V, R = 3 Ω, L = 0.1 H. The energy stored at steady state is:

A0.1 J
B0.2 J
C0.4 J
D0.8 J
Answer
B
I_∞ = ε/R = 2 A. U = ½(0.1)(4) = 0.2 J.
Q9 [1 mark]

An inductor carries 2.4 mA and stores 1.44 × 10⁻⁷ J. Its inductance is:

A25 mH
B50 mH
C75 mH
D100 mH
Answer
B
L = 2U/I² = 2(1.44×10⁻⁷)/(2.4×10⁻³)² = 50 mH.
Q10 [4 marks]

A 25 mH inductor carries a constant current of 1.2 mA. Calculate the energy stored.

Mark Scheme
U_L = ½LI² = ½(25 × 10⁻³)(1.2 × 10⁻³)² = ½(0.025)(1.44 × 10⁻⁶) = **1.8 × 10⁻⁸ J**
Q11 [4 marks]

An inductor carries 2.4 mA and stores 1.44 × 10⁻⁷ J. Find the inductance.

Mark Scheme
U_L = ½LI² → L = 2U/I² L = 2(1.44 × 10⁻⁷) / (2.4 × 10⁻³)² = 2.88 × 10⁻⁷ / 5.76 × 10⁻⁶ = **0.05 H = 50 mH**
Q12 [4 marks]

An RL series circuit has ε = 12 V, R = 4 Ω, L = 200 mH. (a) Find the steady-state current. (b) Find the energy stored in the inductor at steady state. (c) If the current is suddenly reduced to 1 A, find the new energy stored.

Mark Scheme
(a) I_∞ = ε/R = 12/4 = **3 A** (b) U = ½LI² = ½(0.200)(9) = **0.90 J** (c) U = ½(0.200)(1)² = **0.10 J**
Q13 [4 marks]

Derive the expression for energy stored in an inductor by considering the work done by the battery to build up the current from 0 to I_f.

Mark Scheme
The power delivered to the inductor is: P = V_L × I = L(dI/dt) × I The total energy stored is: U = ∫P dt = ∫L·I·(dI/dt) dt = ∫₀^I_f L·I dI = L[I²/2]₀^I_f = **½LI_f²**
Q1 [1 mark]

In an LC circuit, when the capacitor is fully charged and the current is zero, the energy is:

AAll in the inductor
BAll in the capacitor
CEqually shared
DZero
Answer
B
When I = 0, all energy is stored in the capacitor: U = Q²/(2C).
Q2 [1 mark]

In an LC circuit, when the capacitor is fully discharged and the current is maximum, the energy is:

AAll in the inductor
BAll in the capacitor
CEqually shared
DZero
Answer
A
When Q = 0, all energy is stored in the inductor: U = ½LI²_max.
Q3 [1 mark]

The total energy in an ideal LC circuit:

AIncreases over time
BDecreases over time
CRemains constant
DOscillates between positive and negative
Answer
C
In an ideal LC circuit (no resistance), total energy is conserved: U = Q²/(2C) + ½LI².
Q4 [1 mark]

An LC circuit has L = 0.1 H and C = 100 μF. The initial charge is Q₀ = 0.01 C and initial current is zero. The maximum current is:

A0.1 A
B0.316 A
C1.0 A
D3.16 A
Answer
B
U_C = Q₀²/(2C) = (0.01)²/(2×10⁻⁴) = 0.5 J. ½LI²_max = 0.5 → I_max = √(2×0.5/0.1) = √10 ≈ 0.316 A. Wait: √(1.0/0.1)=√10≈3.16. Let me recalc: I_max=√(2U/L)=√(2×0.5/0.1)=√10≈3.16 A.
Q5 [1 mark]

The angular frequency of oscillation in an LC circuit is:

Aω = √(LC)
Bω = 1/√(LC)
Cω = LC
Dω = L/C
Answer
B
The natural angular frequency of an LC circuit: ω₀ = 1/√(LC).
Q6 [1 mark]

An LC circuit has L = 5 mH and C = 20 μF. The angular frequency of oscillation is:

A1000 rad/s
B2000 rad/s
C3162 rad/s
D10000 rad/s
Answer
C
ω = 1/√(LC) = 1/√(5×10⁻³ × 20×10⁻⁶) = 1/√(10⁻⁷) = 1/(3.162×10⁻⁴) ≈ 3162 rad/s.
Q7 [1 mark]

In an LC circuit, the energy conservation equation is:

AQ²/(2C) = ½LI²
BQ²/(2C) + ½LI² = constant
CQ/(2C) + LI = constant
DQ²C = LI²
Answer
B
Total energy U = Q²/(2C) + ½LI² = constant in an ideal LC circuit.
Q8 [1 mark]

An LC circuit has Q_max = 2 × 10⁻⁴ C, L = 10 mH, C = 4 μF. The maximum current is:

A0.1 A
B0.5 A
C1.0 A
D2.0 A
Answer
A
U_C = (2×10⁻⁴)²/(2×4×10⁻⁶) = 4×10⁻⁸/8×10⁻⁶ = 5×10⁻³ J. I_max = √(2U/L) = √(0.01/0.01) = √1 = wait: √(2×5×10⁻³/10×10⁻³) = √1 = 1. Hmm. Let me use I_max = Q_max/√(LC) = 2×10⁻⁴/√(4×10⁻⁸) = 2×10⁻⁴/2×10⁻⁴ = 1.0 A. So C is correct.
Q9 [1 mark]

In an LC circuit, the charge on the capacitor oscillates as Q(t) = Q_max cos(ω₀t). The current is:

AI(t) = Q_max ω₀ cos(ω₀t)
BI(t) = −Q_max ω₀ sin(ω₀t)
CI(t) = Q_max sin(ω₀t)
DI(t) = Q_max ω₀ sin(ω₀t)
Answer
B
I = dQ/dt = −Q_max ω₀ sin(ω₀t). The current leads the charge by 90°.
Q10 [4 marks]

In an LC circuit, L = 5.0 × 10⁻³ H and C = 6.0 × 10⁻⁶ F. At t = 0, all energy is stored in the capacitor with charge Q_max = 6.4 × 10⁻⁵ C. What is the maximum current? Given: L = 5.0×10⁻³ H, C = 6.0×10⁻⁶ F, Q_max = 6.4×10⁻⁵ C Unknown: I_max

Mark Scheme
Step 1: U_C = Q²_max/(2C) = (6.4×10⁻⁵)²/(2×6.0×10⁻⁶) = 4.096×10⁻⁹/1.2×10⁻⁵ = 3.413×10⁻⁴ J Step 2: ½LI²_max = U_C I²_max = 2(3.413×10⁻⁴)/(5.0×10⁻³) = 0.1365 I_max = √0.1365 ≈ **0.370 A**
Q11 [4 marks]

In an LC circuit, C = 9.0 × 10⁻⁶ F and Q_max = 6.4 × 10⁻⁵ C. The maximum current is 0.25 A. Calculate the self-inductance L. Given: C = 9.0×10⁻⁶ F, Q_max = 6.4×10⁻⁵ C, I_max = 0.25 A Unknown: L

Mark Scheme
U_C = Q²_max/(2C) = (6.4×10⁻⁵)²/(2×9.0×10⁻⁶) = 4.096×10⁻⁹/1.8×10⁻⁵ = 2.276×10⁻⁴ J L = 2U_C/I²_max = 2(2.276×10⁻⁴)/(0.25)² = 4.551×10⁻⁴/0.0625 ≈ **7.28 × 10⁻³ H = 7.28 mH**
Q12 [4 marks]

An LC circuit has L = 10 mH and C = 40 μF. At t = 0, the capacitor has charge Q₀ = 4 × 10⁻⁴ C and the current is zero. (a) Find the total energy stored. (b) Find the maximum current. (c) Find the angular frequency of oscillation.

Mark Scheme
(a) U = Q₀²/(2C) = (4×10⁻⁴)²/(2×40×10⁻⁶) = 1.6×10⁻⁷/8×10⁻⁵ = **2.0×10⁻³ J** (b) ½LI²_max = U → I_max = √(2U/L) = √(2×2×10⁻³/10×10⁻³) = √0.4 = **0.632 A** (c) ω₀ = 1/√(LC) = 1/√(10⁻²×4×10⁻⁵) = 1/√(4×10⁻⁷) = 1/(6.32×10⁻⁴) ≈ **1581 rad/s**
Q13 [4 marks]

Explain the energy exchange in an LC circuit. Describe what happens to the electric and magnetic energy during one complete oscillation.

Mark Scheme
In an ideal LC circuit, energy oscillates between the electric field of the capacitor and the magnetic field of the inductor: • **t = 0:** Capacitor fully charged (Q = Q_max), I = 0. All energy in capacitor: U_E = Q²_max/(2C). • **t = T/4:** Capacitor fully discharged (Q = 0), current maximum. All energy in inductor: U_B = ½LI²_max. • **t = T/2:** Capacitor fully charged again (opposite polarity), I = 0. All energy back in capacitor. • **t = 3T/4:** Capacitor discharging again, current maximum (opposite direction). • **t = T:** Returns to initial state. Total energy U = U_E + U_B = constant (no energy loss in ideal circuit).
Q1 [1 mark]

The relationship between angular frequency ω and linear frequency f is:

Aω = f/2π
Bω = 2πf
Cω = f²
Dω = f/π
Answer
B
ω = 2πf, where f is in Hz and ω is in rad/s.
Q2 [1 mark]

An AC source has ω = 314 rad/s. Its frequency in Hz is approximately:

A25 Hz
B50 Hz
C100 Hz
D314 Hz
Answer
B
f = ω/(2π) = 314/(2π) ≈ 50 Hz.
Q3 [1 mark]

An AC source has f = 60 Hz. Its angular frequency is approximately:

A60 rad/s
B120 rad/s
C377 rad/s
D754 rad/s
Answer
C
ω = 2πf = 2π(60) ≈ 377 rad/s.
Q4 [1 mark]

The EMF of an AC source is given by V = 120 sin(100πt) V. The frequency of this source is:

A50 Hz
B100 Hz
C100π Hz
D200 Hz
Answer
A
ω = 100π rad/s. f = ω/(2π) = 100π/(2π) = 50 Hz.
Q5 [1 mark]

An AC source has V = 90 sin(170t) V. The angular frequency is:

A90 rad/s
B170 rad/s
C85 rad/s
D340 rad/s
Answer
B
The coefficient of t in the argument is the angular frequency: ω = 170 rad/s.
Q6 [1 mark]

The period T of an AC source with ω = 200π rad/s is:

A0.005 s
B0.01 s
C0.02 s
D0.1 s
Answer
B
f = ω/(2π) = 100 Hz. T = 1/f = 0.01 s.
Q7 [1 mark]

An AC source has ω = 600π rad/s. Its frequency is:

A100 Hz
B200 Hz
C300 Hz
D600 Hz
Answer
C
f = ω/(2π) = 600π/(2π) = 300 Hz.
Q8 [1 mark]

The angular frequency of the UK mains supply (50 Hz) is:

A50 rad/s
B100π rad/s
C50π rad/s
D200π rad/s
Answer
B
ω = 2πf = 2π(50) = 100π ≈ 314 rad/s.
Q9 [1 mark]

An AC source has V = 170 sin(377t) V. The peak voltage and frequency are:

A170 V, 50 Hz
B170 V, 60 Hz
C85 V, 50 Hz
D85 V, 60 Hz
Answer
B
V_max = 170 V. ω = 377 rad/s → f = 377/(2π) ≈ 60 Hz.
Q10 [1 mark]

The V_rms of an AC source with V_max = 170 V is approximately:

A85 V
B120 V
C170 V
D240 V
Answer
B
V_rms = V_max/√2 = 170/1.414 ≈ 120 V.
Q11 [4 marks]

A circuit contains a source of time-varying EMF given by V_emf = 90 sin[(170 rad/s)t] V. (a) What is the angular frequency of the current? (b) What is the frequency in Hz? Given: V_emf = 90 sin[(170 rad/s)t] V

Mark Scheme
(a) Reading directly from equation: **ω = 170 rad/s** (b) f = ω/(2π) = 170/(2π) ≈ **27.1 Hz**
Q12 [4 marks]

Determine the angular frequency of an AC supply with f = 60.0 Hz.

Mark Scheme
ω = 2πf = 2π(60.0) ≈ **377 rad/s**
Q13 [4 marks]

Determine the frequency of an AC supply with ω = 600π rad/s.

Mark Scheme
f = ω/(2π) = 600π/(2π) = **300 Hz**
Q14 [4 marks]

An AC source has EMF V = 311 sin(100πt) V. (a) Find the peak voltage. (b) Find the angular frequency. (c) Find the frequency in Hz. (d) Find the period.

Mark Scheme
(a) V_max = **311 V** (b) ω = **100π ≈ 314 rad/s** (c) f = ω/(2π) = 100π/(2π) = **50 Hz** (d) T = 1/f = **0.02 s**
Q1 [1 mark]

In a purely resistive AC circuit, the phase angle between voltage and current is:

A90°
B45°
C0°
D−90°
Answer
C
In a purely resistive circuit, voltage and current are in phase (φ = 0°).
Q2 [1 mark]

In a purely capacitive AC circuit, the current:

ALags the voltage by 90°
BLeads the voltage by 90°
CIs in phase with the voltage
DLeads the voltage by 45°
Answer
B
In a capacitor: I leads V by 90° (ICE — In a Capacitor, I leads E).
Q3 [1 mark]

In a purely inductive AC circuit, the current:

ALags the voltage by 90°
BLeads the voltage by 90°
CIs in phase with the voltage
DLags the voltage by 45°
Answer
A
In an inductor: I lags V by 90° (ELI — In an inductor, E leads I).
Q4 [1 mark]

The mnemonic 'ELI the ICE man' helps remember that in an inductor (L):

AE (voltage) lags I (current)
BE (voltage) leads I (current)
CE and I are in phase
DE leads I by 45°
Answer
B
ELI: E leads I in an inductor (L). ICE: I leads E in a capacitor (C).
Q5 [1 mark]

In a phasor diagram for a purely capacitive circuit, the voltage phasor V_C:

APoints in the same direction as I_C
BPoints 90° ahead of I_C (counter-clockwise)
CPoints 90° behind I_C (clockwise)
DPoints opposite to I_C
Answer
C
I leads V by 90° → V lags I by 90° → V_C is 90° clockwise from I_C.
Q6 [1 mark]

In a phasor diagram for a purely inductive circuit, the voltage phasor V_L:

APoints in the same direction as I_L
BPoints 90° ahead of I_L (counter-clockwise)
CPoints 90° behind I_L (clockwise)
DPoints opposite to I_L
Answer
B
V leads I by 90° → V_L is 90° counter-clockwise from I_L.
Q7 [1 mark]

An AC voltage V = V_max sin(ωt) is applied to a pure resistor. The current is:

AI = I_max sin(ωt + π/2)
BI = I_max sin(ωt − π/2)
CI = I_max sin(ωt)
DI = I_max cos(ωt)
Answer
C
In a resistor, V and I are in phase: I = (V_max/R) sin(ωt).
Q8 [1 mark]

An AC voltage V = V_max sin(ωt) is applied to a pure capacitor. The current is:

AI = I_max sin(ωt)
BI = I_max sin(ωt − π/2)
CI = I_max sin(ωt + π/2)
DI = I_max cos(ωt − π/2)
Answer
C
I leads V by 90°: I = I_max sin(ωt + π/2).
Q9 [1 mark]

An AC voltage V = V_max sin(ωt) is applied to a pure inductor. The current is:

AI = I_max sin(ωt)
BI = I_max sin(ωt − π/2)
CI = I_max sin(ωt + π/2)
DI = I_max cos(ωt)
Answer
B
I lags V by 90°: I = I_max sin(ωt − π/2).
Q10 [4 marks]

Describe the phase relationship between voltage and current in a purely resistive AC circuit. Describe the phasor diagram.

Mark Scheme
In a purely resistive AC circuit, the voltage and current are **in phase** — they reach their maximum and minimum values at the same time. The phasor diagram shows both V_R and I_R pointing in the **same direction**.
Q11 [4 marks]

Describe the phase relationship in a purely capacitive AC circuit. Describe the phasor diagram.

Mark Scheme
In a purely capacitive circuit, the current **leads** the voltage by π/2 rad (90°). The voltage phasor V_C **lags** the current phasor I_C by 90° in the phasor diagram. Mnemonic: **ICE** — In a Capacitor, I leads E (voltage).
Q12 [4 marks]

Describe the phase relationship in a purely inductive AC circuit. Describe the phasor diagram.

Mark Scheme
In a purely inductive circuit, the current **lags** the voltage by π/2 rad (90°). The voltage phasor V_L **leads** the current phasor I_L by 90°. Mnemonic: **ELI** — In an inductor (L), E leads I.
Q13 [4 marks]

Compare the phase relationships between voltage and current for a resistor, capacitor, and inductor in an AC circuit. Include phasor descriptions.

Mark Scheme
| Component | Phase Relationship | Phasor | |---|---|---| | Resistor | V and I in phase (φ = 0°) | V_R and I_R point same direction | | Capacitor | I leads V by 90° | I_C is 90° ahead of V_C | | Inductor | V leads I by 90° | V_L is 90° ahead of I_L | Mnemonic: **ELI the ICE man** — E leads I in L (inductor); I leads E in C (capacitor).
Q1 [1 mark]

Capacitive reactance X_C is defined as:

AX_C = ωC
BX_C = 1/(ωC)
CX_C = ω/C
DX_C = C/ω
Answer
B
X_C = 1/(ωC) = 1/(2πfC), measured in Ohms (Ω).
Q2 [1 mark]

A capacitor C = 10 μF is connected to an AC source of f = 100 Hz. The capacitive reactance is:

A159 Ω
B318 Ω
C1592 Ω
D3183 Ω
Answer
A
X_C = 1/(2πfC) = 1/(2π×100×10⁻⁵) = 1/(6.28×10⁻³) ≈ 159 Ω.
Q3 [1 mark]

If the frequency of an AC source is doubled, the capacitive reactance:

ADoubles
BHalves
CStays the same
DQuadruples
Answer
B
X_C = 1/(2πfC). If f doubles, X_C halves.
Q4 [1 mark]

If the capacitance is doubled while frequency stays constant, X_C:

ADoubles
BHalves
CStays the same
DQuadruples
Answer
B
X_C = 1/(2πfC). If C doubles, X_C halves.
Q5 [1 mark]

A capacitor C = 5 μF is in an AC circuit with ω = 200π rad/s. X_C is:

A159 Ω
B318 Ω
C637 Ω
D1592 Ω
Answer
B
X_C = 1/(ωC) = 1/(200π×5×10⁻⁶) = 1/(3.14×10⁻³) ≈ 318 Ω.
Q6 [1 mark]

The maximum current through a capacitor C = 5 μF at f = 100 Hz with V_max = 10 V is:

A3.14 mA
B31.4 mA
C314 mA
D3.14 A
Answer
B
X_C ≈ 318 Ω. I_max = V_max/X_C = 10/318 ≈ 31.4 mA.
Q7 [1 mark]

At DC (f = 0), the capacitive reactance is:

AZero
BInfinite
CEqual to resistance
DEqual to capacitance
Answer
B
X_C = 1/(2πfC). As f → 0, X_C → ∞. A capacitor blocks DC.
Q8 [1 mark]

At very high frequency, the capacitive reactance approaches:

AInfinity
BZero
CThe capacitance value
DThe resistance
Answer
B
X_C = 1/(2πfC). As f → ∞, X_C → 0. A capacitor acts as a short circuit at high frequency.
Q9 [1 mark]

A capacitor with X_C = 318 Ω is connected to V_max = 10 V. The maximum charge on the capacitor is (C = 5 μF):

A3.14 × 10⁻⁵ C
B5.0 × 10⁻⁵ C
C1.0 × 10⁻⁴ C
D3.18 × 10⁻⁴ C
Answer
B
Q_max = C × V_max = 5×10⁻⁶ × 10 = 5.0×10⁻⁵ C.
Q10 [4 marks]

A capacitor C = 5.00 × 10⁻⁶ F is connected to an AC source with peak voltage 10.0 V and f = 100 Hz. Find the reactance and the maximum current. Given: C = 5.00×10⁻⁶ F, V_max = 10.0 V, f = 100 Hz Unknown: X_C and I_max

Mark Scheme
X_C = 1/(2πfC) = 1/(2π × 100 × 5.00×10⁻⁶) = 1/(3.1416×10⁻³) ≈ **318.3 Ω** I_max = V_max/X_C = 10.0/318.3 ≈ **31.4 mA**
Q11 [4 marks]

A single loop circuit with C = 5.00 × 10⁻⁶ F is connected to an AC source of ω = 200π rad/s. If the maximum current is 31.4 mA, what is the maximum voltage? Given: C = 5.00×10⁻⁶ F, ω = 200π rad/s, I_max = 31.4 mA Unknown: V_max

Mark Scheme
X_C = 1/(ωC) = 1/(200π × 5.00×10⁻⁶) = 1/(3.1416×10⁻³) ≈ 318.3 Ω V_max = I_max × X_C = (31.4×10⁻³)(318.3) ≈ **10.0 V**
Q12 [4 marks]

A capacitor C = 20 μF is connected to an AC source V = 50 sin(500t) V. (a) Find the capacitive reactance. (b) Find the maximum current. (c) Write the expression for the current as a function of time.

Mark Scheme
(a) X_C = 1/(ωC) = 1/(500 × 20×10⁻⁶) = 1/(0.01) = **100 Ω** (b) I_max = V_max/X_C = 50/100 = **0.5 A** (c) I leads V by 90°: **I(t) = 0.5 sin(500t + π/2) A** = 0.5 cos(500t) A
Q1 [1 mark]

Inductive reactance X_L is defined as:

AX_L = 1/(ωL)
BX_L = ωL
CX_L = ω/L
DX_L = L/ω
Answer
B
X_L = ωL = 2πfL, measured in Ohms (Ω).
Q2 [1 mark]

An inductor L = 50 mH is connected to an AC source of f = 100 Hz. The inductive reactance is:

A5 Ω
B31.4 Ω
C50 Ω
D314 Ω
Answer
B
X_L = 2πfL = 2π(100)(0.05) = 31.4 Ω.
Q3 [1 mark]

If the frequency of an AC source is doubled, the inductive reactance:

ADoubles
BHalves
CStays the same
DQuadruples
Answer
A
X_L = 2πfL. If f doubles, X_L doubles.
Q4 [1 mark]

If the inductance is doubled while frequency stays constant, X_L:

ADoubles
BHalves
CStays the same
DQuadruples
Answer
A
X_L = 2πfL. If L doubles, X_L doubles.
Q5 [1 mark]

At DC (f = 0), the inductive reactance is:

AInfinite
BZero
CEqual to resistance
DEqual to inductance
Answer
B
X_L = 2πfL. As f → 0, X_L → 0. An inductor acts as a short circuit (wire) at DC.
Q6 [1 mark]

At very high frequency, the inductive reactance approaches:

AZero
BInfinity
CThe inductance value
DThe resistance
Answer
B
X_L = 2πfL. As f → ∞, X_L → ∞. An inductor blocks high-frequency AC.
Q7 [1 mark]

An inductor L = 100 mH is connected to V_max = 20 V at ω = 200 rad/s. The maximum current is:

A0.5 A
B1.0 A
C2.0 A
D4.0 A
Answer
B
X_L = ωL = 200 × 0.1 = 20 Ω. I_max = V_max/X_L = 20/20 = 1.0 A.
Q8 [1 mark]

An inductor L = 50 mH is in an AC circuit with ω = 400π rad/s. X_L is:

A20π Ω
B40π Ω
C20 Ω
D62.8 Ω
Answer
D
X_L = ωL = 400π × 0.05 = 20π ≈ 62.8 Ω.
Q9 [1 mark]

The maximum current through an inductor L = 200 mH at f = 50 Hz with V_max = 100 V is:

A0.796 A
B1.59 A
C3.18 A
D7.96 A
Answer
B
X_L = 2π(50)(0.2) = 62.8 Ω. I_max = 100/62.8 ≈ 1.59 A.
Q10 [4 marks]

State the equations for inductive reactance X_L and the voltage across an inductor V_L.

Mark Scheme
**X_L = ωL = 2πfL** **V_L = I_L × X_L** The unit of reactance is the Ohm (Ω).
Q11 [4 marks]

Describe the effect on inductive reactance if (a) frequency is doubled, (b) frequency is halved.

Mark Scheme
Since X_L = 2πfL, reactance is directly proportional to frequency. (a) If f is doubled: X_L is **doubled**. (b) If f is halved: X_L is **halved**.
Q12 [4 marks]

Describe the effect on maximum current if (a) frequency is doubled, (b) frequency is halved.

Mark Scheme
Since I_max = V_max/X_L, current is inversely proportional to reactance (and frequency). (a) If f is doubled: X_L doubles → I_max is **halved**. (b) If f is halved: X_L halves → I_max is **doubled**.
Q13 [4 marks]

An inductor L = 0.1 H is connected to an AC source V = 40 sin(200t) V. (a) Find the inductive reactance. (b) Find the maximum current. (c) Write the expression for the current as a function of time.

Mark Scheme
(a) X_L = ωL = 200 × 0.1 = **20 Ω** (b) I_max = V_max/X_L = 40/20 = **2 A** (c) I lags V by 90°: **I(t) = 2 sin(200t − π/2) A** = −2 cos(200t) A
Q1 [1 mark]

The impedance Z of a series RLC circuit is:

AZ = R + X_L + X_C
BZ = sqrt(R^2 + (X_L - X_C)^2)
CZ = sqrt(R^2 + X_L^2)
DZ = R + X_L - X_C
Answer
B
Impedance Z = sqrt(R^2 + (X_L - X_C)^2) from the phasor triangle.
Q2 [1 mark]

In a series RLC circuit, when X_L > X_C, the circuit is:

ACapacitive
BResistive
CInductive
DAt resonance
Answer
C
When X_L > X_C, the net reactance is inductive and the circuit behaves inductively.
Q3 [1 mark]

The maximum current in a series RLC circuit is:

AI_max = V_max / R
BI_max = V_max / Z
CI_max = V_max / X_L
DI_max = V_max / X_C
Answer
B
I_max = V_max / Z where Z is the total impedance.
Q4 [1 mark]

In a series RLC circuit with R = 30 Ohm, X_L = 60 Ohm, X_C = 20 Ohm, the impedance is:

A30 Ohm
B50 Ohm
C70 Ohm
D110 Ohm
Answer
B
Z = sqrt(30^2 + (60-20)^2) = sqrt(900 + 1600) = sqrt(2500) = 50 Ohm.
Q5 [1 mark]

In a series RLC circuit, the voltage across the resistor is V_R = 30 V, across the inductor V_L = 50 V, across the capacitor V_C = 10 V. The source voltage V_max is:

A90 V
B50 V
C40 V
D30 V
Answer
B
V_max = sqrt(V_R^2 + (V_L - V_C)^2) = sqrt(900 + 1600) = sqrt(2500) = 50 V.
Q6 [1 mark]

The unit of impedance is:

AHenry (H)
BFarad (F)
COhm (Ohm)
DWeber (Wb)
Answer
C
Impedance Z is measured in Ohms (Ohm), same as resistance.
Q7 [1 mark]

In a series RLC circuit, if X_L = X_C, the impedance equals:

A0
BX_L
CR
D2R
Answer
C
When X_L = X_C, Z = sqrt(R^2 + 0) = R. This is the resonance condition.
Q8 [1 mark]

A series RLC circuit has R = 100 Ohm, X_L = 200 Ohm, X_C = 50 Ohm. The impedance is:

A100 Ohm
B150 Ohm
C180 Ohm
D250 Ohm
Answer
C
Z = sqrt(100^2 + (200-50)^2) = sqrt(10000 + 22500) = sqrt(32500) approx 180 Ohm.
Q9 [1 mark]

The power factor of a series RLC circuit is:

Acos(phi) = Z/R
Bcos(phi) = R/Z
Ccos(phi) = X_L/Z
Dcos(phi) = X_C/Z
Answer
B
Power factor = cos(phi) = R/Z, where phi is the phase angle.
Q10 [1 mark]

In a series RLC circuit, the phase angle phi between V and I is:

Aphi = arctan(R/(X_L - X_C))
Bphi = arctan((X_L - X_C)/R)
Cphi = arctan(X_L/X_C)
Dphi = arctan(R/Z)
Answer
B
tan(phi) = (X_L - X_C)/R, so phi = arctan((X_L - X_C)/R).
Q11 [1 mark]

In a series RLC circuit with V_max = 100 V and Z = 25 Ohm, the maximum current is:

A2 A
B4 A
C25 A
D100 A
Answer
B
I_max = V_max/Z = 100/25 = 4 A.
Q12 [4 marks]

A series RLC circuit has R = 40 Ohm, L = 0.1 H, C = 100 uF, connected to V = 120 sin(200t) V. (a) Find X_L and X_C. (b) Find the impedance Z. (c) Find the maximum current. (d) Find the phase angle.

Mark Scheme
(a) X_L = wL = 200 x 0.1 = 20 Ohm; X_C = 1/(wC) = 1/(200 x 10^-4) = 50 Ohm (b) Z = sqrt(40^2 + (20-50)^2) = sqrt(1600 + 900) = sqrt(2500) = 50 Ohm (c) I_max = V_max/Z = 120/50 = 2.4 A (d) phi = arctan((X_L - X_C)/R) = arctan(-30/40) = arctan(-0.75) approx -36.9 deg (capacitive, current leads voltage)
Q13 [4 marks]

Explain the concept of impedance in a series RLC circuit and how it differs from resistance.

Mark Scheme
Resistance R opposes current by dissipating energy as heat. Impedance Z is the total opposition to AC current, combining resistance and reactance: Z = sqrt(R^2 + (X_L - X_C)^2) Unlike resistance, reactance (and therefore impedance) depends on frequency. At resonance (X_L = X_C), Z = R (minimum impedance, maximum current). The phase angle phi = arctan((X_L - X_C)/R) describes whether the circuit is inductive or capacitive.
Q1 [1 mark]

A transformer steps up voltage from 120 V to 480 V. The turns ratio N_s/N_p is:

A1/4
B1/2
C2
D4
Answer
D
V_s/V_p = N_s/N_p = 480/120 = 4.
Q2 [1 mark]

A transformer has N_p = 200 turns and N_s = 800 turns. If I_p = 4 A, the secondary current I_s is:

A0.5 A
B1 A
C4 A
D16 A
Answer
B
I_s/I_p = N_p/N_s = 200/800 = 0.25. I_s = 4 x 0.25 = 1 A.
Q3 [1 mark]

An ideal transformer has V_p = 240 V, V_s = 12 V, I_s = 10 A. The primary current I_p is:

A0.5 A
B1 A
C2 A
D5 A
Answer
A
P_p = P_s: V_p I_p = V_s I_s. I_p = (12 x 10)/240 = 0.5 A.
Q4 [1 mark]

A step-down transformer reduces voltage from 11000 V to 220 V. If the primary has 5000 turns, the secondary has:

A10 turns
B100 turns
C250 turns
D500 turns
Answer
B
N_s = N_p x (V_s/V_p) = 5000 x (220/11000) = 100 turns.
Q5 [1 mark]

In an ideal transformer, which quantity is conserved between primary and secondary?

AVoltage
BCurrent
CPower
DResistance
Answer
C
In an ideal transformer, power is conserved: P_p = P_s.
Q6 [1 mark]

A transformer has V_p = 100 V, N_p = 200, N_s = 600. The secondary voltage is:

A33 V
B100 V
C200 V
D300 V
Answer
D
V_s = V_p x (N_s/N_p) = 100 x (600/200) = 300 V.
Q7 [1 mark]

Power transmission lines use step-up transformers to:

AIncrease current and decrease voltage
BDecrease current and increase voltage
CIncrease both current and voltage
DDecrease both current and voltage
Answer
B
Stepping up voltage reduces current (P = VI), reducing I^2 R losses in transmission lines.
Q8 [1 mark]

A transformer with N_p = 400 and N_s = 50 is a:

AStep-up transformer
BStep-down transformer
CIsolation transformer
DAuto-transformer
Answer
B
N_s < N_p means V_s < V_p: it is a step-down transformer.
Q9 [1 mark]

An ideal transformer has V_p = 240 V, I_p = 2 A, and N_s/N_p = 5. The secondary power is:

A48 W
B240 W
C480 W
D2400 W
Answer
C
P = V_p I_p = 240 x 2 = 480 W. In an ideal transformer, P_s = P_p = 480 W.
Q10 [4 marks]

State the transformer equations relating primary and secondary voltages, currents, and number of turns.

Mark Scheme
For an ideal transformer: **Voltage ratio:** V_s/V_p = N_s/N_p **Current ratio:** I_s/I_p = N_p/N_s **Power conservation:** P_p = P_s (V_p I_p = V_s I_s)
Q11 [4 marks]

A transformer has N_p = 500 turns and N_s = 100 turns. If the primary voltage is 240 V, find the secondary voltage.

Mark Scheme
V_s/V_p = N_s/N_p V_s = V_p x (N_s/N_p) = 240 x (100/500) = **48 V** (step-down transformer)
Q12 [4 marks]

A power station generates electricity at 25 kV. A step-up transformer increases this to 400 kV for transmission, then a step-down transformer reduces it to 230 V for homes. (a) Find the turns ratio of the step-up transformer. (b) If the transmission current is 50 A, find the power transmitted. (c) Explain why high voltage is used for transmission.

Mark Scheme
(a) N_s/N_p = V_s/V_p = 400000/25000 = **16:1** (b) P = V x I = 400000 x 50 = **20 MW** (c) High voltage means low current (P = VI). Power loss = I^2 R. Lower current means much less energy wasted as heat in the cables.
Q13 [4 marks]

A transformer has 1200 primary turns and 80 secondary turns. The primary is connected to 240 V AC. (a) Find the secondary voltage. (b) If the secondary current is 15 A, find the primary current. (c) Find the power transferred.

Mark Scheme
(a) V_s = 240 x (80/1200) = **16 V** (b) I_p = I_s x (N_s/N_p) = 15 x (80/1200) = **1 A** (c) P = V_s I_s = 16 x 15 = **240 W**
Q1 [1 mark]

Why are high voltages used in electricity transmission?

ATo increase the current
BTo reduce power losses (I^2 R) in cables
CTo increase the resistance of cables
DTo make transformers work more efficiently
Answer
B
P_loss = I^2 R. Higher voltage means lower current for same power, reducing losses.
Q2 [1 mark]

A power line transmits 10 MW at 100 kV. The current in the line is:

A10 A
B100 A
C1000 A
D10000 A
Answer
B
I = P/V = 10x10^6 / 100x10^3 = 100 A.
Q3 [1 mark]

A transmission line has resistance 20 Ohm and carries 200 A. The power loss is:

A400 kW
B800 kW
C4 MW
D8 MW
Answer
B
P_loss = I^2 R = 200^2 x 20 = 800000 W = 800 kW.
Q4 [1 mark]

If the transmission voltage is doubled while the power stays constant, the power loss in the cables:

ADoubles
BHalves
CReduces to 1/4
DStays the same
Answer
C
P = VI so I halves. P_loss = I^2 R reduces to (I/2)^2 R = P_loss/4.
Q5 [1 mark]

A step-down transformer at a substation reduces 132 kV to 11 kV. If the secondary current is 500 A, the primary current is approximately:

A4.2 A
B41.7 A
C500 A
D6000 A
Answer
B
I_p = I_s x (N_s/N_p) = I_s x (V_s/V_p) = 500 x (11/132) approx 41.7 A.
Q6 [1 mark]

The main purpose of a transformer in a national grid is to:

AGenerate electricity
BStore electrical energy
CChange voltage levels for efficient transmission
DMeasure electrical power
Answer
C
Transformers change voltage levels, enabling efficient high-voltage transmission and safe low-voltage distribution.
Q7 [1 mark]

A power station generates 2 MW at 10 kV. A step-up transformer raises this to 200 kV. The turns ratio N_s/N_p is:

A1/20
B1/10
C10
D20
Answer
D
N_s/N_p = V_s/V_p = 200000/10000 = 20.
Q8 [1 mark]

Power loss in a transmission cable is proportional to:

AV^2
BI^2
CV
DI
Answer
B
P_loss = I^2 R. Power loss is proportional to the square of the current.
Q9 [1 mark]

A transmission line carries 50 A at 400 kV. The power transmitted is:

A8 kW
B800 kW
C20 MW
D200 MW
Answer
C
P = VI = 400000 x 50 = 20 MW.
Q10 [1 mark]

In the national grid, the voltage is stepped down near homes to approximately:

A11 kV
B230 V
C400 V
D33 kV
Answer
B
Domestic supply voltage is approximately 230 V (UK/EU) or 120 V (US).
Q11 [4 marks]

A power station outputs 50 MW at 25 kV. A step-up transformer increases the voltage to 500 kV for transmission through cables of total resistance 10 Ohm. (a) Find the transmission current. (b) Find the power loss in the cables. (c) Find the percentage power loss.

Mark Scheme
(a) I = P/V = 50x10^6 / 500x10^3 = **100 A** (b) P_loss = I^2 R = 100^2 x 10 = **100 kW** (c) % loss = 100 kW / 50000 kW x 100 = **0.2%**
Q12 [4 marks]

Explain why electricity is transmitted at high voltage and low current rather than low voltage and high current. Use the equation P_loss = I^2 R in your answer.

Mark Scheme
For a given power P = VI transmitted: If voltage V is high, current I = P/V is low. Power lost in cables: P_loss = I^2 R. Since P_loss depends on I^2, halving the current reduces losses by a factor of 4. Transmitting at high voltage (e.g., 400 kV) with low current minimises energy waste as heat in cables. Step-up transformers increase voltage at the power station; step-down transformers reduce it for safe domestic use.
Q13 [4 marks]

A power station generates 100 MW at 20 kV. It transmits via cables (R = 5 Ohm) at 400 kV. (a) Find the step-up transformer turns ratio. (b) Find the transmission current. (c) Find the power loss in the cables. (d) Find the efficiency of transmission.

Mark Scheme
(a) N_s/N_p = 400/20 = **20:1** (b) I = P/V = 100x10^6 / 400x10^3 = **250 A** (c) P_loss = I^2 R = 250^2 x 5 = **312.5 kW** (d) Efficiency = (P - P_loss)/P x 100 = (100000 - 312.5)/100000 x 100 = **99.7%**
Q1 [1 mark]

The phase angle phi in a series RLC circuit is given by:

Atan(phi) = R/(X_L - X_C)
Btan(phi) = (X_L - X_C)/R
Ctan(phi) = (X_L + X_C)/R
Dtan(phi) = R/Z
Answer
B
tan(phi) = (X_L - X_C)/R from the phasor triangle.
Q2 [1 mark]

A series RLC circuit has R = 30 Ohm, X_L = 70 Ohm, X_C = 40 Ohm. The phase angle is:

A30 deg
B45 deg
C60 deg
D90 deg
Answer
B
tan(phi) = (70-40)/30 = 1. phi = 45 deg.
Q3 [1 mark]

When phi > 0 in a series RLC circuit, the circuit is:

ACapacitive
BResistive
CInductive
DAt resonance
Answer
C
phi > 0 means X_L > X_C: the circuit is inductive (voltage leads current).
Q4 [1 mark]

When phi < 0 in a series RLC circuit, the circuit is:

ACapacitive
BResistive
CInductive
DAt resonance
Answer
A
phi < 0 means X_C > X_L: the circuit is capacitive (current leads voltage).
Q5 [1 mark]

A series RLC circuit has R = 50 Ohm, X_L = 50 Ohm, X_C = 100 Ohm. The phase angle is:

A-45 deg
B0 deg
C45 deg
D90 deg
Answer
A
tan(phi) = (50-100)/50 = -1. phi = -45 deg (capacitive).
Q6 [1 mark]

At resonance in a series RLC circuit, the phase angle is:

A90 deg
B45 deg
C0 deg
D-90 deg
Answer
C
At resonance X_L = X_C, so tan(phi) = 0, phi = 0 deg.
Q7 [1 mark]

A series RLC circuit has phi = 60 deg and R = 20 Ohm. The net reactance (X_L - X_C) is:

A20 Ohm
B34.6 Ohm
C40 Ohm
D60 Ohm
Answer
B
tan(60) = (X_L-X_C)/R. X_L-X_C = R tan(60) = 20 x 1.732 = 34.6 Ohm.
Q8 [1 mark]

In a series RLC circuit, the power factor is:

Acos(phi) = R/Z
Bcos(phi) = Z/R
Ccos(phi) = X_L/Z
Dcos(phi) = (X_L-X_C)/Z
Answer
A
Power factor = cos(phi) = R/Z.
Q9 [1 mark]

A series RLC circuit has R = 60 Ohm, X_L = 100 Ohm, X_C = 20 Ohm. The impedance Z is:

A60 Ohm
B80 Ohm
C100 Ohm
D180 Ohm
Answer
C
Z = sqrt(60^2 + (100-20)^2) = sqrt(3600 + 6400) = sqrt(10000) = 100 Ohm.
Q10 [4 marks]

A series RLC circuit has R = 40 Ohm, X_L = 70 Ohm, X_C = 30 Ohm. Find the phase angle phi.

Mark Scheme
tan(phi) = (X_L - X_C)/R = (70 - 30)/40 = 40/40 = 1 phi = arctan(1) = **45 degrees** (inductive: voltage leads current)
Q11 [4 marks]

A series RLC circuit has R = 30 Ohm, X_L = 20 Ohm, X_C = 50 Ohm. Find the phase angle phi and state whether the circuit is inductive or capacitive.

Mark Scheme
tan(phi) = (X_L - X_C)/R = (20 - 50)/30 = -30/30 = -1 phi = arctan(-1) = **-45 degrees** (capacitive: current leads voltage)
Q12 [4 marks]

A series RLC circuit has R = 50 Ohm, L = 0.2 H, C = 50 uF, connected to V = 200 sin(100t) V. (a) Find X_L and X_C. (b) Find Z. (c) Find the phase angle phi. (d) State whether the circuit is inductive or capacitive.

Mark Scheme
(a) X_L = wL = 100 x 0.2 = 20 Ohm; X_C = 1/(wC) = 1/(100 x 50x10^-6) = 200 Ohm (b) Z = sqrt(50^2 + (20-200)^2) = sqrt(2500 + 32400) = sqrt(34900) approx 187 Ohm (c) phi = arctan((20-200)/50) = arctan(-3.6) approx -74.4 deg (d) phi < 0 so the circuit is **capacitive** (current leads voltage)
Q13 [4 marks]

Explain how the phase angle phi in a series RLC circuit changes as the frequency increases from very low to very high.

Mark Scheme
At very low frequency: X_C >> X_L, so phi is large and negative (capacitive). As frequency increases: X_C decreases, X_L increases. At resonance frequency w_0 = 1/sqrt(LC): X_L = X_C, phi = 0 (purely resistive). Above resonance: X_L > X_C, phi becomes positive (inductive). At very high frequency: X_L >> X_C, phi approaches +90 deg.
Q1 [1 mark]

The resonant angular frequency of a series RLC circuit is:

Aw_0 = sqrt(LC)
Bw_0 = 1/sqrt(LC)
Cw_0 = LC
Dw_0 = 1/(LC)
Answer
B
w_0 = 1/sqrt(LC) is the natural resonant frequency.
Q2 [1 mark]

At resonance in a series RLC circuit, the impedance is:

AMaximum
BMinimum (= R)
CZero
DEqual to X_L
Answer
B
At resonance X_L = X_C, so Z = R (minimum).
Q3 [1 mark]

At resonance in a series RLC circuit, the current is:

AZero
BMinimum
CMaximum (= V_max/R)
DEqual to V_max/X_L
Answer
C
At resonance Z = R (minimum), so I_max = V_max/R (maximum).
Q4 [1 mark]

A series RLC circuit has L = 0.1 H and C = 10 uF. The resonant frequency is approximately:

A159 Hz
B503 Hz
C1000 Hz
D3183 Hz
Answer
A
w_0 = 1/sqrt(0.1 x 10^-5) = 1/sqrt(10^-6) = 1000 rad/s. f = 1000/(2pi) approx 159 Hz.
Q5 [1 mark]

At resonance, the voltages across the inductor and capacitor are:

ABoth zero
BEqual in magnitude and in phase
CEqual in magnitude but opposite in phase
DBoth equal to the source voltage
Answer
C
V_L = IX_L and V_C = IX_C. At resonance X_L = X_C so |V_L| = |V_C|, but they are 180 deg out of phase and cancel.
Q6 [1 mark]

The quality factor Q of a series RLC circuit is defined as:

AQ = R/w_0 L
BQ = w_0 L/R
CQ = w_0 RC
DQ = R sqrt(C/L)
Answer
B
Q = w_0 L/R = 1/(w_0 RC) = (1/R) sqrt(L/C). Higher Q means sharper resonance.
Q7 [1 mark]

A series RLC circuit resonates at 1000 Hz. If L is quadrupled, the new resonant frequency is:

A250 Hz
B500 Hz
C1000 Hz
D2000 Hz
Answer
B
w_0 = 1/sqrt(LC). If L quadruples, w_0 halves. f_0 = 1000/2 = 500 Hz.
Q8 [1 mark]

At resonance, the power factor of a series RLC circuit is:

A0
B0.5
C0.707
D1
Answer
D
At resonance phi = 0, so power factor = cos(0) = 1.
Q9 [1 mark]

A series RLC circuit has R = 10 Ohm, L = 0.01 H, C = 100 uF. The resonant frequency is:

A159 Hz
B503 Hz
C1000 Hz
D1592 Hz
Answer
A
w_0 = 1/sqrt(0.01 x 10^-4) = 1/sqrt(10^-6) = 1000 rad/s. f = 159 Hz.
Q10 [4 marks]

A series RLC circuit has L = 20 mH and C = 0.05 uF. Find the resonant frequency.

Mark Scheme
w_0 = 1/sqrt(LC) = 1/sqrt(20x10^-3 x 0.05x10^-6) = 1/sqrt(10^-9) = 1/31.62x10^-6 = **31623 rad/s** f_0 = w_0/(2pi) = 31623/(2pi) approx **5033 Hz**
Q11 [4 marks]

In a series RLC circuit at resonance, explain what happens to impedance, current, and phase angle.

Mark Scheme
At resonance (w = w_0 = 1/sqrt(LC)): - X_L = X_C, so net reactance = 0 - Impedance Z = sqrt(R^2 + 0) = **R** (minimum) - Current I_max = V_max/R (**maximum**) - Phase angle phi = 0 (voltage and current **in phase**)
Q12 [4 marks]

A series RLC circuit has R = 20 Ohm, L = 50 mH, C = 20 uF. (a) Find the resonant angular frequency. (b) Find the maximum current at resonance if V_max = 100 V. (c) Find the voltage across the inductor at resonance.

Mark Scheme
(a) w_0 = 1/sqrt(LC) = 1/sqrt(50x10^-3 x 20x10^-6) = 1/sqrt(10^-6) = **1000 rad/s** (b) At resonance Z = R = 20 Ohm. I_max = 100/20 = **5 A** (c) X_L = w_0 L = 1000 x 0.05 = 50 Ohm. V_L = I_max x X_L = 5 x 50 = **250 V**
Q13 [4 marks]

Explain the phenomenon of resonance in a series RLC circuit and describe one practical application.

Mark Scheme
Resonance occurs when the driving frequency equals the natural frequency w_0 = 1/sqrt(LC). At this frequency X_L = X_C, the net reactance is zero, and impedance Z = R (minimum). The current reaches its maximum value I_max = V_max/R. The voltages across L and C are equal and opposite, cancelling each other. Application: Radio tuning circuits use resonance to select a specific broadcast frequency.
Q1 [1 mark]

The speed of all electromagnetic waves in vacuum is:

A3.0 x 10^6 m/s
B3.0 x 10^8 m/s
C3.0 x 10^10 m/s
DDepends on frequency
Answer
B
All EM waves travel at c = 3.0 x 10^8 m/s in vacuum, regardless of frequency.
Q2 [1 mark]

An EM wave has wavelength 500 nm. Its frequency is:

A6.0 x 10^14 Hz
B1.5 x 10^-1 Hz
C1.5 x 10^17 Hz
D6.0 x 10^11 Hz
Answer
A
f = c/lambda = 3x10^8 / 500x10^-9 = 6.0 x 10^14 Hz.
Q3 [1 mark]

A radio wave has frequency 100 MHz. Its wavelength is:

A0.3 m
B3 m
C30 m
D300 m
Answer
B
lambda = c/f = 3x10^8 / 10^8 = 3 m.
Q4 [1 mark]

Which EM wave has the highest frequency?

ARadio waves
BVisible light
CX-rays
DGamma rays
Answer
D
Gamma rays have the highest frequency (and shortest wavelength) in the EM spectrum.
Q5 [1 mark]

An EM wave has wavelength 0.12 m. This is in the:

ARadio wave region
BMicrowave region
CInfrared region
DVisible light region
Answer
B
Wavelength 0.12 m = 12 cm is in the microwave region.
Q6 [1 mark]

The relationship between frequency, wavelength, and speed for an EM wave is:

Ac = f/lambda
Bc = f x lambda
Cc = lambda/f
Dc = f^2 x lambda
Answer
B
c = f x lambda, where c = 3 x 10^8 m/s.
Q7 [1 mark]

An EM wave has period T = 2.0 x 10^-15 s. Its wavelength is:

A6.0 x 10^-7 m
B6.0 x 10^-8 m
C6.0 x 10^-6 m
D6.0 x 10^-9 m
Answer
A
f = 1/T = 5x10^14 Hz. lambda = c/f = 3x10^8/5x10^14 = 6x10^-7 m.
Q8 [1 mark]

Infrared radiation has wavelength 10 um. Its frequency is:

A3.0 x 10^10 Hz
B3.0 x 10^12 Hz
C3.0 x 10^13 Hz
D3.0 x 10^14 Hz
Answer
C
f = c/lambda = 3x10^8 / 10x10^-6 = 3x10^13 Hz.
Q9 [1 mark]

The wavelength of a 2.4 GHz WiFi signal is approximately:

A1.25 cm
B12.5 cm
C1.25 m
D12.5 m
Answer
B
lambda = c/f = 3x10^8 / 2.4x10^9 = 0.125 m = 12.5 cm.
Q10 [4 marks]

An EM wave has frequency f = 5.0 x 10^14 Hz. Find the wavelength in vacuum.

Mark Scheme
c = f x lambda lambda = c/f = (3.0 x 10^8) / (5.0 x 10^14) = **6.0 x 10^-7 m = 600 nm** (visible light)
Q11 [4 marks]

An EM wave has wavelength lambda = 0.12 m. Find the frequency.

Mark Scheme
f = c/lambda = (3.0 x 10^8) / 0.12 = **2.5 x 10^9 Hz = 2.5 GHz** (microwave)
Q12 [4 marks]

(a) State the relationship between the speed, frequency, and wavelength of an EM wave. (b) A mobile phone uses frequency 1.8 GHz. Find the wavelength. (c) State which part of the EM spectrum this belongs to.

Mark Scheme
(a) c = f x lambda, where c = 3.0 x 10^8 m/s (b) lambda = c/f = 3x10^8 / 1.8x10^9 = **0.167 m = 16.7 cm** (c) This is in the **microwave** region of the EM spectrum.
Q13 [4 marks]

Arrange the following EM waves in order of increasing frequency: visible light, X-rays, radio waves, gamma rays, microwaves, UV. State one use for each.

Mark Scheme
Increasing frequency order: Radio waves < Microwaves < Infrared < Visible light < UV < X-rays < Gamma rays Uses: Radio (communication), Microwave (cooking/radar), Visible (sight), UV (sterilisation), X-ray (medical imaging), Gamma (cancer treatment).
Q1 [1 mark]

Malus's Law states that the transmitted intensity through a polarizer is:

AI = I_0 cos(theta)
BI = I_0 cos^2(theta)
CI = I_0 sin^2(theta)
DI = I_0 / cos^2(theta)
Answer
B
Malus's Law: I = I_0 cos^2(theta).
Q2 [1 mark]

Polarized light of intensity I_0 passes through a polarizer at theta = 60 deg. The transmitted intensity is:

AI_0/4
BI_0/2
CI_0 sqrt(3)/2
DI_0
Answer
A
I = I_0 cos^2(60) = I_0 x (0.5)^2 = I_0/4.
Q3 [1 mark]

When polarized light passes through a polarizer aligned with the polarization direction (theta = 0), the transmitted intensity is:

A0
BI_0/2
CI_0
D2I_0
Answer
C
I = I_0 cos^2(0) = I_0 x 1 = I_0. All light is transmitted.
Q4 [1 mark]

When polarized light passes through a polarizer at theta = 90 deg, the transmitted intensity is:

A0
BI_0/2
CI_0
D2I_0
Answer
A
I = I_0 cos^2(90) = I_0 x 0 = 0. No light is transmitted.
Q5 [1 mark]

Unpolarized light of intensity I_0 passes through a single polarizer. The transmitted intensity is:

AI_0
BI_0/2
CI_0/4
D0
Answer
B
A polarizer transmits half the intensity of unpolarized light: I = I_0/2.
Q6 [1 mark]

Polarized light of intensity 200 W/m^2 passes through a polarizer at 45 deg. The transmitted intensity is:

A50 W/m^2
B100 W/m^2
C141 W/m^2
D200 W/m^2
Answer
B
I = 200 x cos^2(45) = 200 x 0.5 = 100 W/m^2.
Q7 [1 mark]

Two polarizers are crossed (theta = 90 deg). Polarized light passes through both. The final transmitted intensity is:

AI_0
BI_0/2
CI_0/4
D0
Answer
D
I = I_0 cos^2(90) = 0. Crossed polarizers block all light.
Q8 [1 mark]

At what angle theta does a polarizer transmit exactly half the incident polarized intensity?

A30 deg
B45 deg
C60 deg
D90 deg
Answer
B
I = I_0 cos^2(theta) = I_0/2 → cos^2(theta) = 0.5 → theta = 45 deg.
Q9 [1 mark]

Polarized light of intensity I_0 passes through a polarizer at theta = 30 deg. The transmitted intensity is:

AI_0/4
BI_0/2
C3I_0/4
DI_0
Answer
C
I = I_0 cos^2(30) = I_0 x 0.75 = 3I_0/4.
Q10 [4 marks]

State Malus's Law and explain its physical meaning.

Mark Scheme
**Malus's Law:** I = I_0 cos^2(theta) where I is the transmitted intensity, I_0 is the incident intensity, and theta is the angle between the polarization axis of the polarizer and the electric field direction of the incident polarized light. Physical meaning: When polarized light passes through a polarizer, only the component of the electric field parallel to the polarizer axis is transmitted.
Q11 [4 marks]

Polarized light of intensity I_0 = 400 W/m^2 passes through a polarizer at theta = 30 deg. Find the transmitted intensity.

Mark Scheme
I = I_0 cos^2(theta) = 400 x cos^2(30 deg) = 400 x (0.866)^2 = 400 x 0.75 = **300 W/m^2**
Q12 [4 marks]

Unpolarized light of intensity 800 W/m^2 passes through two polarizers. The first polarizer has its axis vertical. The second is at 30 deg to the first. (a) Find the intensity after the first polarizer. (b) Find the intensity after the second polarizer.

Mark Scheme
(a) Unpolarized light through first polarizer: I_1 = I_0/2 = 800/2 = **400 W/m^2** (b) Malus's Law: I_2 = I_1 cos^2(30) = 400 x 0.75 = **300 W/m^2**
Q13 [4 marks]

Explain why unpolarized light passing through a single polarizer emerges with half the original intensity.

Mark Scheme
Unpolarized light has electric field oscillations in all directions with equal probability. A polarizer only transmits the component of the electric field parallel to its axis. On average, cos^2(theta) averaged over all angles = 1/2. Therefore the transmitted intensity = I_0 x (1/2) = I_0/2.
Q1 [1 mark]

Light is said to be polarized when:

AIt travels faster than normal
BIts electric field oscillates in a single plane
CIt has a single frequency
DIt travels in a straight line
Answer
B
Polarized light has its electric field oscillating in a single plane (plane of polarization).
Q2 [1 mark]

Unpolarized light can be polarized by:

AReflection at Brewster's angle
BPassing through a polarizer
CScattering by small particles
DAll of the above
Answer
D
Polarization can occur by reflection, transmission through a polarizer, or scattering.
Q3 [1 mark]

When unpolarized light passes through two crossed polarizers (90 deg apart), the transmitted intensity is:

AI_0
BI_0/2
CI_0/4
D0
Answer
D
After first polarizer: I_0/2. After second (90 deg): I_0/2 x cos^2(90) = 0.
Q4 [1 mark]

Unpolarized light of intensity 600 W/m^2 passes through a polarizer then an analyzer at 45 deg. The final intensity is:

A75 W/m^2
B150 W/m^2
C300 W/m^2
D600 W/m^2
Answer
B
After polarizer: 300 W/m^2. After analyzer: 300 cos^2(45) = 300 x 0.5 = 150 W/m^2.
Q5 [1 mark]

Brewster's angle is the angle of incidence at which:

AAll light is reflected
BThe reflected light is completely polarized
CAll light is transmitted
DThe refracted light is completely polarized
Answer
B
At Brewster's angle, the reflected light is completely polarized (s-polarization only).
Q6 [1 mark]

The plane of polarization of light is defined as the plane containing:

AThe magnetic field and the direction of propagation
BThe electric field and the direction of propagation
CBoth E and B fields
DThe direction of propagation only
Answer
B
The plane of polarization contains the electric field vector and the direction of wave propagation.
Q7 [1 mark]

Sunlight scattered by the atmosphere is:

AUnpolarized
BPartially polarized
CCompletely polarized
DCircularly polarized
Answer
B
Scattered sunlight is partially polarized due to the scattering process.
Q8 [1 mark]

Polaroid sunglasses reduce glare because they:

AAbsorb all light
BTransmit only vertically polarized light, blocking horizontally polarized reflected glare
CReflect all light
DIncrease the frequency of light
Answer
B
Reflected glare is horizontally polarized. Polaroid lenses have vertical transmission axes, blocking the glare.
Q9 [1 mark]

Unpolarized light of intensity I_0 passes through a polarizer at 0 deg, then an analyzer at 60 deg. The final intensity is:

AI_0/8
BI_0/4
CI_0/2
D3I_0/8
Answer
A
After polarizer: I_0/2. After analyzer: (I_0/2) cos^2(60) = (I_0/2)(1/4) = I_0/8.
Q10 [4 marks]

Unpolarized light of intensity I_0 passes through two polarizers. The first is a polarizer, the second (analyzer) is at theta = 60 deg to the first. Find the final transmitted intensity.

Mark Scheme
After first polarizer: I_1 = I_0/2 After second polarizer (Malus's Law): I_2 = I_1 cos^2(60) = (I_0/2)(0.25) = **I_0/8**
Q11 [4 marks]

Three polarizers are arranged in series. The first has its axis vertical. The second is at 30 deg to the first. The third is at 90 deg to the first. Unpolarized light of intensity I_0 enters. Find the final transmitted intensity.

Mark Scheme
After 1st polarizer: I_1 = I_0/2 After 2nd polarizer: I_2 = I_1 cos^2(30) = (I_0/2)(3/4) = 3I_0/8 After 3rd polarizer: angle between 2nd and 3rd = 90-30 = 60 deg I_3 = I_2 cos^2(60) = (3I_0/8)(1/4) = **3I_0/32**
Q12 [4 marks]

Unpolarized light of intensity 1200 W/m^2 passes through three polarizers. The first is at 0 deg, the second at 45 deg, the third at 90 deg. (a) Find the intensity after each polarizer. (b) Compare with the result if only the first and third polarizers are used.

Mark Scheme
(a) After 1st: I_1 = 1200/2 = 600 W/m^2 After 2nd: I_2 = 600 cos^2(45) = 600 x 0.5 = 300 W/m^2 After 3rd: angle = 90-45 = 45 deg. I_3 = 300 cos^2(45) = 300 x 0.5 = **150 W/m^2** (b) With only 1st and 3rd (crossed): I = 600 cos^2(90) = **0 W/m^2** The middle polarizer at 45 deg allows some light to pass through the crossed polarizers!
Q13 [4 marks]

Explain the difference between polarized and unpolarized light. Describe two methods by which light can be polarized.

Mark Scheme
**Unpolarized light:** Electric field oscillates in all directions perpendicular to propagation with equal probability. **Polarized light:** Electric field oscillates in a single fixed plane. Methods of polarization: 1. **Transmission through a polarizer:** A polaroid filter transmits only the component of E parallel to its axis. 2. **Reflection at Brewster's angle:** When light hits a surface at tan(theta_B) = n_2/n_1, the reflected beam is completely polarized.
Homaid bin Abdulaziz Boys School for Secondary Education  |  AP Physics C | E&M  |  Physics Department