Alternating Current — Class 12 Physics Worksheet | AAI ATC

Before you begin

Formulas you will need

Key formulas I_rms = I₀ ∕ √2 · V_rms = V₀ ∕ √2 · X_L = ωL = 2πfL · X_C = 1 ∕ ωC
Z = √[R² + (X_L − X_C)²] · Power factor = cosφ = R ∕ Z · P = V_rmsI_rms cosφ
Resonance: X_L = X_C, f = 1 ∕ 2π√(LC) · At resonance Z = R
Transformer: V_p ∕ V_s = N_p ∕ N_s · I_p ∕ I_s = N_s ∕ N_p

How to use this worksheet. Resistance and reactance never add arithmetically — they combine as perpendicular vectors, so impedance is the square root of the sum of squares. Remember also that inductive reactance rises with frequency while capacitive reactance falls.

Section A · Level 1

Definitions, phase and reactance

Ten questions. None should take more than twenty seconds.

Q1

The root mean square value of a sinusoidal current of peak value I₀ is:

For a sinusoid the rms value is the peak divided by the square root of two, roughly 0.707 times the peak.

Q2

The average value of a sinusoidal alternating current over one complete cycle is:

The positive and negative halves cancel exactly, so the mean over a full cycle is zero. The value 2I₀ ∕ π is the average over a half cycle only.

Q3

The SI unit of inductance is the:

Inductance is measured in henry, defined as one volt-second per ampere.

Q4

Inductive reactance is given by:

X_L = ωL = 2πfL. It rises with frequency, so an inductor blocks high frequencies.

Q5

Capacitive reactance is given by:

X_C = 1 ∕ ωC. It falls as frequency rises, so a capacitor passes high frequencies easily — the exact opposite of an inductor.

Q6

In a purely resistive AC circuit, the voltage and current are:

A resistor introduces no phase shift, so voltage and current rise and fall together and the power factor is 1.

Q7

In a purely inductive AC circuit, the current:

An inductor opposes changes in current, so the current lags the voltage by a quarter cycle. In a capacitor the current leads instead.

Q8

In a purely capacitive AC circuit, the current:

The capacitor must charge before a voltage appears across it, so the current leads the voltage by a quarter cycle.

Q9

The power factor of a purely inductive circuit is:

The phase angle is 90°, and the cosine of 90° is zero. No net power is consumed over a full cycle.

Q10

A transformer works on the principle of:

Changing flux in the primary induces an emf in the secondary through the shared core. That is mutual induction.

Section B · Level 2

Values, reactances and impedance

Nine questions. Convert the units before you substitute.

Q11

A transformer cannot operate on a direct current supply because:

An emf is induced only when the flux linking the secondary changes. Steady direct current gives constant flux, so nothing is induced.

Q12

The peak value of an alternating current is 10 A. Its rms value is closest to:

I_rms = 10 ∕ √2 = 7.07 A. Option 2 multiplies by √2 instead of dividing.

Q13

The mains supply is quoted as 220 V rms. Its peak voltage is closest to:

V₀ = V_rms × √2 = 220 × 1.414 = 311 V. Quoted mains voltages are always rms values.

Q14

The reactance of a 0.1 H inductor at 50 Hz is closest to:

X_L = 2πfL = 2π × 50 × 0.1 = 31.4 Ω.

Q15

The reactance of a 100 µF capacitor at 50 Hz is closest to:

X_C = 1 ∕ (2πfC) = 1 ∕ (2π × 50 × 100 × 10⁻⁶) = 31.8 Ω. Converting the microfarads is the step to watch.

Q16

A series circuit has R = 3 Ω and a net reactance of 4 Ω. Its impedance is:

Z = √(3² + 4²) = 5 Ω. Resistance and reactance add as perpendicular vectors, never arithmetically.

Q17

Resonance in a series LCR circuit occurs when:

At resonance the two reactances are equal and cancel, leaving only the resistance.

Q18

The resonant frequency of a series LCR circuit is:

Setting ωL = 1 ∕ ωC gives ω = 1 ∕ √(LC), so f = 1 ∕ 2π√(LC).

Q19

In a step-up transformer, the secondary has:

Voltage is proportional to turns, so raising the voltage requires more turns on the secondary. The secondary current falls correspondingly.

Section C · Level 3

Power, resonance and transformers

Ten questions. Each needs a formula plus a piece of reasoning.

Q20

The average power in an AC circuit is given by:

Only the component of current in phase with the voltage delivers power, so the true power is V_rmsI_rms cosφ. The product without the cosine is the apparent power.

Q21

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

The reactances cancel, so the impedance falls to its minimum value of R and the current is at a maximum. In a parallel circuit the behaviour is the reverse.

Q22

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

With Z = R the phase angle is zero, so cosφ = 1 and the circuit behaves as a pure resistance.

Q23

A series LCR circuit has L = 1 H and C = 1 µF. Its resonant frequency is closest to:

f = 1 ∕ 2π√(1 × 10⁻⁶) = 1 ∕ (2π × 10⁻³) ≈ 159 Hz.

Q24

In an ideal transformer, the ratio of primary to secondary current equals:

Since the power is unchanged, raising the voltage must lower the current. The current ratio is therefore the inverse of the turns ratio.

Q25

Energy losses due to eddy currents in a transformer core are reduced by:

Thin laminations separated by insulation break up the circulating current paths, which cuts the eddy current loss sharply.

Q26

A choke coil is preferred over a resistor for controlling current in an AC circuit because it:

An ideal inductor has a phase angle of 90°, so cosφ is nearly zero and it dissipates almost no energy. A resistor doing the same job would waste power as heat.

Q27

As the frequency of the supply increases, the current through a capacitor:

Capacitive reactance is inversely proportional to frequency, so higher frequency means lower reactance and hence more current.

Q28

A wattless current flows when the phase angle between voltage and current is:

At 90° the power factor is zero, so no net energy is transferred over a cycle even though a current is flowing. Purely reactive circuits behave this way.

Q29

An AC generator works on the principle of:

Rotating a coil in a magnetic field changes the flux through it continuously, inducing an alternating emf by Faraday's law.

Section D · Previous year questions

Straight from the AAI ATC papers

Five questions, exactly as they appeared. Four come from the 2023 papers, recovered by optical character recognition of the scanned answer sheets and independently re-solved.

Q30PYQ 2023 · Morning

An AC voltage is applied to a pure capacitor. The current in the capacitor ________ the voltage by ________.

In a pure capacitor the current leads the voltage by a quarter cycle, that is π/2 radians. The capacitor must be charged before a voltage appears across it, so the current is ahead. In a pure inductor the relationship is reversed.

Q31PYQ 2023 · Morning

A circuit consisting of an inductor (L) and a capacitor (C) oscillates at a frequency of 1.0 kHz. If L = 50 mH, then C is close to (consider π² ≈ 10):

For an LC oscillator f = 1 ∕ 2π√(LC), so LC = 1 ∕ 4π²f². Substituting π² = 10 and f = 10³ Hz gives LC = 1 ∕ (4 × 10 × 10⁶) = 2.5 × 10⁻⁸. Dividing by L = 0.05 H gives C = 5 × 10⁻⁷ F, which is 0.5 µF. The instruction to take π² as 10 is a hint that the arithmetic is meant to come out clean.

Q32PYQ 2023 · Evening

Consider an alternating current circuit consisting of a resistance R, an inductor of inductance L and a capacitor of capacitance C in series, driven by a voltage of amplitude V_m, such that the current flowing through the circuit is maximum. The current in the circuit is:

Maximum current means the circuit is at resonance, where X_L and X_C cancel and the impedance falls to R alone. The current is therefore simply V_m ∕ R. Recognising the phrase “current is maximum” as a statement about resonance is the whole question.

Q33PYQ 2023 · Evening

Consider a pure inductor of 20 mH connected to an alternating source of 240 V at a frequency of 100 Hz. The inductive reactance and rms current in the circuit are, respectively (consider π ≈ 3.0):

X_L = 2πfL = 2 × 3 × 100 × 0.020 = 12 Ω. The rms current is then I = V ∕ X_L = 240 ∕ 12 = 20 A. Note that the reactance and the current must both match — option 3 has the right current but the wrong reactance.

Q34PYQ 2025 · Shift 1

The input voltage across the primary coil of a transformer is 200 V and the number of turns in this coil is X. If the output voltage across the secondary coil having 1500 turns is 3000 V, then what will be the value of X?

For an ideal transformer V_p ∕ V_s = N_p ∕ N_s, so 200 ∕ 3000 = X ∕ 1500. That gives X = 1500 × 200 ∕ 3000 = 100 turns. Since the output voltage is fifteen times the input, the primary must have one fifteenth of the secondary turns — a useful sanity check before you divide.

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