CBSE Class 12 Physics Alternating Current MCQs Set 04

Answer: (b) 311 V

Explanation:
1. v_m = √2 × V_rms
2. v_m = 1.414 × 220 ≈ 311 V

Teacher's Note:
1. Multiply by √2 to go from rms to peak; divide by √2 to go back.
2. 156 V is what you get by wrongly dividing by √2.

Answer: (c) 3.54 A and 50 Hz

Explanation:
1. Compare with i = i_m sin ωt: i_m = 5 A and ω = 100π rad s⁻¹.
2. I_rms = \( \frac{5}{\sqrt{2}} \) ≈ 3.54 A and ν = \( \frac{\omega}{2\pi} = \frac{100\pi}{2\pi} \) = 50 Hz
3. Answer: 3.54 A and 50 Hz

Teacher's Note:
1. The number multiplying t is ω, not ν.
2. Divide ω by 2π to get the frequency in hertz.

Answer: (a) zero

Explanation:
1. The current is positive for one half-cycle and equally negative for the next.
2. The two halves cancel, so the average over a full cycle is zero.

Teacher's Note:
1. This is why we use the rms value, not the average, to describe ac.
2. The heating effect does not cancel, because it depends on i².

Answer: (d) 157 Ω

Explanation:
1. X_L = 2πνL = 2 × 3.14 × 50 × 0.5
2. X_L ≈ 157 Ω

Teacher's Note:
1. Do not forget the factor 2π; νL alone gives 25 Ω.
2. X_L is measured in ohm, just like resistance.

Answer: (b) 318 Ω

Explanation:
1. X_C = \( \frac{1}{2\pi\nu C} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} \)
2. X_C = \( \frac{1}{3.14 \times 10^{-3}} \) ≈ 318 Ω

Teacher's Note:
1. Convert μF to F (× 10⁻⁶) before substituting.
2. The first option is 2πνC itself; remember to take its reciprocal.

Answer: (a) X_L doubles and X_C is halved

Explanation:
1. X_L = 2πνL is proportional to ν, so it doubles.
2. X_C = 1/(2πνC) is inversely proportional to ν, so it is halved.
3. Answer: X_L doubles and X_C is halved

Teacher's Note:
1. An inductor resists high frequencies; a capacitor resists low frequencies.
2. This opposite behaviour is what makes resonance possible.

Answer: (c) a pure inductor

Explanation:
1. The current reaches its maximum a quarter of a cycle after the voltage does, so i lags v by 90°.
2. A lag of exactly 90° happens for a pure inductor.

Teacher's Note:
1. Read the graph from left to right: whichever peaks first is leading.
2. For a capacitor the current would peak first; for a resistor both would peak together.

Answer: (d) 50 Ω

Explanation:
1. Z = \( \sqrt{R^2 + (X_L - X_C)^2} \) = \( \sqrt{30^2 + 40^2} \)
2. Z = \( \sqrt{2500} \) = 50 Ω

Teacher's Note:
1. X_L and X_C act in opposite directions, so subtract them first.
2. Adding R, X_L and X_C directly (150 Ω) ignores phase.

Answer: (b) 0.8

Explanation:
1. Z = \( \sqrt{8^2 + (10 - 4)^2} \) = \( \sqrt{100} \) = 10 Ω
2. Power factor cos φ = \( \frac{R}{Z} = \frac{8}{10} \) = 0.8

Teacher's Note:
1. 0.6 is sin φ, not cos φ.
2. Since X_L > X_C, the current lags the voltage.

Answer: (a) 159 Hz

Explanation:
1. ν₀ = \( \frac{1}{2\pi\sqrt{LC}} \), with LC = 0.1 × 10 × 10⁻⁶ = 10⁻⁶, so √(LC) = 10⁻³ s
2. ν₀ = \( \frac{1}{2\pi \times 10^{-3}} \) ≈ 159 Hz

Teacher's Note:
1. 1000 is the angular frequency ω₀ in rad s⁻¹, not the frequency in Hz.
2. The value of R does not affect the resonant frequency.

Answer: (c) The impedance equals the resistance and the current is maximum.

Explanation:
1. At resonance X_L = X_C, so Z = \( \sqrt{R^2 + 0} \) = R, the least possible value.
2. With minimum impedance the current is maximum, the phase angle is zero and the power factor is 1.
3. Answer: the impedance equals the resistance and the current is maximum

Teacher's Note:
1. Remember: at resonance Z is minimum, I is maximum and cos φ = 1.
2. The voltages across L and C are equal and opposite, so they cancel.

Answer: (d) I sin φ

Explanation:
1. The component of I along V is I cos φ; it consumes power.
2. The component at 90° to V is I sin φ; its average power is zero.
3. Wattless current = I sin φ

Teacher's Note:
1. The word 'wattless' means it carries no average power.
2. In a pure inductor or capacitor, φ = 90°, so the whole current is wattless.

Answer: (b) 0.2 A

Explanation:
1. For an ideal transformer, \( \frac{I_s}{I_p} = \frac{N_p}{N_s} \)
2. I_s = 2 × \( \frac{1}{10} \) = 0.2 A

Teacher's Note:
1. A step-up transformer raises voltage but lowers current in the same ratio.
2. Power in = power out, so V and I change in opposite ways.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For dc, ω = 0, so X_C → ∞ and no steady current flows. For ac, X_C is finite, so current flows. The Assertion is correct.
2. The Reason is a correct statement of the formula.
3. The infinite reactance at ω = 0 is exactly why dc is blocked, so the Reason explains the Assertion.

Teacher's Note:
1. The higher the frequency, the more easily ac passes through a capacitor.
2. Charge does not actually cross the gap; the plates charge and discharge.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. P = VI cos φ with φ = π/2 gives cos φ = 0, so P = 0. The Assertion is correct.
2. The current lags the voltage by π/2 in a pure inductor, so the Reason is correct.
3. It is this phase difference that makes cos φ zero, so the Reason explains the Assertion.

Teacher's Note:
1. Energy stored in the magnetic field in one quarter-cycle is returned in the next.
2. A real coil has some resistance and does consume a little power.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Steady dc produces a constant flux, so no emf is induced in the secondary. The Assertion is correct.
2. N_s > N_p for a step-up transformer, so the Reason is correct.
3. The number of turns does not explain why dc fails; the reason is that the flux must change. So the Reason is not the correct explanation.

Teacher's Note:
1. A transformer works on mutual induction, which needs a changing current.
2. Choose (b) when both are true but the Reason talks about something else.

Answer: (c) A is true, but R is false.

Explanation:
1. At resonance X_L = X_C, so tan φ = 0 and φ = 0. The Assertion is correct.
2. At resonance Z = R, which is the minimum impedance, not the maximum. The Reason is false.

Teacher's Note:
1. Minimum impedance means maximum current at resonance.
2. Do not confuse series resonance with any 'maximum opposition' idea.

Answer: (d) A is false, but R is true.

Explanation:
1. Since X_L = ωL increases with frequency, an inductor opposes high-frequency ac more. The Assertion is false.
2. X_L = ωL is the correct formula, so the Reason is true.

Teacher's Note:
1. An inductor behaves like a plain wire (zero reactance) for dc.
2. Inductors are used as chokes to block high-frequency signals.

Answer: (c) A is true, but R is false.

Explanation:
1. For the same power P = VI, a high V means a small I, so the loss I²R in the lines is small. The Assertion is correct.
2. High voltage makes the current small, not large, and a large current would increase the loss. The Reason is false.

Teacher's Note:
1. Line loss depends on I², so halving the current cuts the loss to one-fourth.
2. Step-up transformers at the station and step-down transformers near users make this possible.

Answer: (c) 50 Ω

Explanation:
1. Z = \( \sqrt{R^2 + (X_L - X_C)^2} \) = \( \sqrt{40^2 + (60 - 30)^2} \)
2. Z = \( \sqrt{1600 + 900} \) = 50 Ω

Teacher's Note:
1. Subtract the reactances first: 60 − 30 = 30 Ω.
2. Then use Pythagoras with R = 40 Ω.

Answer: (a) 4 A

Explanation:
1. I = \( \frac{V}{Z} = \frac{200}{50} \) = 4 A (rms)

Teacher's Note:
1. AC meters read rms values, and V = 200 V is already rms.
2. 5 A would be obtained by wrongly dividing by R alone.

Answer: (d) 0.8, current lags the voltage

Explanation:
1. cos φ = \( \frac{R}{Z} = \frac{40}{50} \) = 0.8
2. Since X_L (60 Ω) > X_C (30 Ω), the circuit is inductive and the current lags.
3. Answer: 0.8, current lags the voltage

Teacher's Note:
1. Inductive circuit: current lags; capacitive circuit: current leads.
2. 0.6 = 30/50 is sin φ.

Answer: (b) 640 W

Explanation:
1. P = VI cos φ = 200 × 4 × 0.8 = 640 W
2. Check: P = I²R = 16 × 40 = 640 W

Teacher's Note:
1. Only the resistor dissipates power; L and C store and return energy.
2. 800 W = VI is the apparent power, not the true power.

Answer: (a) 5 A

Explanation:
1. At resonance X_L = X_C, so Z = R = 40 Ω.
2. I = \( \frac{200}{40} \) = 5 A

Teacher's Note:
1. At resonance the current is the largest possible for that R.
2. The new capacitor must have X_C = 60 Ω, i.e. half the old capacitance.

Answer: (b) 60

Explanation:
1. \( \frac{N_s}{N_p} = \frac{V_s}{V_p} \) ⇒ N_s = 1200 × \( \frac{11}{220} \)
2. N_s = 60 turns

Teacher's Note:
1. A step-down transformer has fewer turns in the secondary.
2. 24000 comes from turning the ratio upside down.

Answer: (d) 0.1 A

Explanation:
1. For an ideal transformer, V_pI_p = V_sI_s
2. I_p = \( \frac{11 \times 2}{220} \) = 0.1 A

Teacher's Note:
1. The low-voltage side carries the larger current.
2. Power in = power out for an ideal transformer.

Answer: (c) 88%

Explanation:
1. Output power = V_sI_s = 11 × 2 = 22 W
2. η = \( \frac{22}{25} \) × 100% = 88%

Teacher's Note:
1. The missing 3 W is lost as heat in the windings and the core.
2. Efficiency is always output divided by input.

Answer: (a) reduce heating due to currents induced in the core

Explanation:
1. The changing flux induces currents in the iron core itself, which heat it and waste energy.
2. Thin insulated sheets break up the paths of these currents, so the loss is reduced.

Teacher's Note:
1. Laminating the core does not change the turns ratio or the frequency.
2. Soft iron is used because it is easy to magnetise and demagnetise.

Answer: (d) zero

Explanation:
1. A steady current produces a constant flux in the core.
2. With no change in flux, no emf is induced, so the secondary voltage is zero.

Teacher's Note:
1. 0.6 V would be the answer only for a 12 V ac supply.
2. The primary may also overheat, because its resistance is small.