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Alternating Current Set-2 MCQ Online Practice Test | Class 12 | Quiz Tweek

Alternating Current Set-2

Practice Important MCQs for Alternating Current — Physics, Class 12.

Mastery of Alternating Current (AC) is essential for Class 12 physics and board examinations because the chapter ties together phasor methods, impedance of circuit elements, resonance phenomena and power calculations — topics that frequently appear in numerical and reasoning-based questions. Competitive exams (JEE/NEET) emphasize multi-step application of these concepts (complex impedance, power factor correction, Q‑factor, bandwidth), so practice with varied contexts builds the algebraic and conceptual agility required.

This set focuses on problems that blend quantitative calculation, phasor reasoning and interpretation of frequency/graph data (resonance curves, phase shifts, instantaneous vs. average power). Work through the multi-concept items and assertion–reason pairs to strengthen exam‑level problem solving and avoid common misconceptions (reactive voltages, series vs. parallel compensation, instantaneous power sign changes).

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Q1. A sinusoidal source $v(t)=200\sqrt{2}\sin(\omega t)\ \text{V}$ (so $V_{\rm rms}=200\ \text{V}$ ) is applied to a series circuit containing $R=50\ \Omega$ and an inductor whose reactance at that frequency is $X_L=50\ \Omega$ . Calculate (i) the RMS current in the circuit and (ii) the average power dissipated in the resistor.

(A)

Irms = $2.00\ \text{A}$ , $P_{\rm avg}=200\ \text{W}$

(B)

Irms = $4.00\ \text{A}$ , $P_{\rm avg}=800\ \text{W}$

(C)

Irms = $2.828\ \text{A}$ , $P_{\rm avg}=400\ \text{W}$

(D)

Irms = $1.414\ \text{A}$ , $P_{\rm avg}=100\ \text{W}$

Q2. In a series RLC circuit driven at its resonance frequency with $R\ll X_L,X_C$ , it is often observed that the magnitudes of voltages across the inductor and capacitor are much larger than the source voltage though they cancel in phasor sum. Which statement best explains this observation?

(A)

At resonance $X_L=X_C$ so each component must have voltage equal to the source voltage in magnitude.

(B)

At resonance the net impedance $\approx R$ is small, so large current $I\approx V_0/R$ flows; since $V_L=I X_L$ and $X_L$ (≈ $X_C$ ) can be large, individual $V_L$ and $V_C$ may exceed the source magnitude even while their phasor sum cancels.

(C)

At resonance the inductive and capacitive reactances individually become $0$ , so the circuit current becomes nearly zero and the component voltages remain small.

(D)

The source voltage is multiplied by the circuit at resonance in the same way as an ideal transformer, increasing $V_L$ and $V_C$ .

Q3. The instantaneous voltage and current for a load are given by $v(t)=220\sqrt{2}\sin(\omega t)\ \text{V}$ and $i(t)=10\sqrt{2}\sin(\omega t+\pi/3)\ \text{A}$ . From these waveforms (phase lead of current by $\pi/3$ ), the average real power absorbed by the load is

(A)

$2200\ \text{W}$

(B)

$1100/\sqrt{2}\ \text{W}$

(C)

$600\ \text{W}$

(D)

$1100\ \text{W}$

Q4. Assertion (A): The average power absorbed by an ideal capacitor connected to a sinusoidal source over one complete cycle is zero.
Reason (R): In an ideal capacitor the current leads the voltage by $\pi/2$ , so the instantaneous power $p(t)$ is symmetric with equal positive and negative areas over a cycle.

(A)

(A) Both A and R are true and R is the correct explanation of A.

(B)

(B) Both A and R are true but R is not the correct explanation of A.

(C)

(D)

(D) A is false but R is true.

Q5. A plant draws $P=1200\ \text{W}$ from a $200\ \text{V}$ (rms) AC supply with lagging power factor $\cos\phi_1=0.60$ . A capacitor bank is added in parallel to improve the power factor to $\cos\phi_2=0.90$ (lagging). Determine approximately (i) the reactive power $Q_C$ (in VAR) that the capacitor must supply and (ii) the new supply current $I_{\rm new}$ (rms, in A).

(A)

$Q_C=1600\ \text{VAR}$ , $I_{\rm new}=10.0\ \text{A}$

(B)

$Q_C\approx 581.5\ \text{VAR}$ , $I_{\rm new}\approx 6.67\ \text{A}$

(C)

$Q_C\approx 1.02\times10^{3}\ \text{VAR (capacitive)}$ , $I_{\rm new}\approx 6.67\ \text{A}$

(D)

$Q_C\approx 420\ \text{VAR}$ , $I_{\rm new}\approx 8.00\ \text{A}$

Q6. A series $R$ – $L$ circuit is connected to $v(t)=120\sqrt{2}\sin(\omega t)\ \text{V}$ . The measured RMS current is $I_{\rm rms}=2.0\ \text{A}$ and the current lags the applied voltage by $30^\circ$ . Determine the numerical values of $R$ and $X_L$ (in $\Omega$ ).

(A)

$R=30\ \Omega,\ X_L=52\ \Omega$

(B)

$R\approx 52\ \Omega,\ X_L=30\ \Omega$

(C)

$R=60\ \Omega,\ X_L=0\ \Omega$

(D)

$R\approx 52\ \Omega,\ X_L\approx 60\ \Omega$

Q7. A series RLC circuit shows resonance at $f_0=1.00\ \text{kHz}$ . The amplitude of current falls to $I_0/\sqrt{2}$ at $f_1=950\ \text{Hz}$ and $f_2=1050\ \text{Hz}$ . If the inductance is $L=0.10\ \text{H}$ , estimate (i) the quality factor $Q$ and (ii) the series resistance $R$ of the circuit (use $Q=\dfrac{f_0}{f_2-f_1}$ and $Q=\dfrac{\omega_0 L}{R}$ ).

(A)

$Q=10,\ R\approx 62.8\ \Omega$

(B)

$Q=5,\ R\approx 125.7\ \Omega$

(C)

$Q=20,\ R\approx 31.4\ \Omega$

(D)

$Q=1,\ R\approx 1256.6\ \Omega$

Q8. An inductive load has impedance $Z=10+j100\ \Omega$ (i.e. $R=10\ \Omega,\ X_L=100\ \Omega$ ) across a $220\ \text{V}$ (rms) supply. A capacitor with reactance $X_C=100\ \Omega$ is inserted in series with the load (same frequency). Estimate the RMS supply current before and after adding the capacitor, and the RMS voltage across the inductor after adding the capacitor.

(A)

Before: $2.19\ \text{A}$ ; After: $2.19\ \text{A}$ ; $V_L\approx 220\ \text{V}$

(B)

Before: $22.0\ \text{A}$ ; After: $2.19\ \text{A}$ ; $V_L\approx 2200\ \text{V}$

(C)

Before: $2.19\ \text{A}$ ; After: $22.0\ \text{A}$ ; $V_L\approx 220\ \text{V}$

(D)

Before: $2.19\ \text{A}$ ; After: $22.0\ \text{A}$ ; $V_L\approx 2200\ \text{V}$

Q9. Assertion (A): The power factor of a purely inductive AC circuit is zero.
Reason (R): The inductive reactance $X_L=\omega L$ increases with frequency $\omega$ .

(A)

(A) Both A and R are true and R is the correct explanation of A.

(B)

(B) Both A and R are true but R is not the correct explanation of A.

(C)

(D)

(D) A is false but R is true.

Q10. A series circuit has $R=2\ \Omega$ , $L=0.10\ \text{H}$ and $C=10\ \mu\text{F}$ . The source is $v(t)=100\sqrt{2}\sin(\omega t)\ \text{V}$ and the circuit is driven at its resonance angular frequency $\omega_0$ . Calculate (i) the quality factor $Q$ and (ii) the peak (amplitude) voltage across the inductor at resonance.

(A)

$Q=5,\ V_{L,\text{peak}}\approx 707.1\ \text{V}$

(B)

$Q=20,\ V_{L,\text{peak}}\approx 1414\ \text{V}$

(C)

$Q=50,\ V_{L,\text{peak}}=5000\sqrt{2}\ \text{V}\ (\approx7071\ \text{V})$

(D)

$Q=500,\ V_{L,\text{peak}}\approx 70\,710\ \text{V}$

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