iCON Education - AC Circuits

Q51 DPT AC Currents MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Aug 2026

Concept:

From the image given below, if $V_{1} = V_{0} \sin(\omega t)$ and $V_{2} = V_{0} \sin\left(\omega t + \frac{\pi}{3}\right)$, then find the resultant of the two voltages.

A.

$V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

B.

$\sqrt{2} V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

C.

$\sqrt{3} V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

D.

$2 V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

Q52 DPT AC Currents MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Aug 2026

Concept:

From the image given below, if $V_{1} = V_{0} \sin(\omega t)$ and $V_{2} = V_{0} \sin\left(\omega t + \frac{\pi}{3}\right)$, then find the resultant of the two voltages.

A.

$V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

B.

$\sqrt{2} V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

C.

$\sqrt{3} V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

D.

$2 V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

Q53 DPT AC Currents MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Aug 2026

Concept:

A voltage $V = 60 \sin(\pi t)$ volt is applied across a $20~\Omega$ resistor. What will an ac ammeter in series with the resistor read?

A.

$1.5~A$

B.

$2.12~A$

C.

$3.0~A$

D.

$4.24~A$

Q54 DPT AC Currents MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Aug 2026

Concept:

An alternating current source $E = 100 \sin(1000t)$ volt is connected through an inductor of $10~\mu H$, then write down the equation of current.

A.

$10000 \sin(1000t + \frac{\pi}{2})~\text{A}$

B.

$1000 \sin(1000t - \frac{\pi}{2})~\text{A}$

C.

$10000 \sin(1000t - \frac{\pi}{2})~\text{A}$

D.

$100 \sin(1000t - \frac{\pi}{2})~\text{A}$

Correct Answer: C

Explanation:

From the given voltage equation $E = 100 \sin(1000t)$, the peak emf is $E_{0} = 100~\text{V}$ and angular frequency is $\omega = 1000~\text{rad/s}$.
The inductance is given as $L = 10~\mu\text{H} = 10 \times 10^{-6}~\text{H}$.
The inductive reactance $X_{L}$ is calculated as $\omega L = 1000 \times 10 \times 10^{-6} = 0.01~\Omega$.
The peak current $I_{0}$ is $\frac{E_{0}}{X_{L}} = \frac{100}{0.01} = 10000~\text{A}$.
Since current lags voltage by $\pi/2$ in a pure inductor, the equation of current is $I = 10000 \sin(1000t - \frac{\pi}{2})~\text{A}$.

Q55 DPT AC Currents MCQ

iCON Education HYD, 79930 92826, 73309 72826 29 Aug 2026

Concept:

An alternating voltage $E = 200 \sqrt{2} \sin(100t)$ volt is applied to a $2~H$ inductor through an a.c. ammeter. What will be the reading of the ammeter?

A.

$1.41~A$

B.

$1.0~A$

C.

$0.7~A$

D.

$2.0~A$

Q56 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

A capacitor of $50$ pF is connected to an a.c. source of frequency $1$ kHz. Calculate its reactance.

A.

$\frac{10^6}{\pi} \Omega$

B.

$\frac{10^7}{\pi} \Omega$

C.

$\frac{10^5}{\pi} \Omega$

D.

$\frac{10^8}{\pi} \Omega$

Q57 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

In given circuit applied voltage $V = 50\sqrt{2} \sin (100\pi t)$ volt and ammeter reading is $2$ A then calculate value of $L$ from the image given below

A.

$\frac{1}{2\pi}$ H

B.

$\frac{1}{4\pi}$ H

C.

$\frac{1}{\pi}$ H

D.

$\frac{1}{3\pi}$ H

Correct Answer: B

Explanation:

$V_{rms} = I_{rms} X_{L}$
Reading of ammeter $= I_{rms}$
$X_{L} = \frac{V_{rms}}{I_{rms}} = \frac{V_{0}}{\sqrt{2} I_{rms}} = \frac{50\sqrt{2}}{\sqrt{2} \times 2} = 25 \Omega$
$\Rightarrow L = \frac{X_{L}}{\omega} = \frac{25}{100\pi} = \frac{1}{4\pi}$ H

Q58 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

An inductor of $\frac{10}{\pi}$ mH is connected across an ac-source of $50$ V, $50$ Hz. Find out impedance, root mean square current, peak current, and the equation of current.

A.

$Z = 1 \Omega, I_{rms} = 50$ A, $I_{0} = 50\sqrt{2}$ A, $I = 50\sqrt{2}\sin(100\pi t - \frac{\pi}{2})$

B.

$Z = 2 \Omega, I_{rms} = 25$ A, $I_{0} = 25\sqrt{2}$ A, $I = 25\sqrt{2}\sin(100\pi t - \frac{\pi}{2})$

C.

$Z = 1 \Omega, I_{rms} = 10$ A, $I_{0} = 10\sqrt{2}$ A, $I = 10\sqrt{2}\sin(100\pi t - \frac{\pi}{2})$

D.

$Z = 0.5 \Omega, I_{rms} = 100$ A, $I_{0} = 100\sqrt{2}$ A, $I = 100\sqrt{2}\sin(100\pi t - \frac{\pi}{2})$

Correct Answer: A

Explanation:

Given $L = \frac{10}{\pi}$ mH, $f = 50$ Hz
(i) Impedance $Z = \omega L = 2\pi f L = 2\pi \times 50 \times \frac{10}{\pi} \times 10^{-3} = 1 \Omega$
(ii) $I_{rms} = \frac{V_{rms}}{Z} = \frac{50}{1} = 50$ A
(iii) $I_{0} = I_{rms}\sqrt{2} = 50\sqrt{2}$ A
(iv) Equation of current $I = 50\sqrt{2}\sin(100\pi t - \frac{\pi}{2}) = -50\sqrt{2}\cos(100\pi t)$

Q59 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

Across two different inductors same alternating source is connected. For this $Z$ versus $\omega$ graph is given. Find out the ratio $\frac{L_{1}}{L_{2}}$ from the image given below

A.

$3$

B.

$\sqrt{3}$

C.

$\frac{1}{\sqrt{3}}$

D.

$\frac{1}{3}$

Q60 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

Calculate the impedance of the circuit shown from the image given below

A.

$30 \Omega$

B.

$40 \Omega$

C.

$50 \Omega$

D.

$70 \Omega$

Q61 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

When $10$ V, dc is applied across a coil current through it is $2.5$ A, if $10$ V, $50$ Hz A.C. is applied current reduces to $2$ A. Calculate reactance of the coil.

A.

$5 \Omega$

B.

$4 \Omega$

C.

$3 \Omega$

D.

$2 \Omega$

Correct Answer: C

Explanation:

For $10$ V D.C.: $V = IR$
Resistance of coil $R = \frac{10}{2.5} = 4 \Omega$
For $10$ V A.C.: $V = IZ \Rightarrow Z = \frac{V}{I} = \frac{10}{2} = 5 \Omega$
$Z = \sqrt{R^2 + X_{L}^2} = 5 \Rightarrow R^2 + X_{L}^2 = 25 \Rightarrow X_{L}^2 = 5^2 - 4^2 \Rightarrow X_{L} = 3 \Omega$

Q62 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

When an alternating voltage of $220$ V is applied across a device X, a current of $0.5$ A flows through the circuit and is in phase with the applied voltage. When the same voltage is applied across another device Y, the same current again flows through the circuit but it leads the applied voltage by $\frac{\pi}{2}$ radians. Calculate the current flowing in the circuit when same voltage is applied across the series combination of X and Y.

A.

$0.25$ A

B.

$0.35$ A

C.

$0.50$ A

D.

$0.70$ A

Correct Answer: B

Explanation:

X is resistor and Y is a capacitor
Since the current in the two devices is the same ($0.5$ A at $220$ V)
When R and C are in series across the same voltage then
$R = X_{C} = \frac{220}{0.5} = 440 \Omega$
$I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{rms}}{\sqrt{R^2 + X_{C}^2}} = \frac{220}{\sqrt{(440)^2 + (440)^2}} = \frac{220}{440\sqrt{2}} = 0.35$ A

Q63 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

Find out the required inductance to put in series of bulb ($10$ W, $60$ V) to run it safely across an alternating supply of $100$ V, $60$ Hz from the image given below

A.

$\frac{2}{\pi}$ H

B.

$\frac{3}{\pi}$ H

C.

$\frac{4}{\pi}$ H

D.

$\frac{5}{\pi}$ H

Correct Answer: C

Explanation:

For $RL$-circuit
$V = \sqrt{V_{R}^2 + V_{L}^2}$
$V_{L} = \sqrt{(100)^2 - (60)^2} = 80$ V
$P = VI \Rightarrow I = \frac{P}{V} = \frac{10\text{ W}}{60\text{ V}} = \frac{1}{6}$ A (in bulb)
$X_{L} = \frac{V_{L}}{I} = \frac{80}{\frac{1}{6}} = 480 \Omega$
$\therefore L = \frac{X_{L}}{\omega} = \frac{480}{2\pi(60)} = \frac{4}{\pi}$ H

Q64 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

Find out the required inductance to put in series of bulb ($10$ W, $60$ V) to run it safely across an alternating supply of $100$ V, $60$ Hz from the image given below

A.

$\frac{2}{\pi}$ H

B.

$\frac{3}{\pi}$ H

C.

$\frac{4}{\pi}$ H

D.

$\frac{5}{\pi}$ H

Correct Answer: C

Explanation:

For $RL$-circuit
$V = \sqrt{V_{R}^2 + V_{L}^2}$
$V_{L} = \sqrt{(100)^2 - (60)^2} = 80$ V
$P = VI \Rightarrow I = \frac{P}{V} = \frac{10\text{ W}}{60\text{ V}} = \frac{1}{6}$ A (in bulb)
$X_{L} = \frac{V_{L}}{I} = \frac{80}{\frac{1}{6}} = 480 \Omega$
$\therefore L = \frac{X_{L}}{\omega} = \frac{480}{2\pi(60)} = \frac{4}{\pi}$ H

Q65 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

A voltage $V = 60 \sin \pi t$ volt is applied across a $20 \Omega$ resistor. What will an ac ammeter in series with the resistor read?

A.

$1.5 \sqrt{2} A$

B.

$2.12 A$

C.

$3.0 A$

D.

$1.0 A$

Q66 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

An alternating current source $E = 100 \sin(1000t)$ volt is connected through an inductor of $10 \, \mu\text{H}$ then write down the equation of current.

A.

$10000 \sin(1000t - \frac{\pi}{2}) \, \text{A}$

B.

$5000 \sin(1000t - \frac{\pi}{2}) \, \text{A}$

C.

$1000 \sin(1000t - \frac{\pi}{2}) \, \text{A}$

D.

$10000 \cos(1000t) \, \text{A}$

Correct Answer: A

Explanation:

The given alternating voltage equation is $E = 100 \sin(1000t)$
Comparing with $E = E_{0} \sin(\omega t)$, the peak voltage $E_{0} = 100 \, \text{V}$ and angular frequency $\omega = 1000 \, \text{rad/s}$
Inductance $L = 10 \, \mu\text{H} = 10 \times 10^{-6} \, \text{H}$
Inductive reactance $X_{L} = \omega L = 1000 \times 10 \times 10^{-6} = 0.01 \, \Omega$
Peak current $I_{0} = \frac{E_{0}}{X_{L}} = \frac{100}{0.01} = 10000 \, \text{A}$
Since current lags voltage by $\frac{\pi}{2}$ in a pure inductive circuit, the equation of current is $I = I_{0} \sin(\omega t - \frac{\pi}{2}) = 10000 \sin(1000t - \frac{\pi}{2}) \, \text{A}$

Q67 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

An alternating voltage $E = 200 \sqrt{2} \sin(100t)$ volt is applied to a $2 \, \text{H}$ inductor through an ac ammeter. What will be the reading of the ammeter?

A.

$2.0 \, \text{A}$

B.

$0.5 \, \text{A}$

C.

$1.0 \, \text{A}$

D.

$1.41 \, \text{A}$

Correct Answer: C

Explanation:

The given alternating voltage equation is $E = 200 \sqrt{2} \sin(100t)$
Comparing with $E = E_{0} \sin(\omega t)$, the peak voltage $E_{0} = 200 \sqrt{2} \, \text{V}$ and angular frequency $\omega = 100 \, \text{rad/s}$
The root mean square voltage is $E_{rms} = \frac{E_{0}}{\sqrt{2}} = \frac{200 \sqrt{2}}{\sqrt{2}} = 200 \, \text{V}$
Inductance $L = 2 \, \text{H}$
Inductive reactance $X_{L} = \omega L = 100 \times 2 = 200 \, \Omega$
The reading of the ac ammeter is the rms current $I_{rms} = \frac{E_{rms}}{X_{L}} = \frac{200}{200} = 1.0 \, \text{A}$

Q68 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

A $15.0 \, \mu\text{F}$ capacitor is connected to a $220 \, \text{V}, 50 \, \text{Hz}$ source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current?

A.

$X_{C} = 212.2 \, \Omega$, $I_{rms} = 1.04 \, \text{A}$, $I_{0} = 1.47 \, \text{A}$; reactance becomes half and current is doubled

B.

$X_{C} = 106.1 \, \Omega$, $I_{rms} = 2.08 \, \text{A}$, $I_{0} = 2.94 \, \text{A}$; reactance becomes double and current is halved

C.

$X_{C} = 424.4 \, \Omega$, $I_{rms} = 0.52 \, \text{A}$, $I_{0} = 0.74 \, \text{A}$; reactance remains same and current remains same

D.

$X_{C} = 212.2 \, \Omega$, $I_{rms} = 1.04 \, \text{A}$, $I_{0} = 1.47 \, \text{A}$; reactance becomes double and current is doubled

Correct Answer: A

Explanation:

Given capacitance $C = 15.0 \, \mu\text{F} = 15 \times 10^{-6} \, \text{F}$, voltage $V_{rms} = 220 \, \text{V}$, and frequency $f = 50 \, \text{Hz}$
Capacitive reactance $X_{C} = \frac{1}{2 \pi f C} = \frac{1}{2 \times \pi \times 50 \times 15 \times 10^{-6}} = 212.2 \, \Omega$
RMS current $I_{rms} = \frac{V_{rms}}{X_{C}} = \frac{220}{212.2} = 1.04 \, \text{A}$
Peak current $I_{0} = \sqrt{2} I_{rms} = \sqrt{2} \times 1.04 = 1.47 \, \text{A}$
When frequency is doubled ($f' = 2f$), since $X_{C} \propto \frac{1}{f}$, the capacitive reactance becomes half ($106.1 \, \Omega$), and since current is inversely proportional to $X_{C}$, the current in the circuit is doubled ($2.08 \, \text{A}$).

Q69 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

A $60 \, \mu\text{F}$ capacitor is connected to a $110 \, \text{V}, 60 \, \text{Hz}$ a.c. supply. Determine the rms value of the current in the circuit.

A.

$1.25 \, \text{A}$

B.

$2.49 \, \text{A}$

C.

$3.10 \, \text{A}$

D.

$4.20 \, \text{A}$

Correct Answer: B

Explanation:

Given capacitance $C = 60 \, \mu\text{F} = 60 \times 10^{-6} \, \text{F}$, rms voltage $V_{rms} = 110 \, \text{V}$, and frequency $f = 60 \, \text{Hz}$
Capacitive reactance $X_{C} = \frac{1}{2 \pi f C} = \frac{1}{2 \times \pi \times 60 \times 60 \times 10^{-6}} = 44.21 \, \Omega$
The rms value of the current in the circuit is $I_{rms} = \frac{V_{rms}}{X_{C}} = \frac{110}{44.21} = 2.49 \, \text{A}$

Q70 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

The given graphs (a) and (b) represent the variation of the opposition offered by the circuit element to the flow of alternating current, with frequency of the applied emf from the image given below. Identify the circuit element corresponding to each graph.

A.

(a) Resistor, (b) Inductor

B.

(a) Inductor, (b) Capacitor

C.

(a) Capacitor, (b) Resistor

D.

(a) Resistor, (b) Capacitor

Q71 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

When a series combination of inductance and resistance are connected with a $10 \, \text{V}, 50 \, \text{Hz}$ a.c. source, a current of $1 \, \text{A}$ flows in the circuit. The voltage leads the current by a phase angle of $\frac{\pi}{3}$ radian. Calculate the values of resistance and inductive reactance.

A.

$R = 5 \, \Omega$, $X_{L} = 8.66 \, \Omega$

B.

$R = 10 \, \Omega$, $X_{L} = 5 \, \Omega$

C.

$R = 8.66 \, \Omega$, $X_{L} = 5 \, \Omega$

D.

$R = 5 \, \Omega$, $X_{L} = 5 \, \Omega$

Q72 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 03 Sep 2026

Concept:

From the image given below, the current in the shown circuit is found to be $i = 4\sin(314t - \frac{\pi}{4})$ A. Find the value of inductance.

A.

$0.5$ H

B.

$0.78$ H

C.

$1.0$ H

D.

$0.25$ H

Q73 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 04 Sep 2026

Concept:

A current of $4$ A flows in a coil when connected to a $12$ V dc source. If the same coil is connected to a $12$ V, $50$ rad/s a.c. source, a current of $2.4$ A flows in the circuit. Determine the inductance of the coil.

A.

$40$ mH

B.

$60$ mH

C.

$80$ mH

D.

$100$ mH

Q74 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

A current of 4A flows in a coil when connected to a 12V dc source. If the same coil is connected to a 12V, 50 rad/s a.c. source, a current of 2.4A flows in the circuit. Determine the inductance of the coil.

A.

0.04 H

B.

0.08 H

C.

0.12 H

D.

0.16 H

Correct Answer: B

Explanation:

Step 1: For the DC source, the resistance of the coil is calculated as $R = V / I_{dc} = 12 / 4 = 3 \Omega$.
Step 2: For the AC source, the impedance of the coil is calculated as $Z = V / I_{ac} = 12 / 2.4 = 5 \Omega$.
Step 3: The inductive reactance $X_L$ is found using the relation $Z = \sqrt{R^2 + X_L^2}$. Thus, $5 = \sqrt{3^2 + X_L^2} \implies 25 = 9 + X_L^2 \implies X_L = 4 \Omega$.
Step 4: The inductance $L$ is calculated using $X_L = \omega L$. Given $\omega = 50$ rad/s, $4 = 50 \times L \implies L = 4 / 50 = 0.08$ H.

Q75 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

An alternating voltage E=200 sin(300t) volt is applied across a series combination of R=10 $\Omega$ and inductance of 800mH. Calculate the impedance of the circuit.

A.

200.5 $\Omega$

B.

220.2 $\Omega$

C.

240.2 $\Omega$

D.

260.5 $\Omega$

Correct Answer: C

Explanation:

Step 1: From the given alternating voltage equation $E = 200 \sin(300t)$, the angular frequency is $\omega = 300$ rad/s.
Step 2: The given resistance is $R = 10 \Omega$ and inductance is $L = 800$ mH = $0.8$ H.
Step 3: Calculate the inductive reactance $X_L = \omega L = 300 \times 0.8 = 240 \Omega$.
Step 4: Calculate the impedance of the circuit using $Z = \sqrt{R^2 + X_L^2} = \sqrt{10^2 + 240^2} = \sqrt{100 + 57600} = \sqrt{57700} \approx 240.2 \Omega$.

Q76 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

A coil of reactance 100 $\Omega$ and resistance 100 $\Omega$ is connected to a 240 V, 50 Hz a.c. supply.
(a) What is the maximum current in the coil?
(b) What is the time lag between the voltage maximum and the current maximum?

A.

(a) 2.4 A, (b) 2.5 ms

B.

(a) 3.4 A, (b) 3.5 ms

C.

(a) 2.4 A, (b) 5.0 ms

D.

(a) 1.2 A, (b) 2.5 ms

Correct Answer: A

Explanation:

Step 1: Calculate the impedance $Z = \sqrt{R^2 + X_L^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2} \Omega$.
Step 2: Calculate the RMS current $I_{rms} = V_{rms} / Z = 240 / (100\sqrt{2})$ A.
Step 3: The maximum current is $I_0 = \sqrt{2} I_{rms} = \sqrt{2} \times (240 / (100\sqrt{2})) = 240 / 100 = 2.4$ A.
Step 4: Calculate the phase angle $\phi$ using $\tan \phi = X_L / R = 100 / 100 = 1$, which gives $\phi = \pi / 4$ rad.
Step 5: Calculate the time lag $\Delta t = \phi / (2 \pi f) = (\pi / 4) / (2 \pi \times 50) = 1 / 400 = 0.0025$ s = $2.5$ ms.

Q77 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

An inductance has a resistance of 100 $\Omega$. When a.c. signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45°. Calculate the self inductance of the coil.

A.

10.5 mH

B.

15.9 mH

C.

20.2 mH

D.

25.4 mH

Correct Answer: B

Explanation:

Step 1: The given values are resistance $R = 100 \Omega$, frequency $f = 1000$ Hz, and phase angle $\phi = 45^\circ$.
Step 2: Use the phase angle formula $\tan \phi = X_L / R$. Since $\phi = 45^\circ$, $\tan 45^\circ = 1$, which means $X_L = R = 100 \Omega$.
Step 3: Use the formula for inductive reactance $X_L = 2 \pi f L$ to find the self-inductance $L$.
Step 4: Substitute the values to get $100 = 2 \pi \times 1000 \times L \implies L = 100 / 2000\pi = 1 / (20\pi) \approx 0.0159$ H = $15.9$ mH.

Q78 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

Match the following circuit components connected across an a.c. source of angular frequency $\omega = 200$ rad/s with the phase difference between current and source voltage from the image given below.
Column I:
(A) Series combination of $10 \Omega$ and $500 \mu F$
(B) Pure inductor of $5 H$
(C) Pure capacitor of $500 \mu F$
(D) Series combination of $3 \mu F$ and $4 H$
(E) Series combination of $5 H$ and $1 k\Omega$
Column II:
(p) $\frac{\pi}{2}$
(q) $\frac{\pi}{6}$
(r) $\frac{\pi}{4}$
(s) $\frac{\pi}{3}$
(t) None of the above

A.

A-p, B-q, C-r, D-s, E-t

B.

A-r, B-p, C-p, D-p, E-r

C.

A-q, B-r, C-s, D-p, E-r

D.

A-s, B-p, C-p, D-r, E-q

Correct Answer: B

Explanation:

For (A), $X_C = \frac{1}{200 \times 500 \times 10^{-6}} = 10 \Omega$.
Since $R = 10 \Omega$, $\tan\phi = \frac{X_C}{R} = 1$, so $\phi = \frac{\pi}{4}$, hence A matches r.
For (B), a pure inductor has $R = 0$, so the phase difference is $\frac{\pi}{2}$, hence B matches p.
For (C), a pure capacitor has $R = 0$, so the phase difference is $\frac{\pi}{2}$, hence C matches p.
For (D), $X_L = 200 \times 4 = 800 \Omega$ and $X_C = \frac{1}{200 \times 3 \times 10^{-6}} = \frac{5000}{3} \Omega$.
Since the circuit is purely reactive and has non-zero net reactance, the phase difference is $\frac{\pi}{2}$, hence D matches p.
For (E), $X_L = 200 \times 5 = 1000 \Omega$ and $R = 1000 \Omega$.
Thus $\tan\phi = \frac{X_L}{R} = 1$, so $\phi = \frac{\pi}{4}$, hence E matches r.
Therefore the correct matching is A-r, B-p, C-p, D-p, E-r.

Q79 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

For the given LCR-series circuit, find out applied source voltage of ac from the image given below (with $V_R = 40$ V, $V_L = 50$ V, $V_C = 20$ V).

A.

40 V

B.

50 V

C.

70 V

D.

110 V

Q80 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

A capacitor, a resistor and a inductor are connected in series to an ac-source of 110 V and frequency 60 Hz. Find reading of voltmeter $V_3$ and ammeter in the given LCR-series circuit from the image given below (with $V_1 = 300$ V, $V_2 = 300$ V, $R = 220 \Omega$).

A.

$V_3 = 220$ V, $I = 0.25$ A

B.

$V_3 = 0$ V, $I = 1$ A

C.

$V_3 = 110$ V, $I = 1$ A

D.

$V_3 = 110$ V, $I = 0.5$ A

Q81 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

Find out the impedance of given circuit from the image given below (with $R = 4 \Omega$, $X_L = 9 \Omega$, $X_C = 6 \Omega$).

A.

3 $\Omega$

B.

4 $\Omega$

C.

7 $\Omega$

D.

5 $\Omega$

Q82 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

Find out reading of A.C. ammeter and also calculate the potential difference across resistance and capacitor from the image given below (with $E = 100 \sin \omega t$ volt, $R = 10 \Omega$, $X_C = 10 \Omega$).

A.

Ammeter = 10 A, $V_R = V_C = 100$ V

B.

Ammeter = 5 A, $V_R = V_C = 50$ V

C.

Ammeter = 5 A, $V_R = V_C = 25$ V

D.

Ammeter = 7.07 A, $V_R = V_C = 70.7$ V

Q83 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

In LCR circuit with an AC source $R = 300 \Omega$, $C = 20 \mu F$, $L = 1.0$ H, $E_{rms} = 50$ V and $f = \frac{50}{\pi}$ Hz. Find RMS current in the circuit.

A.

0.5 A

B.

1.0 A

C.

0.1 A

D.

0.2 A

Q84 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

In given circuit find the reading of ammeter and voltmeter from the image given below (with $E = 50 \sin 100\pi t$ volt, and resistors $10 \Omega$, $20 \Omega$, $10 \Omega$).

A.

Ammeter = 2.5 A, Voltmeter = 25 V

B.

Ammeter = 5.0 A, Voltmeter = 50 V

C.

Ammeter = 1.25 A, Voltmeter = 12.5 V

D.

Ammeter = 3.5 A, Voltmeter = 35 V

Q85 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

For the circuit shown in figure, write down the instantaneous current through each element from the image given below (with $V = V_0 \sin \omega t$ applied across parallel R, L, C).

A.

$i_R = \frac{V_0}{R} \sin \omega t$, $i_L = -\frac{V_0}{\omega L} \cos \omega t$, $i_C = V_0 \omega C \cos \omega t$

B.

$i_R = \frac{V_0}{R} \cos \omega t$, $i_L = \frac{V_0}{\omega L} \sin \omega t$, $i_C = -V_0 \omega C \sin \omega t$

C.

$i_R = \frac{V_0}{R} \sin \omega t$, $i_L = \frac{V_0}{\omega L} \cos \omega t$, $i_C = V_0 \omega C \sin \omega t$

D.

$i_R = \frac{V_0}{R} \sin \omega t$, $i_L = -\frac{V_0}{\omega L} \sin \omega t$, $i_C = V_0 \omega C \cos \omega t$

Q86 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

A variable frequency 230V alternating voltage source is connected across a series combination of $L = 5.0$ H, $C = 80 \mu$F and $R = 40 \Omega$. Calculate (a) The angular frequency of the source at resonance. (b) Amplitude of current at resonance frequency.

A.

(a) 50 rad/s, (b) 8.13 A

B.

(a) 100 rad/s, (b) 5.75 A

C.

(a) 50 rad/s, (b) 5.75 A

D.

(a) 100 rad/s, (b) 8.13 A

Correct Answer: A

Explanation:

(a) Angular frequency at resonance $\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5.0 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{400 \times 10^{-6}}} = \frac{1}{2 \times 10^{-2}} = 50$ rad/s.
(b) At resonance, impedance $Z = R = 40 \Omega$.
RMS current $I_{rms} = \frac{V_{rms}}{R} = \frac{230}{40} = 5.75$ A.
Amplitude of current $I_m = \sqrt{2} I_{rms} = 1.414 \times 5.75 \approx 8.13$ A.

Q87 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

A coil of resistance R and inductance L is connected in series with a capacitor C and complete combination is connected to a.c. voltage. Circuit resonates when angular frequency of supply is $\omega = \omega_0$. Find out relation between $\omega_0$, L and C and phase difference between V and I at resonance.

A.

$\omega_0 = \frac{1}{LC}$, $\phi = 45°$

B.

$\omega_0 = \sqrt{LC}$, $\phi = 90°$

C.

$\omega_0 = \frac{1}{\sqrt{LC}}$, $\phi = 0°$

D.

$\omega_0 = \frac{L}{C}$, $\phi = 60°$

Q88 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

Find the phase difference between voltage and current in series LCR circuit at half power frequencies.

A.

$\frac{\pi}{6}$

B.

$\frac{\pi}{3}$

C.

$\frac{\pi}{4}$

D.

$\frac{\pi}{2}$

Q89 DPT AC Circuits MCQ

iCON Education HYD, 79930 92826, 73309 72826 05 Sep 2026

Concept:

A series LCR circuit with $L = 0.12$ H, $C = 480$ nF, $R = 23 \Omega$ is connected to a 230 V variable frequency supply. Find –
(a) Source frequency for which current is maximum.
(b) Q–factor of the given circuit.

A.

(a) $6.63 \times 10^2$ Hz, (b) Q = 21.7

B.

(a) $3.31 \times 10^2$ Hz, (b) Q = 10.85

C.

(a) $6.63 \times 10^2$ Hz, (b) Q = 10.85

D.

(a) $1.33 \times 10^3$ Hz, (b) Q = 43.4

Physics-II Chapters

AC Circuits

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