iCON Education - Alternating Current JEE Advanced Test

Candidate Name

Student

Exam Name

JEE Advanced

Test Name

Alternating Current

Subject

Physics

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Question 1: Palette 1/16

JEE Advanced 2025 MCQ

A circuit with an electrical load having impedance $Z$ is connected with an AC source as shown in the diagram. The source voltage varies in time as $V(t)=300 \sin (400 t) \mathrm{V}$, where $t$ is time in s . List-I shows various options for the load. The possible currents $i(t)$ in the circuit as a function of time are given in List-II.

JEE Advanced 2025 Paper 1 Online Physics - Alternating Current Question 4 English 1

Choose the option that describes the correct match between the entries in List-I to those in ListII.

List–I	List–II
(P)	(1)
(Q)	(2)
(R)	(3)
(S)	(4)
	(5)

$ \mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1$

$\mathrm{P} \rightarrow 1, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 3$

$\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1$

$\mathrm{P} \rightarrow 1, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 5$

Question 2: Palette 2/16

JEE Advanced 2024 MCQ

The circuit shown in the figure contains an inductor $L$, a capacitor $C_0$, a resistor $R_0$ and an ideal battery. The circuit also contains two keys $\mathrm{K}_1$ and $\mathrm{K}_2$. Initially, both the keys are open and there is no charge on the capacitor. At an instant, key $\mathrm{K}_1$ is closed and immediately after this the current in $R_0$ is found to be $I_1$. After a long time, the current attains a steady state value $I_2$. Thereafter, $\mathrm{K}_2$ is closed and simultaneously $\mathrm{K}_1$ is opened and the voltage across $C_0$ oscillates with amplitude $V_0$ and angular frequency $\omega_0$.

JEE Advanced 2024 Paper 1 Online Physics - Alternating Current Question 6 English

Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

List-I	List-II
(P) The value of $I_1$ in Ampere is	(1) $0$
(Q) The value of $I_2$ in Ampere is	(2) $2$
(R) The value of $\omega_0$ in kilo-radians/s is	(3) $4$
(S) The value of $V_0$ in Volt is	(4) $20$
	(5) $200$

$\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 5$

$\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 5$

$\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 4$

$\mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 5 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 4$

Question 3: Palette 3/16

JEE Advanced 2023 MCQ

A series LCR circuit is connected to a $45 \sin (\omega t)$ Volt source. The resonant angular frequency of the circuit is $10^5 ~\mathrm{rad}~ \mathrm{s}^{-1}$ and current amplitude at resonance is $I_0$. When the angular frequency of the source is $\omega=8 \times 10^4 ~\mathrm{rad} ~\mathrm{s}^{-1}$, the current amplitude in the circuit is $0.05 I_0$. If $L=50 ~\mathrm{mH}$, match each entry in List-I with an appropriate value from List-II and choose the correct option.

List - I	List - II
(P) $I_0$ in $\mathrm{mA}$	(1) 44.4
(Q) The quality factor of the circuit	(2) 18
(R) The bandwidth of the circuit in $\mathrm{rad}~ \mathrm{s}^{-1}$	(3) 400
(S) The peak power dissipated at resonance in Watt	(4) 2250
	(5) 500

$P \rightarrow 2, Q \rightarrow 3, R \rightarrow 5, S \rightarrow 1$

$P \rightarrow 3, Q \rightarrow 1, R \rightarrow 4, S \rightarrow 2$

$P \rightarrow 4, Q \rightarrow 5, R \rightarrow 3, S \rightarrow 1$

$P \rightarrow 4, Q \rightarrow 2, R \rightarrow 1, S \rightarrow 5$

Question 5: Palette 5/16

JEE Advanced 2017 MSQ

In the circuit shown, $L = 1\,\mu H,C = 1\,\mu F\,$ and $R = 1\,k\Omega .$ They are connected in series with an a.c. source $V = {V_0}\sin \omega t$ as shown.

Which of the following options is/are correct?

JEE Advanced 2017 Paper 1 Offline Physics - Alternating Current Question 19 English

JEE Advanced 2017 Paper 1 Offline Physics - Alternating Current Question 19 English

The current will be in phase with the voltage if $\omega = {10^4}$ $rad.{s^{ - 1}}$

The frequency at which the current will be in phase with the voltage is independent of $R$

At $\omega \sim 0$ the current flowing through the circuit becomes nearly zero

At $\omega > > {10^6}rad.{s^{ - 1}},$ the circuit behaves like a capacitor

Question 7: Palette 7/16

JEE Advanced 2013 MCQ

In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1 : 10. If the power of the consumers has to be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is

200 : 1

150 : 1

100 : 1

50 : 1

Question 8: Palette 8/16

JEE Advanced 2012 MSQ

In the given circuit, the AC source has $\omega$ = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

IIT-JEE 2012 Paper 2 Offline Physics - Alternating Current Question 13 English

The current through the circuit, I is 0.3 A.

The current through the circuit, I is 0.3$\sqrt2$ A.

The voltage across 100 $\Omega$ resistor = 10$\sqrt2$ V.

The voltage across 50 $\Omega$ resistor = 10 V.

Question 9: Palette 9/16

JEE Advanced 2011 MSQ

A series RC-current is connected to AC voltage source. Consider two cases : (A) When C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current I_R through the resistor and voltage V_C across the capacitor are compared in the two cases. Which of the following is/are true?

$I_R^A > I_R^B$

$I_R^A < I_R^B$

$V_C^A > V_C^B$

$V_C^A < V_C^B$

Question 10: Palette 10/16

JEE Advanced 2010 MCQ

You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column II. When a current I (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage $V_1$ and $V_2$ (indicated in circuits) are related as shown in Column I. Match the two :

IIT-JEE 2010 Paper 2 Offline Physics - Alternating Current Question 5 English 1

IIT-JEE 2010 Paper 2 Offline Physics - Alternating Current Question 5 English 2

(A)→(R), (S), (T); (B)→(Q), (R), (S), (T); (C)→(P), (Q); (D)→(Q), (R), (S), (T)

(A)→(R), (S); (B)→(Q), (R), (S), (T); (C)→(P), (Q); (D)→(Q), (R), (T)

(A)→(R), (S), (T); (B)→(Q), (R), (S); (C)→(P), (Q); (D)→(Q), (R), (S)

(A)→(S), (T); (B)→(Q), (R), (S), (T); (C)→(P); (D)→(Q), (R), (S), (T)

Question 11: Palette 11/16

JEE Advanced 2010 MCQ

An AC voltage source of variable angular frequency $\omega$ and fixed amplitude V₀ is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When $\omega$ is increased

the bulb glows dimmer.

the bulb glows brighter.

total impedance of the circuit is unchanged.

total impedance of the circuit increases.

Question 12: Palette 12/16

JEE Advanced 2007 MCQ

STATEMENT 1

A vertical iron rod has a coil of wire wound over it at the bottom end. An alternating current flows in the coil. The rod goes through a conducting ring as shown in the figure. The ring can float at a certain height above the coil.

IIT-JEE 2007 Paper 2 Offline Physics - Alternating Current Question 8 English

Because

STATEMENT 2

In the above situation, a current is induced in the ring which interacts with the horizontal component of the magnetic field to produce an average force in the upward direction.

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

Statement-1 is True, Statement-2 is False.

Statement-1 is False, Statement-2 is True.

Question 13: Palette 13/16

JEE Advanced 2006 MCQ

Match the following columns.

	Column I		Column II
(A)	Dielectric ring uniformly charged.	(P)	Time independent electrostatic field out of system.
(B)	Dielectric ring uniformly charged rotating with angular velocity $\omega$.	(Q)	Magnetic field.
(C)	Constant current in ring $io$	(R)	Induced electric field.
(D)	$i=i_0\cos\omega t$	(S)	Magnetic moment.

A $\to$ (P); B $\to$ (Q, S); C $\to$ (Q, S); D $\to$ (Q, R, S)

A $\to$ (P); B $\to$ (S); C $\to$ (Q, S); D $\to$ (R, S)

A $\to$ (P); B $\to$ (Q); C $\to$ (Q, S); D $\to$ (S)

A $\to$ (P); B $\to$ (S); C $\to$ (Q); D $\to$ (Q, R, S)

Question 14: Palette 14/16

JEE Advanced 2006 MCQ

Initially, the capacitor was uncharged. Now, switch $S_1$ is closed and $S_2$ is kept open. If time constant of this circuit is $\tau$, then

after time interval $\tau$, charge on the capacitor is $\frac{\mathrm{CV}}{2}$.

after time interval $2 \tau$, charge on the capacitor is $\mathrm{CV}\left(1-e^{-2}\right)$.

the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged.

after time interval $2 \tau$, charge on the capacitor is $\mathrm{CV}\left(1-e^{-1}\right)$.

Question 15: Palette 15/16

JEE Advanced 2006 MCQ

After the capacitor gets fully charged, $\mathrm{S}_1$ is opened and $S_2$ is closed so that the inductor is connected in series with the capacitor. Then,

at $t=0$, the energy stored in the circuit is purely in the form of magnetic energy.

at any time $t>0$, the current in the circuit is in the same direction.

at $t>0$, there is no exchange of energy between the inductor and the capacitor.

at any time $t>0$, the instantaneous current in the circuit may $\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}$.

Question 16: Palette 16/16

JEE Advanced 2006 MCQ

If the total charge stored in the LC circuit.is $\mathrm{Q}_0$, then for $t \geq 0$,

the charge on the capacitor is

$ \mathrm{Q}=\mathrm{Q}_0 \cos \left(\frac{\pi}{2}+\frac{t}{\sqrt{\mathrm{LC}}}\right) $

the charge on the capacitor is

$ \mathrm{Q}=\mathrm{Q}_0 \cos \left(\frac{\pi}{2}-\frac{1}{\sqrt{\mathrm{LC}}}\right) . $

the charge on the capacitor is

$ \mathrm{Q}=-\mathrm{LC} \frac{d^2 \mathrm{Q}}{d t^2} . $

the charge on the capacitor is

$ \mathrm{Q}=-\frac{1}{\sqrt{\mathrm{LC}}} \frac{d^2 \mathrm{Q}}{d t^2} . $

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