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Alternating Currents — JEE Previous Year Questions

Every Alternating Currents question asked in JEE Main and JEE Advanced across the last 186 papers — 140 questions, each with its correct answer. Free to read, no account needed.

Questions

140

Papers it appeared in

108/186

Appearance rate

58%

All 140 Alternating Currents questions

Most recent papers first.

Q1·PhysicsSingle correctJEE Main 2026
A LCR series circuit driven with Erms=90E_{rms} = 90Erms​=90 V at frequency fd=30f_d = 30fd​=30 Hz has resistance R=80R = 80R=80 Ω, an inductance with inductive reactance XL=20.0X_L = 20.0XL​=20.0 Ω and capacitance with capacitive reactance XC=80.0X_C = 80.0XC​=80.0 Ω. The power factor of the circuit is _______.
  1. (A)0.8
  2. (B)0.64
  3. (C)0.9
  4. (D)0.5

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correctJEE Main 2026
An a.c. source of angular frequency ω\omegaω is connected across a resistor RRR and a capacitor CCC in series. The current is observed as III. Now the frequency of the source is changed to ω/4\omega/4ω/4, (keeping the voltage unchanged) the current is found to be I/3I/3I/3. The ratio of resistance to reactance at frequency ω\omegaω is
  1. (A)67\sqrt{\frac{6}{7}}76​​
  2. (B)35\sqrt{\frac{3}{5}}53​​
  3. (C)78\sqrt{\frac{7}{8}}87​​
  4. (D)34\sqrt{\frac{3}{4}}43​​

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsNumericalJEE Main 2026
A series LCR circuit with R=20R = 20R=20 Ω, L=1.6L = 1.6L=1.6 H and C=40C = 40C=40 μF is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is ________ Ω.

Correct answer: 200

Step-by-step solution →
Q4·PhysicsNumericalJEE Main 2026
An inductor of 10 mH, capacitor of 0.1 μ0.1\ \mu0.1 μF and a resistor of 100 Ω100\ \Omega100 Ω are connected in series across an a.ca.ca.c power supply 220 V, 70 Hz. The power factor of the given circuit is 0.5. The difference in the inductive reactance and capacitance reactance is 3α Ω\sqrt{3}\alpha\ \Omega3​α Ω. The value of α\alphaα is ______.

Correct answer: 100

Step-by-step solution →
Q5·PhysicsSingle correctJEE Main 2026
The figure given below shows an LCRLCRLCR series circuit with two switches S1S_1S1​ and S2S_2S2​. When switch S1S_1S1​ is closed keeping S2S_2S2​ open, the phase difference (ϕ)(\phi)(ϕ) between the current and source voltage is 30∘30^\circ30∘ and phase difference is 60∘60^\circ60∘ when S2S_2S2​ is closed keeping S1S_1S1​ open. The value of (3L1−L2)(3L_1 - L_2)(3L1​−L2​) is ________ H.
  1. (A)92\frac{9}{2}29​
  2. (B)29\frac{2}{9}92​
  3. (C)13\frac{1}{3}31​
  4. (D)333

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsNumericalJEE Main 2026
An inductor stores 16 J of magnetic field energy and dissipates 32 W of thermal energy due to its resistance when an a.c. current of 2 A (rms) and frequency 50 Hz flows through it. The ratio of inductive reactance to its resistance is _________. (π = 3.14)

Correct answer: 314

Step-by-step solution →
Q7·PhysicsSingle correctJEE Main 2026
The electric current in the circuit is given as i=i0(t/T)i = i_{0}(t/T)i=i0​(t/T). The r.m.s current for the period t=0t = 0t=0 to t=Tt = Tt=T is __________
  1. (A)i02\frac{i_{0}}{\sqrt{2}}2​i0​​
  2. (B)i0i_{0}i0​
  3. (C)i06\frac{i_{0}}{\sqrt{6}}6​i0​​
  4. (D)i03\frac{i_{0}}{\sqrt{3}}3​i0​​

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correctJEE Main 2026
For the series LCR circuit connected with 220 V, 50 Hz a.c source as shown in the figure, the power factor is α10\dfrac{\alpha}{10}10α​. The value of α is_______
  1. (A)4
  2. (B)10
  3. (C)6
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q9·PhysicsNumericalJEE Main 2026
Using a variable-frequency a.c. voltage source, the maximum current measured in the given LCR circuit is 50 mA for V=5sin(100t). The values of L and R are shown in the figure. The capacitance of the capacitor (C) used is ______ μF.

Correct answer: 50

Step-by-step solution →
Q10·PhysicsSingle correctJEE Main 2026
A capacitor C is first charged fully with potential difference of V0V_0V0​ and disconnected from the battery. The charged capacitor is connected across an inductor having inductance L. In t s 25% of the initial energy in the capacitor is transferred to the inductor. The value of t is________s.
  1. (A)πLC3\frac{\pi\sqrt{LC}}{3}3πLC​​
  2. (B)πLC6\frac{\pi\sqrt{LC}}{6}6πLC​​
  3. (C)πLC2\frac{\pi\sqrt{LC}}{2}2πLC​​
  4. (D)πLC2\pi\sqrt{\frac{LC}{2}}π2LC​​

Correct answer: (B)

Step-by-step solution →
Q11·PhysicsSingle correctJEE Advanced 2025
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(400t) 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. Choose the option that describes the correct match between the entries in List-I to those in ListII.
List-IList-II
P.see figure1.see figure
Q.see figure2.see figure
R.see figure3.see figure
S.see figure4.see figure
5.see figure
  1. (A)(P) → (3), (Q) → (5), (R) → (2), (S) → (1)
  2. (B)(P) → (1), (Q) → (5), (R) → (2), (S) → (3)
  3. (C)(P) → (3), (Q) → (4), (R) → (2), (S) → (1)
  4. (D)(P) → (1), (Q) → (4), (R) → (2), (S) → (5)

Correct answer: (A)

Step-by-step solution →
Q12·PhysicsIntegerJEE Main 2025
For ac circuit shown in figure, R = 100 kΩ\OmegaΩ and C = 100 pF and the phase difference between VinV_{in}Vin​ and (VB−VA)(V_B-V_A)(VB​−VA​) is 90∘90^\circ90∘. The input signal frequency is 10x10^x10x rad/sec, where xxx is ______.

Correct answer: 5

Step-by-step solution →
Q13·PhysicsSingle correctJEE Main 2025
An ac current is represented as i=52+10cos⁡(650πt+π6)i=5\sqrt{2}+10\cos\left(650\pi t+\dfrac{\pi}{6}\right)i=52​+10cos(650πt+6π​) Amp. The r.m.s value of the current is
  1. (A)50 Amp
  2. (B)100 Amp
  3. (C)10 Amp
  4. (D)525\sqrt{2}52​ Amp

Correct answer: (C)

Step-by-step solution →
Q14·PhysicsIntegerJEE Main 2025
An inductor of reactance 100 Ω\OmegaΩ, a capacitor of reactance 50 Ω\OmegaΩ, and a resistor of resistance 50 Ω\OmegaΩ are connected in series with an AC source of 10 V, 50 Hz. Average power dissipated by the circuit is __________ W.

Correct answer: 1

Step-by-step solution →
Q15·PhysicsSingle correctJEE Main 2025
An alternating current is represented by the equation, i=1002sin⁡(100πt)i=100\sqrt{2}\sin(100\pi t)i=1002​sin(100πt) ampere. The RMS value of current and the frequency of the given alternating current are
  1. (A)1002100\sqrt{2}1002​ A, 100 Hz
  2. (B)1002\dfrac{100}{\sqrt{2}}2​100​ A, 100 Hz
  3. (C)100 A, 50 Hz
  4. (D)50250\sqrt{2}502​ A, 50 Hz

Correct answer: (C)

Step-by-step solution →
Q16·PhysicsSingle correctJEE Main 2025
An electric bulb rated as 100 W-220 V is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is:
  1. (A)0.64 A
  2. (B)0.45 A
  3. (C)2.2 A
  4. (D)0.32 A

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correctJEE Main 2025
An alternating current is given by I=IAsin⁡ωt+IBcos⁡ωtI=I_A\sin\omega t+I_B\cos\omega tI=IA​sinωt+IB​cosωt. The r.m.s. current will be :-
  1. (A)IA2+IB2\sqrt{I_A^2+I_B^2}IA2​+IB2​​
  2. (B)IA2+IB22\dfrac{\sqrt{I_A^2+I_B^2}}{2}2IA2​+IB2​​​
  3. (C)IA2+IB22\sqrt{\dfrac{I_A^2+I_B^2}{2}}2IA2​+IB2​​​
  4. (D)∣IA+IB∣2\dfrac{|I_A+I_B|}{\sqrt2}2​∣IA​+IB​∣​

Correct answer: (C)

Step-by-step solution →
Q18·PhysicsIntegerJEE Main 2025
In a series LCR circuit, a resistor of 300 Ω300\ \Omega300 Ω, a capacitor of 252525 nF and an inductor of 100100100 mH are used. For maximum current in the circuit, the angular frequency of the ac source is __________ ×104\times10^4×104 radians s−1\mathrm{s^{-1}}s−1.

Correct answer: 2

Step-by-step solution →
Q19·PhysicsSingle correctJEE Main 2025
A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is I0I_0I0​. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be
  1. (A)I0I_0I0​
  2. (B)I02\frac{I_0}{2}2I0​​
  3. (C)I02\frac{I_0}{\sqrt{2}}2​I0​​
  4. (D)2I02I_02I0​

Correct answer: (B)

Step-by-step solution →
Q20·PhysicsSingle correctJEE Advanced 2024
The circuit shown in the figure contains an inductor LLL, a capacitor C0C_{0}C0​, a resistor R0R_{0}R0​ and an ideal battery. The circuit also contains two keys K1K_{1}K1​ and K2K_{2}K2​. Initially, both the keys are open and there is no charge on the capacitor. At an instant, key K1K_{1}K1​ is closed and immediately after this the current in R0R_{0}R0​ is found to be I1I_{1}I1​. After a long time, the current attains a steady state value I2I_{2}I2​. Thereafter, K2K_{2}K2​ is closed and simultaneously K1K_{1}K1​ is opened and the voltage across C0C_{0}C0​ oscillates with amplitude V0V_{0}V0​ and angular frequency ω0\omega_{0}ω0​. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
List-IList-II
P.The value of I1I_{1}I1​ in Ampere is1.0
Q.The value of I2I_{2}I2​ in Ampere is2.2
R.The value of ω0\omega_{0}ω0​ in kilo-radians/s3.4
S.The value of V0V_{0}V0​ in Volt is4.20
5.200
  1. (A)P →\to→ 1; Q →\to→ 3; R →\to→ 2; S →\to→ 5
  2. (B)P →\to→ 1; Q →\to→ 2; R →\to→ 3; S →\to→ 5
  3. (C)P →\to→ 1; Q →\to→ 3; R →\to→ 2 S →\to→ 4
  4. (D)P →\to→ 2 Q →\to→ 5 R →\to→ 3 S →\to→ 4

Correct answer: (A)

Step-by-step solution →
Q21·PhysicsNumericalJEE Main 2024
A capacitor of reactance 43 Ω4\sqrt{3}\,\Omega43​Ω and a resistor of resistance 4 Ω4\,\Omega4Ω are connected in series with an ac source of peak value 828\sqrt{2}82​ V. The power dissipation in the circuit is _______ W.

Correct answer: 4

Step-by-step solution →
Q22·PhysicsSingle correctJEE Main 2024
A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb:
  1. (A)increases
  2. (B)remains same
  3. (C)becomes zero
  4. (D)decreases

Correct answer: (A)

Step-by-step solution →
Q23·PhysicsNumericalJEE Main 2024
When a coil is connected across a 20 V dc supply, it draws a current of 5 A. When it is connected across 20 V, 50 Hz ac supply, it draws a current of 4 A. The self inductance of the coil is ______ mH. (Take π=3\pi=3π=3)

Correct answer: 10

Step-by-step solution →
Q24·PhysicsSingle correctJEE Main 2024
A coil of negligible resistance is connected in series with a 90 Ω\OmegaΩ resistor across a 120 V, 60 Hz supply. A voltmeter reads 36 V across the resistance. Inductance of the coil is:
  1. (A)0.76 H
  2. (B)2.86 H
  3. (C)0.286 H
  4. (D)0.91 H

Correct answer: (A)

Step-by-step solution →
Q25·PhysicsSingle correctJEE Main 2024
A LCR circuit is at resonance for a capacitor CCC, inductance LLL and resistance RRR. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:
  1. (A)Zero
  2. (B)double
  3. (C)same
  4. (D)halved

Correct answer: (B)

Step-by-step solution →
Q26·PhysicsNumericalJEE Main 2024
An alternating emf E=1102sin⁡100tE=110\sqrt{2}\sin 100tE=1102​sin100t volt is applied to a capacitor of 2 μ\muμF, the rms value of current in the circuit is ________ mA.

Correct answer: 22

Step-by-step solution →
Q27·PhysicsNumericalJEE Main 2024
For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 2.52.52.5 nF. If resistance of 200 Ω200\,\Omega200Ω and 100100100 mH inductor is being used in the given circuit. The frequency of ac source is ______ ×103\times 10^{3}×103 Hz. (given π2=10\pi^{2} = 10π2=10)

Correct answer: 10

Step-by-step solution →
Q28·PhysicsSingle correctJEE Main 2024
Given below are two statements: Statement I: In an LCR series circuit, current is maximum at resonance. Statement II: Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to the same voltage source. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I is true but Statement II is false
  2. (B)Statement I is false but Statement II is true
  3. (C)Both Statement I and Statement II are true
  4. (D)Both Statement I and Statement II are false

Correct answer: (C)

Step-by-step solution →
Q29·PhysicsNumericalJEE Main 2024
When a dc voltage of 100 V is applied to an inductor, a current of 5 A flows through it. When an ac voltage of 200 V peak value is connected to the inductor, its inductive reactance is found to be 203 Ω20\sqrt{3}\,\Omega203​Ω. The power dissipated in the circuit is _______ W.

Correct answer: 250

Step-by-step solution →
Q30·PhysicsNumericalJEE Main 2024
An ac source V=502sin⁡(100t)V = 50\sqrt{2}\sin(100t)V=502​sin(100t) V is connected in a series LCR circuit with L=1L = 1L=1 H, R=300 ΩR = 300\ \OmegaR=300 Ω and C=20 μFC = 20\ \mu\text{F}C=20 μF. The rms potential difference across the capacitor is ___ V.

Correct answer: 50

Step-by-step solution →
Q31·PhysicsSingle correctJEE Main 2024
A series LCR circuit is subjected to an AC signal of 200 V, 50 Hz. If the voltage across the inductor (L=10L=10L=10 mH) is 31.4 V, then the current in this circuit is __________:
  1. (A)68 A
  2. (B)63 A
  3. (C)10 A
  4. (D)10 mA

Correct answer: (C)

Step-by-step solution →
Q32·PhysicsSingle correctJEE Main 2024
Match List-I (type of AC circuit) with List-II (the phasor diagram of current I and voltage V). Choose the correct answer from the options given below:
List-IList-II
A.Purely capacitive circuitI.see figure
B.Purely inductive circuitII.see figure
C.LCR series at resonanceIII.see figure
D.LCR series circuitIV.see figure
  1. (A)A-I, B-IV, C-III, D-II
  2. (B)A-IV, B-I, C-III, D-II
  3. (C)A-IV, B-I, C-II, D-III
  4. (D)A-I, B-IV, C-II, D-III

Correct answer: (D)

Step-by-step solution →
Q33·PhysicsNumericalJEE Main 2024
A alternating current at any instant is given by i=[6+56sin⁡(100πt+π3)] Ai=\left[6+\sqrt{56}\sin\left(100\pi t+\dfrac\pi3\right)\right]\,Ai=[6+56​sin(100πt+3π​)]A. The rms value of the current is ___ A.

Correct answer: 8

Step-by-step solution →
Q34·PhysicsSingle correctJEE Main 2024
In an a.c. circuit, the instantaneous current is zero, when the instantaneous voltage is maximum. In this case, the source may be connected to: A. pure inductor. B. pure capacitor. C. pure resistor. D. combination of an inductor and capacitor. Choose the correct answer from the options given below:
  1. (A)A, B and C only
  2. (B)B, C and D only
  3. (C)A and B only
  4. (D)A, B and D only

Correct answer: (D)

Step-by-step solution →
Q35·PhysicsSingle correctJEE Main 2024
In series LCR circuit, the capacitance is changed from CCC to 4C4C4C. To keep the resonance frequency unchanged, the new inductance should be:
  1. (A)reduced by 14L\dfrac14 L41​L
  2. (B)increased by 2L2L2L
  3. (C)reduced by 34L\dfrac34 L43​L
  4. (D)increased to 4L4L4L

Correct answer: (C)

Step-by-step solution →
Q36·PhysicsSingle correctJEE Main 2024
A transformer has an efficiency of 80%80\%80% and works at 10 V and 4 kW. If the secondary voltage is 240 V, then the current in the secondary coil is:
  1. (A)1.59 A
  2. (B)13.33 A
  3. (C)1.33 A
  4. (D)15.1 A

Correct answer: (B)

Step-by-step solution →
Q37·PhysicsSingle correctJEE Main 2024
An AC voltage V=20sin⁡200πtV=20\sin 200\pi tV=20sin200πt is applied to a series LCR circuit which drives a current I=10sin⁡(200πt+π3)I=10\sin\left(200\pi t+\dfrac{\pi}{3}\right)I=10sin(200πt+3π​). The average power dissipated is:
  1. (A)21.6 W
  2. (B)200 W
  3. (C)173.2 W
  4. (D)50 W

Correct answer: (D)

Step-by-step solution →
Q38·PhysicsSingle correctJEE Main 2024
A series L,RL,RL,R circuit connected with an ac source E=(25sin⁡1000t) VE=(25\sin1000t)\,VE=(25sin1000t)V has a power factor of 12\dfrac{1}{\sqrt2}2​1​. If the source of emf is changed to E=(20sin⁡2000t) VE=(20\sin2000t)\,VE=(20sin2000t)V, the new power factor of the circuit will be:
  1. (A)12\dfrac{1}{\sqrt2}2​1​
  2. (B)13\dfrac{1}{\sqrt3}3​1​
  3. (C)15\dfrac{1}{\sqrt5}5​1​
  4. (D)17\dfrac{1}{\sqrt7}7​1​

Correct answer: (C)

Step-by-step solution →
Q39·PhysicsSingle correctJEE Main 2024
An alternating voltage V(t)=220sin⁡100πtV(t)=220\sin 100\pi tV(t)=220sin100πt volt is applied to a purely resistive load of 50 Ω50\,\Omega50Ω. The time taken for the current to rise from half of the peak value to the peak value is:
  1. (A)5 ms
  2. (B)3.3 ms
  3. (C)7.2 ms
  4. (D)2.2 ms

Correct answer: (B)

Step-by-step solution →
Q40·PhysicsNumericalJEE Main 2024
A power transmission line feeds input power at 2.3 kV to a step down transformer with its primary winding 3000 turns. The output power is delivered at 230 V to the primary of the transformer is 5A and its efficiency is 90%. The winding of transformer is made of copper. The output current of transformer is ______ A.

Correct answer: 45

Step-by-step solution →
Q41·PhysicsSingle correctJEE Main 2024
Primary coil of a transformer is connected to 220 V220\,V220V ac. Primary and secondary turns of the transformer are 100100100 and 101010 respectively. Secondary coil is connected to two series resistance shown in figure. The output voltage (V0)(V_0)(V0​) is:
  1. (A)7 V7\,V7V
  2. (B)15 V15\,V15V
  3. (C)44 V44\,V44V
  4. (D)2 V2\,V2V

Correct answer: (A)

Step-by-step solution →
Q42·PhysicsSingle correctJEE Main 2024
Primary side of a transformer is connected to 230 V, 50 Hz supply. Turns ratio of primary to secondary winding is 10:110:110:1. Load resistance connected to secondary side is 46 Ω\OmegaΩ. The power consumed in it is :
  1. (A)12.5 W
  2. (B)10.0 W
  3. (C)11.5 W
  4. (D)12.0 W

Correct answer: (C)

Step-by-step solution →
Q43·PhysicsNumericalJEE Main 2024
A series LCR circuit with L=100πL=\dfrac{100}{\pi}L=π100​ mH, C=10−3πC=\dfrac{10^{-3}}{\pi}C=π10−3​ F and R=10 ΩR=10\,\OmegaR=10Ω, is connected across an ac source of 220 V, 50 Hz supply. The power factor of the circuit would be ________.

Correct answer: 1.00

Step-by-step solution →
Q44·PhysicsSingle correctJEE Advanced 2023
A series LCR circuit is connected to a 45 sin⁡(ωt)\sin(\omega t)sin(ωt) Volt source. The resonant angular frequency of the circuit is 10510^5105 rad s−1s^{-1}s−1 and current amplitude at resonance is I0I_0I0​. When the angular frequency of the source is ω=8×104\omega = 8 \times 10^4ω=8×104 rad s−1s^{-1}s−1, the current amplitude in the circuit is 0.05 I00.05\ I_00.05 I0​. If L = 50 mH, match each entry in List-I with an appropriate value from List-II and choose the correct option.
List-IList-II
P.I0I_0I0​ in mA1.44.4
Q.The quality factor of the circuit2.18
R.The bandwidth of the circuit in rad s−1s^{-1}s−13.400
S.The peak power dissipated at resonance in Watt.4.2250
5.500
  1. (A)P →\rightarrow→ 2, Q →\rightarrow→ 3, R →\rightarrow→ 5, S →\rightarrow→ 1
  2. (B)P →\rightarrow→ 3, Q →\rightarrow→ 1, R →\rightarrow→ 4, S →\rightarrow→ 2
  3. (C)P →\rightarrow→ 4, Q →\rightarrow→ 5, R →\rightarrow→ 3, S →\rightarrow→ 1
  4. (D)P →\rightarrow→ 4, Q →\rightarrow→ 2, R →\rightarrow→ 1, S →\rightarrow→ 5

Correct answer: (B)

Step-by-step solution →
Q45·PhysicsSingle correctJEE Main 2023
Given below are two statements: Statement I: An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor. Statement II: An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor. In the light of above statements, choose the correct answer from the options given below
  1. (A)Both Statement I and Statement II are false
  2. (B)Statement I is true but Statement II is false
  3. (C)Both Statement I and Statement II are true
  4. (D)Statement I is false but Statement II is true

Correct answer: (A)

Step-by-step solution →
Q46·PhysicsSingle correctJEE Main 2023
Given below are two statements: Statement I: When the frequency of an a.c. source in a series LCR circuit increases, the current in the circuit first increases, attains a maximum value and then decreases. Statement II: In a series LCR circuit, the value of power factor at resonance is one. In the light of given statements, choose the most appropriate answer from the options given below:
  1. (A)Statement I is false but Statement II is true.
  2. (B)Both Statement I and Statement II are false.
  3. (C)Statement I is correct but Statement II is false.
  4. (D)Both Statement I and Statement II are true.

Correct answer: (D)

Step-by-step solution →
Q47·PhysicsSingle correctJEE Main 2023
As per the given graph choose the correct representation for curve A and curve B. {Where XCX_CXC​ = reactance of pure capacitive circuit connected with A.C. source, XLX_LXL​ = reactance of pure inductive circuit connected with A.C. source, RRR = impedance of pure resistive circuit connected with A.C. source, ZZZ = Impedance of the LCR series circuit}
  1. (A)A=XC,B=RA = X_C, B = RA=XC​,B=R
  2. (B)A=XL,B=ZA = X_L, B = ZA=XL​,B=Z
  3. (C)A=XC,B=XLA = X_C, B = X_LA=XC​,B=XL​
  4. (D)A=XL,B=RA = X_L, B = RA=XL​,B=R

Correct answer: (C)

Step-by-step solution →
Q48·PhysicsSingle correctJEE Main 2023
Given below are two statements: Statement I: Maximum power is dissipated in a circuit containing an inductor, a capacitor and a resistor connected in series with an AC source, when resonance occurs. Statement II: Maximum power is dissipated in a circuit containing pure resistor due to zero phase difference between current and voltage. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I is false but Statement II is true
  2. (B)Statement I is true but Statement II is false
  3. (C)Both Statement I and Statement II are true
  4. (D)Both Statement I and Statement II are false

Correct answer: (C)

Step-by-step solution →
Q49·PhysicsNumericalJEE Main 2023
An oscillating LC circuit consists of a 75 mH75\,\text{mH}75mH inductor and a 1.2 μF1.2\,\mu\text{F}1.2μF capacitor. If the maximum charge to the capacitor is 2.7 μC2.7\,\mu\text{C}2.7μC. The maximum current in the circuit will be ______ mA\text{mA}mA.

Correct answer: 9

Step-by-step solution →
Q50·PhysicsNumericalJEE Main 2023
A series combination of a resistor of resistance 100 Ω\OmegaΩ, an inductor of inductance 1 H and a capacitor of capacitance 6.25 μ\muμF is connected to an ac source. The quality factor of the circuit will be ____.

Correct answer: 4

Step-by-step solution →
Q51·PhysicsNumericalJEE Main 2023
An ideal transformer with purely resistive load operates at 12 kV12\,kV12kV on the primary side. It supplies electrical energy to a number of nearby houses at 120 V120\,V120V. The average rate of energy consumption in the houses served by the transformer is 60 kW60\,kW60kW. The value of resistive load (Rs)(R_s)(Rs​) required in the secondary circuit will be _____ mΩm\OmegamΩ.

Correct answer: 240

Step-by-step solution →
Q52·PhysicsSingle correctJEE Main 2023
A capacitor of capacitance 150.0 μF\mu FμF is connected to an alternating source of emf given by E=36sin⁡(120πt)E = 36\sin(120\pi t)E=36sin(120πt) V. The maximum value of current in the circuit is approximately equal to:
  1. (A)2 A2\,A2A
  2. (B)12 A\frac{1}{\sqrt{2}}\,A2​1​A
  3. (C)2 A\sqrt{2}\,A2​A
  4. (D)22 A2\sqrt{2}\,A22​A

Correct answer: (A)

Step-by-step solution →
Q53·PhysicsNumericalJEE Main 2023
A series LCR circuit is connected to an ac source of 220220220 V, 505050 Hz. The circuit contains a resistance R=100 ΩR=100\,\OmegaR=100Ω and an inductor of inductive reactance XL=79.6 ΩX_L=79.6\,\OmegaXL​=79.6Ω. The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be _______ μ\muμF.

Correct answer: 40

Step-by-step solution →
Q54·PhysicsSingle correctJEE Main 2023
An alternating voltage source V=260sin⁡(628t)V=260\sin(628t)V=260sin(628t) is connected across a pure inductor of 555 mH. Inductive reactance in the circuit is:
  1. (A)0.318 Ω0.318\,\Omega0.318Ω
  2. (B)6.28 Ω6.28\,\Omega6.28Ω
  3. (C)3.14 Ω3.14\,\Omega3.14Ω
  4. (D)0.5 Ω0.5\,\Omega0.5Ω

Correct answer: (C)

Step-by-step solution →
Q55·PhysicsNumericalJEE Main 2023
A series LCR circuit consists of R=80 ΩR=80\,\OmegaR=80Ω, XL=100 ΩX_L=100\,\OmegaXL​=100Ω and XC=40 ΩX_C=40\,\OmegaXC​=40Ω. The input voltage is 2500cos⁡(100πt)2500\cos(100\pi t)2500cos(100πt) V. The amplitude of current, in the circuit, is _________ A.

Correct answer: 25

Step-by-step solution →
Q56·PhysicsSingle correctJEE Main 2023
If RRR, XLX_LXL​ and XCX_CXC​ represent resistance, inductive reactance and capacitive reactance. Then which of the following is dimensionless :
  1. (A)RXLXC\frac{R}{X_L X_C}XL​XC​R​
  2. (B)RXLXC\frac{R}{\sqrt{X_L X_C}}XL​XC​​R​
  3. (C)RXLXCR\frac{X_L}{X_C}RXC​XL​​
  4. (D)RXLXCRX_L X_CRXL​XC​

Correct answer: (B)

Step-by-step solution →
Q57·PhysicsNumericalJEE Main 2023
An inductor of 0.50.50.5 mH, a capacitor of 20 μF20\,\mu\text{F}20μF and resistance of 20 Ω20\,\Omega20Ω are connected in series with a 220220220 V ac source. If the current is in phase with the emf, the amplitude of current of the circuit is x\sqrt{x}x​ A. The value of xxx is-

Correct answer: 242

Step-by-step solution →
Q58·PhysicsSingle correctJEE Main 2023
In a series AC circuit with inductive reactance XL=200 ΩX_L=200\,\OmegaXL​=200Ω, capacitive reactance XC=100 ΩX_C=100\,\OmegaXC​=100Ω and resistance R=100 ΩR=100\,\OmegaR=100Ω connected to a source of Vrms=2002V_{rms}=200\sqrt{2}Vrms​=2002​ V, the rms value of current (IrmsI_{rms}Irms​) through the resistor R is:
  1. (A)222\sqrt{2}22​ A
  2. (B)222 A
  3. (C)202020 A
  4. (D)12\dfrac{1}{2}21​ A

Correct answer: (B)

Step-by-step solution →
Q59·PhysicsSingle correctJEE Main 2023
In a series LR circuit with XL=RX_L=RXL​=R, power factor is P1P_1P1​. If a capacitor of capacitance C with XC=XLX_C=X_LXC​=XL​ is added to the circuit the power factor becomes P2P_2P2​. The ratio of P1P_1P1​ to P2P_2P2​ will be:
  1. (A)1:31:31:3
  2. (B)1:21:21:2
  3. (C)1:21:\sqrt21:2​
  4. (D)1:11:11:1

Correct answer: (C)

Step-by-step solution →
Q60·PhysicsNumericalJEE Main 2023
An inductor of inductance 2 μH2\,\mu H2μH is connected in series with a resistance, a variable capacitor and an AC source of frequency 7 kHz7\,kHz7kHz. The value of capacitance for which maximum current is drawn into the circuit is 1x F\dfrac{1}{x}\,Fx1​F, where the value of xxx is _____. (Take π=227)\left(\text{Take }\pi=\dfrac{22}{7}\right)(Take π=722​)

Correct answer: 3872

Step-by-step solution →
Q61·PhysicsSingle correctJEE Main 2023
For the given figures, choose the correct options:
  1. (A)At resonance, current in (b) is less than that in (a)
  2. (B)The rms current in circuit (b) can never be larger than that in (a)
  3. (C)The rms current in circuit (b) is always equal to that in (a)
  4. (D)The rms current in circuit (b) can be larger than that in (a)

Correct answer: (B)

Step-by-step solution →
Q62·PhysicsNumericalJEE Main 2023
A series LCR circuit is connected to an AC source of 220 V, 50 Hz. The circuit contains a resistance R=80 ΩR = 80\,\OmegaR=80Ω, an inductor of inductive reactance XL=70 ΩX_L = 70\,\OmegaXL​=70Ω, and a capacitor of capacitive reactance XC=130 ΩX_C = 130\,\OmegaXC​=130Ω. The power factor of circuit is x10\frac{x}{10}10x​. The value of xxx is:

Correct answer: 8

Step-by-step solution →
Q63·PhysicsSingle correctJEE Main 2023
In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes xxx times its initial resonant frequency ω0\omega_0ω0​. The value of xxx is:
  1. (A)444
  2. (B)116\dfrac{1}{16}161​
  3. (C)161616
  4. (D)14\dfrac{1}{4}41​

Correct answer: (D)

Step-by-step solution →
Q64·PhysicsNumericalJEE Main 2023
An LCR series circuit of capacitance 62.5 nF and resistance of 50 Ω50\,\Omega50Ω, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in the circuit, the value of inductance is _______ mH. (Take π2=10\pi^2=10π2=10)

Correct answer: 100

Step-by-step solution →
Q65·PhysicsNumericalJEE Main 2023
In the circuit shown in the figure, the ratio of the quality factor and the band width is _______ S.

Correct answer: 10

Step-by-step solution →
Q66·PhysicsNumericalJEE Advanced 2022
Consider an LC circuit, with inductance L=0.1 H and capacitance C=10−3C = 10^{-3}C=10−3 F, kept on a plane. The area of the circuit is 1 m2^22. It is placed in a constant magnetic field of strength B0B_0B0​ which is perpendicular to the plane of the circuit. At time t = 0, the magnetic field strength starts increasing linearly as B=B0+βtB = B_0 + \beta tB=B0​+βt with β=0.04Ts−1\beta = 0.04 Ts^{-1}β=0.04Ts−1. The maximum magnitude of the current in the circuit is__________ mA .

Correct answer: 4.00

Step-by-step solution →
Q67·PhysicsSingle correctJEE Main 2022
A circuit element X when connected to an a.c. supply of peak voltage 100 V gives a peak current of 5 A which is in phase with the voltage. A second element Y when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by π2\frac{\pi}{2}2π​. If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere ?
  1. (A)102\frac{10}{\sqrt{2}}2​10​
  2. (B)52\frac{5}{\sqrt{2}}2​5​
  3. (C)525\sqrt{2}52​
  4. (D)52\frac{5}{2}25​

Correct answer: (D)

Step-by-step solution →
Q68·PhysicsNumericalJEE Main 2022
A capacitor of capacitance 500 μF is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. The maximum current in LC circuit will be _______ A.

Correct answer: 10

Step-by-step solution →
Q69·PhysicsSingle correctJEE Main 2022
An alternating emf E = 440 sin 100πt is applited to a circuit containing an inductance of 2π\frac{\sqrt{2}}{\pi}π2​​ H. If an a.c. ammeter is connected in the circuit, its reading will be :
  1. (A)4.4 A
  2. (B)1.55 A
  3. (C)2.2 A
  4. (D)3.11 A

Correct answer: (C)

Step-by-step solution →
Q70·PhysicsSingle correctJEE Main 2022
A transformer operating at primary voltage 8 kV and secondary voltage 160 V serves a load of 80 kW. Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be
  1. (A)800 Ω\OmegaΩ and 1.06 Ω\OmegaΩ
  2. (B)10 Ω\OmegaΩ and 500 Ω\OmegaΩ
  3. (C)800 Ω\OmegaΩ and 0.32 Ω\OmegaΩ
  4. (D)1.06 Ω\OmegaΩ and 500 Ω\OmegaΩ

Correct answer: (C)

Step-by-step solution →
Q71·PhysicsSingle correctJEE Main 2022
The equation of current in a purely inductive circuit is 5sin⁡(49πt−30°)5\sin\left(49\pi t - 30°\right)5sin(49πt−30°). If the inductance is 30 mH then the equation for the voltage across the inductor, will be : {Let π=227}\left\{\text{Let } \pi = \frac{22}{7}\right\}{Let π=722​}
  1. (A)1.47sin⁡(49πt−30°)1.47\sin(49\pi t - 30°)1.47sin(49πt−30°)
  2. (B)1.47sin⁡(49πt+60°)1.47\sin(49\pi t + 60°)1.47sin(49πt+60°)
  3. (C)23.1sin⁡(49πt−30°)23.1\sin(49\pi t - 30°)23.1sin(49πt−30°)
  4. (D)23.1sin⁡(49πt+60°)23.1\sin(49\pi t + 60°)23.1sin(49πt+60°)

Correct answer: (D)

Step-by-step solution →
Q72·PhysicsNumericalJEE Main 2022
The frequencies at which the current amplitude in an LCR series circuit becomes 12\frac{1}{\sqrt{2}}2​1​ times its maximum value, are 212 rad s−1^{-1}−1 and 232 rad s−1^{-1}−1. The value of resistance in the circuit is R = 5Ω. The self inductance in the circuit is ________ mH.

Correct answer: 250

Step-by-step solution →
Q73·PhysicsSingle correctJEE Main 2022
A series LCR circuit has L = 0.01 H, R = 10 Ω and C = 1 μF and it is connected to ac voltage of amplitude (VmV_mVm​) 50 V. At frequency 60% lower than resonant frequency, the amplitude of current will be approximately :
  1. (A)466 mA
  2. (B)312 mA
  3. (C)238 mA
  4. (D)196 mA

Correct answer: (C)

Step-by-step solution →
Q74·PhysicsSingle correctJEE Main 2022
A direct current of 4 A and an alternating current of peak value 4 A flow through resistance of 3 Ω and 2 Ω respectively. The ratio of heat produced in the two resistances in same interval of time will be :
  1. (A)3 : 2
  2. (B)3 : 1
  3. (C)3 : 4
  4. (D)4 : 3

Correct answer: (B)

Step-by-step solution →
Q75·PhysicsNumericalJEE Main 2022
To light, a 50 W, 100 V lamp is connected, in series with a capacitor of capacitance 50πx μF\frac{50}{\pi\sqrt{x}}\,\mu\mathrm{F}πx​50​μF, with 200 V, 50Hz AC source. The value of x will be ____ .

Correct answer: 3

Step-by-step solution →
Q76·PhysicsSingle correctJEE Main 2022
In a series LR circuit XL_{L}L​ = R and power factor of the circuit is P1_{1}1​. When capacitor with capacitance C such that XL_{L}L​ = XC_{C}C​ is put in series, the power factor becomes P2_{2}2​. The ratio P1P2\frac{P_{1}}{P_{2}}P2​P1​​ is
  1. (A)12\frac{1}{2}21​
  2. (B)12\frac{1}{\sqrt{2}}2​1​
  3. (C)32\frac{\sqrt{3}}{\sqrt{2}}2​3​​
  4. (D)2 : 1

Correct answer: (B)

Step-by-step solution →
Q77·PhysicsNumericalJEE Main 2022
The effective current I in the given circuit at very high frequencies will be _____A

Correct answer: 44

Step-by-step solution →
Q78·PhysicsSingle correctJEE Main 2022
To increase the resonant frequency in series LCR circuit,
  1. (A)Source frequency should be increased
  2. (B)Another resistance should be added in series with the first resistance.
  3. (C)Another capacitor should be added in series with the first capacitor
  4. (D)The source frequency should be decreased

Correct answer: (C)

Step-by-step solution →
Q79·PhysicsSingle correctJEE Main 2022
The rms value of conduction current in a parallel plate capacitor is 6.9 μ\muμA. The capacity of this capacitor, if it is connected to 230 V ac supply with an angular frequency of 600 rad/s, will be :
  1. (A)5 pF
  2. (B)50 pF
  3. (C)100 pF
  4. (D)200 pF

Correct answer: (B)

Step-by-step solution →
Q80·PhysicsNumericalJEE Main 2022
An inductor of 0.5 mH, a capacitor of 200 μF and a resistor of 2 Ω are connected in series with a 220 V ac source. If the current is in phase with the emf, the frequency of ac source will be___×102\times 10^2×102 Hz.

Correct answer: 5

Step-by-step solution →
Q81·PhysicsSingle correctJEE Main 2022
For a series LCR circuit, I vs ω curve is shown : (a) To the left of ωr\omega_rωr​, the circuit is mainly capacitive. (b) To the left of ωr\omega_rωr​, the circuit is mainly inductive. (c) At ωr\omega_rωr​, impedance of the circuit is equal to the resistance of the circuit. (d) At ωr\omega_rωr​, impedance of the circuit is 0. Choose the most appropriate answer from the options given below :
  1. (A)(a) and (d) only
  2. (B)(b) and (d) only
  3. (C)(a) and (c) only
  4. (D)(b) and (c) only

Correct answer: (C)

Step-by-step solution →
Q82·PhysicsNumericalJEE Main 2022
A telegraph line of length loo km has a capacity of 0.010.010.01 μF/km and it carries an alternating current at 0.50.50.5 kilo cycle per second. If minimum impedance is required, then the value of the inductance that needs to be introduced in series is _________ mH. (if π=10)\left(\text{if } \pi = \sqrt{10}\right)(if π=10​)

Correct answer: 100

Step-by-step solution →
Q83·PhysicsSingle correctJEE Main 2022
If L, C and R are the self inductance, capacitance and resistance respectively, which of the following does not have the dimension of time ?
  1. (A)RC
  2. (B)LR\dfrac{L}{R}RL​
  3. (C)LC\sqrt{LC}LC​
  4. (D)LC\dfrac{L}{C}CL​

Correct answer: (D)

Step-by-step solution →
Q84·PhysicsSingle correctJEE Main 2022
The current flowing through an ac circuit is given by I=5sin⁡(120πt) AI = 5\sin(120\pi t)\,AI=5sin(120πt)A How long will the current take to reach the peak value starting from zero?
  1. (A)160s\frac{1}{60}s601​s
  2. (B)60s
  3. (C)1120s\frac{1}{120}s1201​s
  4. (D)1240s\frac{1}{240}s2401​s

Correct answer: (D)

Step-by-step solution →
Q85·PhysicsNumericalJEE Main 2022
A 220 V, 50 Hz AC source is connected to a 25 V, 5 W lamp and an additional resistance R in series (as shown in figure) to run the lamp at its peak brightness, then the value of R (in ohm) will be _______ .

Correct answer: 975

Step-by-step solution →
Q86·PhysicsNumericalJEE Main 2022
A 110 V , 50 Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance 55 Ω, at resonance in the circuit, will be ............... A.

Correct answer: 0

Step-by-step solution →
Q87·PhysicsSingle correctJEE Main 2022
A sinusoidal voltage V(t) = 210 sin 3000t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 μF and R = 100Ω. The phase difference (Φ) between the applied voltage and resultant current will be :
  1. (A)tan⁡−1(0.17)\tan^{-1}(0.17)tan−1(0.17)
  2. (B)tan⁡−1(9.46)\tan^{-1}(9.46)tan−1(9.46)
  3. (C)tan⁡−1(0.30)\tan^{-1}(0.30)tan−1(0.30)
  4. (D)tan⁡−1(13.33)\tan^{-1}(13.33)tan−1(13.33)

Correct answer: (A)

Step-by-step solution →
Q88·PhysicsSingle correctJEE Main 2022
If wattless current flows in the AC circuit, then the circuit is
  1. (A)Purely Resistive circuit
  2. (B)Purely Inductive circuit
  3. (C)LCR series circuit
  4. (D)RC series circuit only

Correct answer: (B)

Step-by-step solution →
Q89·PhysicsNumericalJEE Main 2022
In a series LCR circuit, the inductance, capacitance and resistance are L = 100mH, C = 100μF and R = 10Ω respectively. They are connected to an AC source of voltage 220V and frequency of 50 Hz. The approximate value of current in the circuit will be____ A.

Correct answer: 22

Step-by-step solution →
Q90·PhysicsSingle correctJEE Main 2022
A resistance of 40 Ω\OmegaΩ is connected to a source of alternating current rated 220 V, 50 Hz. Find the time taken by the current to change from its maximum value to rms value :
  1. (A)2.5 ms
  2. (B)1.25 ms
  3. (C)2.5 s
  4. (D)0.25 s

Correct answer: (A)

Step-by-step solution →
Q91·PhysicsSingle correctJEE Main 2022
Given below are two statements : Statement-I : The reactance of an ac circuit is zero. It is possible that the circuit contains a capacitor and an inductor. Statement-II : In ac circuit, the average poser delivered by the source never becomes zero. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Both Statement I and Statement II are true.
  2. (B)Both Statement I and Statement II are false.
  3. (C)Statement I is true but Statement II in false.
  4. (D)Statement I is false but Statement II is true.

Correct answer: (C)

Step-by-step solution →
Q92·PhysicsNumericalJEE Main 2022
As shown in the figure an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is aπ\dfrac{\sqrt{a}}{\pi}πa​​ A. The value of a is ________.

Correct answer: 242

Step-by-step solution →
Q93·PhysicsNumericalJEE Advanced 2021
In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C µF across a 200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V. Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude of the phase-angle (in degrees) between the current and the supply voltage is φ. Assume, π3≈5\pi\sqrt{3} \approx 5π3​≈5. The value of φ\varphiφ is ____.

Correct answer: 60.00

Step-by-step solution →
Q94·PhysicsNumericalJEE Advanced 2021
In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C µF across a 200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V. Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude of the phase-angle (in degrees) between the current and the supply voltage is φ. Assume, π3≈5\pi\sqrt{3} \approx 5π3​≈5. The value of C is ____.

Correct answer: 100.00

Step-by-step solution →
Q95·PhysicsSingle correctJEE Main 2021
In an ac circuit, an inductor, a capacitor and a resistor are connected in series with XL=R=XCX_{L}=R=X_{C}XL​=R=XC​. Impedance of this circuit is :
  1. (A)2R22R^{2}2R2
  2. (B)Zero
  3. (C)R
  4. (D)R2R\sqrt{2}R2​

Correct answer: (C)

Step-by-step solution →
Q96·PhysicsNumericalJEE Main 2021
At very high frequencies, the effective impendance of the given circuit will be_______Ω\OmegaΩ.

Correct answer: 2

Step-by-step solution →
Q97·PhysicsNumericalJEE Main 2021
The alternating current is given by i={42sin⁡(2πTt)+10}Ai = \left\{\sqrt{42}\sin\left(\frac{2\pi}{T}t\right) + 10\right\}Ai={42​sin(T2π​t)+10}A The r.m.s. value of this current is ........ A.

Correct answer: 11

Step-by-step solution →
Q98·PhysicsNumericalJEE Main 2021
An ac circuit has an inductor and a resistor of resistance R in series, such that XLX_LXL​ = 3R. Now, a capacitor is added in series such that XCX_CXC​ = 2R. The ratio of new power factor with the old power factor of the circuit is 5\sqrt{5}5​ : x . The value of x is _____ .

Correct answer: 1

Step-by-step solution →
Q99·PhysicsSingle correctJEE Main 2021
A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance R = 3 kΩ\OmegaΩ, an inductor of inductive reactance XLX_LXL​ = 250 πΩ\pi\OmegaπΩ and an unknown capacitor. The value of capacitance to maximize the average power should be : (Take π2\pi^2π2 = 10)
  1. (A)4 μ\muμF
  2. (B)25 μ\muμF
  3. (C)400 μ\muμF
  4. (D)40 μ\muμF

Correct answer: (A)

Step-by-step solution →
Q100·PhysicsSingle correctJEE Main 2021
A 100 Ω resistance, a 0.1 μF capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value if inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz.
  1. (A)7.03 × 10−5^{-5}−5 H
  2. (B)70.3 H
  3. (C)70.3 mH
  4. (D)0.70 H

Correct answer: (B)

Step-by-step solution →
Q101·PhysicsSingle correctJEE Main 2021
A 0.07 H inductor and a 12Ω\OmegaΩ resistor are connected in series to a 220 V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [ Take π\piπ as 227\frac{22}{7}722​ ]
  1. (A)88 A and tan⁡−1(116)\tan^{-1}\left(\frac{11}{6}\right)tan−1(611​)
  2. (B)8.8 A and tan⁡−1(116)\tan^{-1}\left(\frac{11}{6}\right)tan−1(611​)
  3. (C)0.88 A and tan⁡−1(116)\tan^{-1}\left(\frac{11}{6}\right)tan−1(611​)
  4. (D)8.8 A and tan⁡−1(611)\tan^{-1}\left(\frac{6}{11}\right)tan−1(116​)

Correct answer: (B)

Step-by-step solution →
Q102·PhysicsNumericalJEE Main 2021
Two circuits are shown in the figure (a) & (b). At a frequency of ________ rad /s the average power dissipated in one cycle will be same in both the circuits.

Correct answer: 500

Step-by-step solution →
Q103·PhysicsSingle correctJEE Main 2021
A 10 Ω10\,\Omega10Ω resistance is connected across 220 V −-− 50 Hz AC supply. The time taken by the current to change from its maximum value to the rms value is :
  1. (A)2.5 ms
  2. (B)4.5 ms
  3. (C)3.0 ms
  4. (D)1.5 ms

Correct answer: (A)

Step-by-step solution →
Q104·PhysicsSingle correctJEE Main 2021
In a circuit consisting of a capacitance and a generator with alternating emf Eg=Eg0sin⁡ωtE_g=E_{g_0}\sin\omega tEg​=Eg0​​sinωt, VCV_CVC​ and ICI_CIC​ are the voltage and current. Correct phase diagram for such circuit is :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q105·PhysicsSingle correctJEE Main 2021
Match List −-− I with List −-− II : Choose the correct answer from the options given below :
List-IList-II
a.ωL>1ωC\omega L>\frac{1}{\omega C}ωL>ωC1​i.Current is in phase with emf
b.ωL=1ωC\omega L=\frac{1}{\omega C}ωL=ωC1​ii.Current lags behind the applied emf
c.ωL<1ωC\omega L<\frac{1}{\omega C}ωL<ωC1​iii.Maximum current occurs
d.Resonant frequencyiv.Current leads the emf
  1. (A)(a) −-−(iv); (b) −-−(iii); (c) −-−(ii); (d) −-−(i)
  2. (B)(a) −-−(ii); (b) −-−(i); (c) −-−(iv); (d) −-−(iii)
  3. (C)(a) −-−(iii); (b) −-−(i); (c) −-−(iv); (d) −-−(ii)
  4. (D)(a) −-−(ii); (b) −-−(i); (c) −-−(iii); (d) −-−(iv)

Correct answer: (B)

Step-by-step solution →
Q106·PhysicsSingle correctJEE Main 2021
For a series LCR circuit with R=100 ΩR = 100\,\OmegaR=100Ω, L=0.5 mHL = 0.5\,mHL=0.5mH and C=0.1 pFC = 0.1\,pFC=0.1pF connected across 220V−50Hz220V - 50Hz220V−50Hz AC supply, the phase angle between current and supplied voltage and the nature of the circuit is :
  1. (A)≈90∘\approx 90^\circ≈90∘, predominantly inductive circuit
  2. (B)0∘0^\circ0∘, resonance circuit
  3. (C)≈90∘\approx 90^\circ≈90∘, predominantly capacitive circuit
  4. (D)0∘0^\circ0∘, resistive circuit

Correct answer: (C)

Step-by-step solution →
Q107·PhysicsNumericalJEE Main 2021
In an LCR series circuit, an inductor 30 mH and a resistor 1 Ω are connected to an Ac source of angular frequency 300 rad /s. The value of capacitance for which, the current leads the voltage by 45° is 1x×10−3\frac{1}{x} \times 10^{-3}x1​×10−3 F. Then the value of x is ____________.

Correct answer: 3

Step-by-step solution →
Q108·PhysicsNumericalJEE Main 2021
A series LCR circuit of R=5Ω,L=20 mH\mathrm{R} = 5\Omega, \mathrm{L} = 20\,\mathrm{mH}R=5Ω,L=20mH and C=0.5 μF\mathrm{C} = 0.5\,\mu\mathrm{F}C=0.5μF is connected across an AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is _________ ×102\times 10^{2}×102 W.

Correct answer: 125

Step-by-step solution →
Q109·PhysicsSingle correctJEE Main 2021
In a series LCR resonance circuit, if we change the resistance only, from a lower to higher value :
  1. (A)The bandwidth of resonance circuit will increase.
  2. (B)The resonance frequency will increase.
  3. (C)The quality factor will increase.
  4. (D)The quality factor and the resonance frequency will remain constant.

Correct answer: (A)

Step-by-step solution →
Q110·PhysicsSingle correctJEE Main 2021
An AC source rated 220 V, 50 Hz is connected to a resistor. The time taken by the current to change from its maximum to the rms value is :
  1. (A)2.5 ms
  2. (B)25 ms
  3. (C)2.5 s
  4. (D)0.25 ms

Correct answer: (A)

Step-by-step solution →
Q111·PhysicsSingle correctJEE Main 2021
In a series LCR circuit, the inductive reactance (XLX_LXL​) is 10 Ω and the capacitive reactance (XCX_CXC​) is 4 Ω. The resistance (R) in the circuit is 6 Ω. The power factor of the circuit is :
  1. (A)12\frac{1}{2}21​
  2. (B)122\frac{1}{2\sqrt{2}}22​1​
  3. (C)12\frac{1}{\sqrt{2}}2​1​
  4. (D)32\frac{\sqrt{3}}{2}23​​

Correct answer: (C)

Step-by-step solution →
Q112·PhysicsSingle correctJEE Main 2021
An AC current is given by I=I1sin⁡ωt+I2cos⁡ωtI = I_1 \sin\omega t + I_2 \cos\omega tI=I1​sinωt+I2​cosωt. A hot wire ammeter will give a reading :
  1. (A)I12−I222\sqrt{\frac{I_1^2 - I_2^2}{2}}2I12​−I22​​​
  2. (B)I12+I222\sqrt{\frac{I_1^2 + I_2^2}{2}}2I12​+I22​​​
  3. (C)I1+I22\frac{I_1 + I_2}{\sqrt{2}}2​I1​+I2​​
  4. (D)I1+I222\frac{I_1 + I_2}{2\sqrt{2}}22​I1​+I2​​

Correct answer: (B)

Step-by-step solution →
Q113·PhysicsSingle correctJEE Main 2021
What happens to the inductive reactance and the current in a purely inductive circuit if the frequency is halved ?
  1. (A)Both, inductive reactance and current will be halved.
  2. (B)Inductive reactance will be halved and current will be doubled.
  3. (C)Inductive reactance will be doubled and current will be halved.
  4. (D)Both, inducting reactance and current will be doubled.

Correct answer: (B)

Step-by-step solution →
Q114·PhysicsSingle correctJEE Main 2021
Match List-I with List-II Choose the most appropriate answer from the options given below :
List-IList-II
a.Phase difference between current and voltage in a purely resistive AC circuiti.π2\frac{\pi}{2}2π​ ; current leads voltage
b.Phase difference between current and voltage in a pure inductive AC circuitii.zero
c.Phase difference between current and voltage in a pure capacitive AC circuitiii.π2\frac{\pi}{2}2π​ ; current lags voltage
d.Phase difference between current and voltage in an LCR series circuitiv.tan⁡−1(XC−XLR)\tan^{-1}\left(\frac{X_{C} - X_{L}}{R}\right)tan−1(RXC​−XL​​)
  1. (A)(a)−(i),(b)−(iii),(c)−(iv),(d)−(ii)
  2. (B)(a)−(ii),(b)−(iv),(c)−(iii),(d)−(i)
  3. (C)(a)−(ii),(b)−(iii),(c)−(iv),(d)−(i)
  4. (D)(a)−(ii),(b)−(iii),(c)−(i),(d)−(iv)

Correct answer: (D)

Step-by-step solution →
Q115·PhysicsNumericalJEE Main 2021
A sinusoidal voltage of peak value 250 V is applied to a series LCR circuit, in which R = 8Ω, L = 24 mH and C = 60µF. The value of power dissipated at resonant condition is 'x' kW. The value of x to the nearest integer is ________.

Correct answer: 4

Step-by-step solution →
Q116·PhysicsSingle correctJEE Main 2021
An RC circuit as shown in the figure is driven by a AC source generating a square wave. The output wave pattern monitored by CRO would look close to :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q117·PhysicsSingle correctJEE Main 2021
For the given circuit, comment on the type of transformer used :
  1. (A)Auxilliary transformer
  2. (B)Auto transformer
  3. (C)Step-up transformer
  4. (D)Step down transformer

Correct answer: (C)

Step-by-step solution →
Q118·PhysicsSingle correctJEE Main 2021
An alternating current is given by the equation i=i1_{1}1​sinωt + i2_{2}2​cosωt. The rms current will be :
  1. (A)12(i12+i22)12\frac{1}{2}\left(i_{1}^{2} + i_{2}^{2}\right)^{\frac{1}{2}}21​(i12​+i22​)21​
  2. (B)12(i12+i22)12\frac{1}{\sqrt{2}}\left(i_{1}^{2} + i_{2}^{2}\right)^{\frac{1}{2}}2​1​(i12​+i22​)21​
  3. (C)12(i1+i2)2\frac{1}{\sqrt{2}}\left(i_{1} + i_{2}\right)^{2}2​1​(i1​+i2​)2
  4. (D)12(i1+i2)\frac{1}{\sqrt{2}}\left(i_{1} + i_{2}\right)2​1​(i1​+i2​)

Correct answer: (B)

Step-by-step solution →
Q119·PhysicsSingle correctJEE Main 2021
Find the peak current and resonant frequency of the following circuit (as shown in figure)
  1. (A)0.2 A and 100 Hz
  2. (B)2 A and 50 Hz
  3. (C)2 A and 100 Hz
  4. (D)0.2 A and 50 Hz

Correct answer: (D)

Step-by-step solution →
Q120·PhysicsNumericalJEE Main 2021
In a series LCR resonant circuit, the quality factor is measured as 100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be ____________.

Correct answer: 282.84

Step-by-step solution →
Q121·PhysicsSingle correctJEE Main 2021
An LCR circuit contains resistance of 110Ω and a supply of 220 V at 300 rad/s angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by 45o^{o}o. If on the other hand, only inductor is removed the current leads by 45o^{o}o with the applied voltage. The rms current flowing in the circuit will be:
  1. (A)2.5A
  2. (B)2A
  3. (C)1A
  4. (D)1.5 A

Correct answer: (B)

Step-by-step solution →
Q122·PhysicsSingle correctJEE Main 2021
The angular frequency of alterlating current in a L-C-R circuit is 100 rad/s. The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.
  1. (A)0.8 H and 250 μ\muμF
  2. (B)0.8 H and 150 μ\muμF
  3. (C)1.33 H and 250 μ\muμF
  4. (D)1.33 H and 150 μ\muμF

Correct answer: (A)

Step-by-step solution →
Q123·PhysicsNumericalJEE Main 2021
A transmitting station releases waves of wavelength 960 m. A capacitor of 256 μ\muμF is used in the resonant circuit. The self inductance of coil necessary for resonance is ______ ×10−8\times 10^{-8}×10−8H.

Correct answer: 10

Step-by-step solution →
Q124·PhysicsNumericalJEE Main 2021
A common transistor radio set requires 12 V (D.C.) for its operation. The D.C. source is constructed by using a transformer and a rectifier circuit, which are operated at 220 V (A.C.) on standard domestic A.C. supply. The number of turns of secondary coil are 24, then the number of turns of primary are ________.

Correct answer: 440

Step-by-step solution →
Q125·PhysicsNumericalJEE Main 2021
A resonance circuit having inductance and resistance 2×10−42 \times 10^{-4}2×10−4 H and 6.28 Ω\OmegaΩ respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is__________. [π\piπ = 3.14]

Correct answer: 2000

Step-by-step solution →
Q126·PhysicsNumericalJEE Main 2021
A series LCR circuit is designed to resonate at an angular frequency ω0=105 rad/s\omega_{0} = 10^{5}\,\mathrm{rad/s}ω0​=105rad/s. The circuit draws 16W power from 120 V source at resonance. The value of resistance ‘R’ in the circuit is ________ Ω\OmegaΩ.

Correct answer: 900

Step-by-step solution →
Q127·PhysicsNumericalJEE Main 2020
In a series LR circuit, power of 400 W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R. Taking the value of C as (n3π)\left(\frac{n}{3\pi}\right)(3πn​) μF, then value of 'n' is ______.

Correct answer: 400.00

Step-by-step solution →
Q128·PhysicsSingle correctJEE Main 2020
A 750 Hz, 20 V (rms) source is connected to a resistance of 100 Ω an inductance of 0.1803 H and a capacitance of 10 μF all in series. The time in which the resistance (heat capacity 2 J/°C) will get heated by 10°C. (assume no loss of heat to the surroundings) is close to:
  1. (A)348 s
  2. (B)365 s
  3. (C)418 s
  4. (D)245 s

Correct answer: (A)

Step-by-step solution →
Q129·PhysicsSingle correctJEE Main 2020
In LC circuit the inductance L = 40 mH and capacitance C = 100 μF. If a voltage V(t) = 10 sin (314 t) is applied to the circuit, the current in the circuit is given as
  1. (A)5.2 cos 314 t
  2. (B)0.52 sin 314 t
  3. (C)0.52 cos 314 t
  4. (D)10 cos 314 t

Correct answer: (C)

Step-by-step solution →
Q130·PhysicsSingle correctJEE Main 2020
A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the connect equivalence would be:
  1. (A)L ↔\leftrightarrow↔ 1b\frac{1}{b}b1​, C ↔\leftrightarrow↔ 1m\frac{1}{m}m1​, R ↔\leftrightarrow↔ 1k\frac{1}{k}k1​
  2. (B)L ↔\leftrightarrow↔ k, C ↔\leftrightarrow↔ b, R ↔\leftrightarrow↔ m
  3. (C)L ↔\leftrightarrow↔ m, C ↔\leftrightarrow↔ 1k\frac{1}{k}k1​, R ↔\leftrightarrow↔ b
  4. (D)L ↔\leftrightarrow↔ m, C ↔\leftrightarrow↔ k, R ↔\leftrightarrow↔ b

Correct answer: (C)

Step-by-step solution →
Q131·PhysicsSingle correctJEE Main 2019
A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW. If the current in the secondary coils is 10 A, then the input voltage and current in the primary coil are:
  1. (A)440 V and 5 A
  2. (B)440 and 20 A
  3. (C)220 V and 20 A
  4. (D)220 V and 10 A

Correct answer: (A)

Step-by-step solution →
Q132·PhysicsSingle correctJEE Main 2019
A circuit connected to an ac source of emf e=e0sin⁡(1000t)e = e_0\sin(1000t)e=e0​sin(1000t) with t in seconds, gives a phase difference of π4\dfrac{\pi}{4}4π​ between the emf e and current i. Which of the following circuits will exhibit this?
  1. (A)RC circuit with R=1kΩR = 1k\OmegaR=1kΩ and C=1μFC = 1\mu FC=1μF
  2. (B)RL circuit with R=1kΩR = 1k\OmegaR=1kΩ and L=10mHL = 10mHL=10mH
  3. (C)RL circuit with R=1kΩR = 1k\OmegaR=1kΩ and L=1mHL = 1mHL=1mH
  4. (D)RC circuit with R=1kΩR = 1k\OmegaR=1kΩ and C=10μFC = 10\mu FC=10μF

Correct answer: (D)

Step-by-step solution →
Q133·PhysicsSingle correctJEE Main 2019
An alternating voltage v(t)=220sin⁡100πtv(t) = 220 \sin 100\pi tv(t)=220sin100πt volt is applied to a purely resistive load of 50 Ω50\,\Omega50Ω. The time taken for the current to rise from half of the peak value of the peak value is:
  1. (A)2.2 ms
  2. (B)3.3 ms
  3. (C)5 ms
  4. (D)7.2 ms

Correct answer: (B)

Step-by-step solution →
Q134·PhysicsSingle correctJEE Main 2019
A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be:
  1. (A)50 A
  2. (B)45 A
  3. (C)35 A
  4. (D)25 A

Correct answer: (B)

Step-by-step solution →
Q135·PhysicsSingle correctJEE Main 2019
A series AC circuit containing an inductor (20 mH), a capacitor (120 μF\mu FμF) and a resistor (60 Ω\OmegaΩ) is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is:
  1. (A)5.65×1025.65 \times 10^{2}5.65×102 J
  2. (B)2.26×1032.26 \times 10^{3}2.26×103 J
  3. (C)5.17×1025.17 \times 10^{2}5.17×102 J
  4. (D)3.39×1033.39 \times 10^{3}3.39×103 J

Correct answer: (C)

Step-by-step solution →
Q136·PhysicsMultiple correctJEE Advanced 2017
In the circuit shown L = 1 μ\muμH, C = 1 μ\muμF and R = 1 kΩk\OmegakΩ. They are connected in series with an a.c. source V = V0V_{0}V0​ sin ωt\omega tωt as shown. Which of the following options is/are correct?
  1. (A)The frequency at which the current will be in the phase with the voltage is independent of R.
  2. (B)At ω∼0\omega \sim 0ω∼0 the current flowing through the circuit becomes nearly zero.
  3. (C)At ω>>106\omega >> 10^{6}ω>>106 rad.s−1rad.s^{-1}rad.s−1, the circuit behave like a capacitor.
  4. (D)The current will be in phase with the voltage if ω=104\omega = 10^{4}ω=104 rad. s−1s^{-1}s−1

Correct answer: (A), (B)

Step-by-step solution →
Q137·PhysicsMultiple correctJEE Advanced 2017
The instantaneous voltages at three terminals marked X, Y and Z are given by VX=V0sin⁡ωtV_{X} = V_{0}\sin\omega tVX​=V0​sinωt VY=V0sin⁡ ⁣(ωt+2π3)V_{Y} = V_{0}\sin\!\left(\omega t + \dfrac{2\pi}{3}\right)VY​=V0​sin(ωt+32π​) and VZ=V0sin⁡ ⁣(ωt+4π3)V_{Z} = V_{0}\sin\!\left(\omega t + \dfrac{4\pi}{3}\right)VZ​=V0​sin(ωt+34π​). An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be
  1. (A)VXYrms=V032V_{XY}^{rms} = V_{0}\sqrt{\dfrac{3}{2}}VXYrms​=V0​23​​
  2. (B)VYZrms=V012V_{YZ}^{rms} = V_{0}\sqrt{\dfrac{1}{2}}VYZrms​=V0​21​​
  3. (C)VXYrms=V0V_{XY}^{rms} = V_{0}VXYrms​=V0​
  4. (D)independent of the choice of the two terminals

Correct answer: (A), (D)

Step-by-step solution →
Q138·PhysicsMultiple correctJEE Advanced 2014
At time t=0t = 0t=0, terminal AAA in the circuit shown in the figure is connected to BBB by a key and an alternating current I(t)=I0cos⁡(ωt)I(t) = I_0 \cos (\omega t)I(t)=I0​cos(ωt), with I0=1 AI_0 = 1 \text{ A}I0​=1 A and ω=500 rad/s\omega = 500 \text{ rad/s}ω=500 rad/s starts flowing in it with the initial direction shown in the figure. At t=7π6ωt = \frac{7\pi}{6\omega}t=6ω7π​, the key is switched from BBB to DDD. Now onwards only AAA and DDD are connected. A total charge QQQ flows from the battery to charge the capacitor fully. If C=20μFC = 20 \mu\text{F}C=20μF, R=10ΩR = 10 \OmegaR=10Ω and the battery is ideal with emf of 50 V50 \text{ V}50 V, identify the correct statement(s).
  1. (A)Magnitude of the maximum charge on the capacitor before t=7π6ωt = \frac{7\pi}{6\omega}t=6ω7π​ is 1×10−3 C1 \times 10^{-3} \text{ C}1×10−3 C.
  2. (B)The current in the left part of the circuit just before t=7π6ωt = \frac{7\pi}{6\omega}t=6ω7π​ is clockwise.
  3. (C)Immediately after AAA is connected to DDD, the current in RRR is 10 A10 \text{ A}10 A.
  4. (D)Q=2×10−3 CQ = 2 \times 10^{-3} \text{ C}Q=2×10−3 C.

Correct answer: (C), (D)

Step-by-step solution →
Q139·PhysicsSingle correctJEE Advanced 2013
A thermal power plant produces electric power of 600 kW600\ \mathrm{kW}600 kW at 4000 V4000\ \mathrm{V}4000 V, which is to be transported to a place 20 km20\ \mathrm{km}20 km away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with the power factor unity. All the currents and voltage mentioned are rms values. If the direct transmission method with a cable of resistance 0.4 Ω km−10.4\ \Omega\ \mathrm{km^{-1}}0.4 Ω km−1 is used, the power dissipation (in %) during transmission is
  1. (A)20
  2. (B)30
  3. (C)40
  4. (D)50

Correct answer: (B)

Step-by-step solution →
Q140·PhysicsSingle correctJEE Advanced 2013
A thermal power plant produces electric power of 600 kW600\ \mathrm{kW}600 kW at 4000 V4000\ \mathrm{V}4000 V, which is to be transported to a place 20 km20\ \mathrm{km}20 km away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with the power factor unity. All the currents and voltage mentioned are rms values. 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 1:101 : 101:10. If the power to the consumers has to be supplied at 200 V200\ \mathrm{V}200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is
  1. (A)200:1200 : 1200:1
  2. (B)150:1150 : 1150:1
  3. (C)100:1100 : 1100:1
  4. (D)50:150 : 150:1

Correct answer: (A)

Step-by-step solution →

Alternating Currents — frequently asked

How many questions from Alternating Currents appear in JEE?

Alternating Currents has appeared in 108 of the last 186 JEE Main and JEE Advanced papers — about 58% of them — contributing 140 questions in total across those papers.

Is Alternating Currents an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 58% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Alternating Currents questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Physics chapters

  • Properties of Solids and Liquids 344
  • Current Electricity 309
  • Rotational Motion 266
  • Geometrical Optics 259
  • Kinematics 255
  • Magnetic Field of Current 211
  • Electromagnetic Waves 208
  • Thermodynamics 201

All 28 Physics chapters →

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