Alternating Current JEE Main previous year questions with solutions

5 solved JEE Main questions on Alternating Current, free to read — no sign-in needed. The full chapter has 169 questions; sign in to attempt the remaining 164 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Main 2026 (06 Apr, Shift 1)AC Circuits
   A LCR series circuit driven with $E_{rms} = 90$ V at frequency $f_d = 30$ Hz has resistance $R = 80\,\Omega$, an inductance with inductive reactance $X_L = 20.0\,\Omega$ and capacitance with capacitive reactance $X_C = 80.0\,\Omega$. The power factor of the circuit is _______.
   1. A.$0.8$
   2. B.$0.64$
   3. C.$0.9$
   4. D.$0.5$
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   Answer: (A)

   The impedance of the LCR series circuit is given by $Z = \sqrt{R^2 + (X_C - X_L)^2}$. Substituting the given values: $Z = \sqrt{80^2 + (80 - 20)^2}$ $Z = \sqrt{80^2 + 60^2}$ $Z = \sqrt{6400 + 3600} = \sqrt{10000} = 100\,\Omega$ The power factor of the circuit is given by $\cos \phi = \dfrac{R}{Z}$. $\cos \phi = \dfrac{80}{100} = 0.8$ Answer: $0.8$
2. Q2JEE Main 2025 (07 Apr, Shift 2)Alternating Current, Voltage and Power
   An inductor of reactance $100 \Omega$, a capacitor of reactance $50 \Omega$, and a resistor of resistance $50 \Omega$ are connected in series with an AC source of 10 V , 50 Hz. Average power dissipated by the circuit is ________ W.
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   Answer: 1

   $\begin{aligned} & \mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi \\ & \mathrm{P}=\mathrm{V}_{\mathrm{rms}} \times \frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{z}} \times \frac{\mathrm{R}}{\mathrm{z}} \\ & \mathrm{P}=\mathrm{V}_{\mathrm{rms}}^2 \times \frac{\mathrm{R}}{\mathrm{z}^2} \\ & \mathrm{z}=\sqrt{\mathrm{R}^2+\left(\mathrm{x}_{\mathrm{L}}-\mathrm{x}_{\mathrm{C}}\right)^2} \\ & \mathrm{Z}=50 \sqrt{2} \Omega \\ & \mathrm{P}=100 \times \frac{50}{2500 \times 2}=1 \mathrm{~W}\end{aligned}$
3. Q3JEE Main 2024 (29 Jan, Shift 1)Charging and Discarging of capacitors
   A capacitor of capacitance $100\mu F$ is charged to a potential of $12V$ and connected to a $6.4mH$ inductor to produce oscillations. The maximum current in the circuit would be :
   1. A.$3.2A$
   2. B.$1.5A$
   3. C.$2.0A$
   4. D.$1.2A$
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   Answer: (B)

   By energy conservation, it follows that $\dfrac{1}{2}C{V}^{2}=\dfrac{1}{2}L{I}_{\max }^{2}...\left(1\right)$ Equation (1) implies that $\begin{matrix}{I}_{\max } & = & \sqrt{\dfrac{C}{L}}V \\ & = & \sqrt{\dfrac{100\times {10}^{-6}}{6.4\times {10}^{-3}}}\times 12 \\ & = & \dfrac{12}{8} \\ & = & 1.5A\end{matrix}$
4. Q4JEE Main 2023 (25 Jan, Shift 2)Quality and Power Factor
   A series LCR circuit is connected to an AC source of $220V$, $50Hz$. The circuit contains a resistance $R=80\Omega$, an inductor of inductive reactance ${X}_{L}=70\Omega$, and a capacitor of capacitive reactance ${X}_{C}=130\Omega$. The power factor of circuit is $\dfrac{x}{10}$. The value of $x$ is:
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   Answer: 8

   Power factor in LCR circuit is given by, $\cos \phi =\dfrac{R}{Z}$, where, $Z$ is the total impedance of the circuit. So, $\cos \phi =\dfrac{R}{\sqrt{{R}^{2}+{\left({X}_{C}-{X}_{L}\right)}^{2}}}$ Putting the values, we have $\Rightarrow \cos \phi =\dfrac{80}{\sqrt{(80{)}^{2}+(60{)}^{2}}}$ $\Rightarrow \cos \phi =\dfrac{80}{100}\Rightarrow \dfrac{8}{10}$ Thus, the value of $x=8$.
5. Q5JEE Main 2022 (26 Jul, Shift 1)LR Circuit
   In a series $LR$ circuit ${X}_{L}=R$ and power factor of the circuit is ${P}_{1}$. When capacitor with capacitance $C$ such that ${X}_{L}={X}_{C}$ is put in series, the power factor becomes ${P}_{2}$. The ratio $\dfrac{{P}_{1}}{{P}_{2}}$ is
   1. A.$\dfrac{1}{2}$
   2. B.$\dfrac{1}{\sqrt{2}}$
   3. C.$\dfrac{\sqrt{3}}{\sqrt{2}}$
   4. D.$2:1$
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   Answer: (B)

   In case of $L-R$ circuit $Z=\sqrt{{{X}_{L}}^{2}+{R}^{2}}$ Therefore, power factor ${P}_{1}=\cos \phi =\dfrac{R}{Z}$ As ${X}_{L}=R$ $\Rightarrow Z=\sqrt{2}R$ $\Rightarrow {P}_{1}=\dfrac{R}{\sqrt{2}R}\Rightarrow {P}_{1}=\dfrac{1}{\sqrt{2}}$ In case of $L-C-R$ circuit $Z=\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}$ According to the question ${X}_{L}={X}_{C}$ $\Rightarrow Z=R$ $\Rightarrow {P}_{2}=\cos \phi =\dfrac{R}{R}=1$ $\Rightarrow \dfrac{{P}_{1}}{{P}_{2}}=\dfrac{1}{\sqrt{2}}$