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 186 questions; sign in to attempt the remaining 181 in the exam simulator.

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 (30 Jan, Shift 1)Quality and Power Factor
   A series $LR$ circuit connected with an ac source $E=(25\sin 1000t)V$ has a power factor of $\dfrac{1}{\sqrt{2}}$. If the source of emf is changed to $E=(20\sin 2000t)V$, the new power factor of the circuit will be :
   1. A.$\dfrac{1}{\sqrt{2}}$
   2. B.$\dfrac{1}{\sqrt{3}}$
   3. C.$\dfrac{1}{\sqrt{5}}$
   4. D.$\dfrac{1}{\sqrt{7}}$
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   Answer: (C)

   For $LR$ circuit, $\cos \phi =\dfrac{R}{\sqrt{{\left({X}_{L}\right)}^{2}+{R}^{2}}}=\dfrac{1}{\sqrt{2}}$ $\Rightarrow R={X}_{L}$ Now, ${\omega }^{'}=2\omega$ Therefore, ${\left({X}_{L}\right)}^{'}={\omega }^{'}L=2\omega L=2{X}_{L}$ Hence, $\cos {\phi }^{'}=\dfrac{R}{\sqrt{{\left(2{X}_{L}\right)}^{2}+{R}^{2}}}=\dfrac{1}{\sqrt{5}}$
4. Q4JEE Main 2023 (06 Apr, Shift 1)Transformers
   <p>An ideal transformer with purely resistive load operates at $12kV$ on the primary side. It supplies electrical energy to a number of nearby houses at $120V.$ The average rate of energy consumption in the houses served by the transformer is $60kW.$ The value of resistive load $({R}_{s})$ required in the secondary circuit will be_________m$\Omega.$</p>
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   Answer: 240

   <p>The given data is $P=60\times {10}^{3}W$<span class="mord mathnormal" style="margin-right: 0.13889em;">P</span> ${V}_{s}=120V$ A transformer has a primary coil and a secondary coil. The formula for the power delivered is given by $P=\dfrac{{{V}^{2}}_{s}}{{R}_{s}}$<span class="mord mathnormal" style="margin-right: 0.13889em;">P</span> $\Rightarrow 60\times {10}^{3}=\dfrac{120\times 120}{{R}_{s}} \Rightarrow {R}_{s}=\dfrac{120\times 120}{60\times {10}^{3}}=240m\Omega$</p>
5. Q5JEE Main 2026 (05 Apr, Shift 2)AC Circuits
   A series LCR circuit with $R = 20\ \Omega$, $L = 1.6\text{ H}$ and $C = 40\ \mu\text{F}$ is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is _______ $\Omega$.
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   Answer: 200

   Given $L = 1.6\text{ H}$ and $C = 40\ \mu\text{F} = 40 \times 10^{-6}\text{ F}$. At resonance, the angular frequency is $\omega_r = \dfrac{1}{\sqrt{LC}}$. The inductive reactance at resonance is $X_L = \omega_r L = \dfrac{L}{\sqrt{LC}} = \sqrt{\dfrac{L}{C}}$. Substituting the given values: $X_L = \sqrt{\dfrac{1.6}{40 \times 10^{-6}}}$ $X_L = \sqrt{\dfrac{1.6}{4 \times 10^{-5}}}$ $X_L = \sqrt{4 \times 10^4}$ $X_L = 200\ \Omega$ Answer: $200$