Alternating Current JEE Main PYQs (2020–2026)

Solved Alternating Current PYQs

3 recent questions with answers. The full set of 150 is free to practise.

1. JEE Main 2026MediumAC Voltage and Current: Peak, RMS, and Average Values
   Q1

   The electric current in the circuit is given as $=i_o(t/T)$ . The r.m.s current for the period $t=0$ to $t=T$ is $\_\_\_\_$ .

   • A$\frac{i_o}{\sqrt{2}}$
   • B$\frac{i_o}{\sqrt{3}}$
   • C$i_o$
   • D$\frac{i_o}{\sqrt{6}}$

   Show answer & solution

   Correct answer: (B)

   The root mean square (r.m.s) value of a time-varying current $i(t)$ over a period $T$ is defined by the formula :
   
   $$i_{rms}=\sqrt{\frac{1}{T}{\int }_0^T{i}^{2}dt}$$
   
   We are given $i=i_o\left(\frac{t}{T}\right)$ .
   
   So, using the rms formula,
   
   $$i_{rms}^2=\frac{1}{T}{\int }_0^T\left(i_o^2\frac{t^2}{T^2}\right)dt$$
   
   Since $i_o$ and $T$ are constants, we can move them outside the integral :
   
   $$i_{rms}^2=\frac{i_o^2}{T^3}{\int }_0^T{t}^{2}dt=\frac{i_o^2}{T^3}{\left[\frac{t^3}{3}\right]}_0^T=\frac{i_o^2}{T^3}\times \frac{T^3}{3}$$
   
   $$\Rightarrow i_{rms}=\sqrt{\frac{i_o^2}{3}}$$
   
   $$\Rightarrow i_{rms}=\frac{i_o}{\sqrt{3}}$$
   
   Hence, the correct option is (B).
2. JEE Main 2026EasyResonance in Series LCR Circuits
   Q2

   A series LCR circuit with $R=20\Omega ,L=1.6H$ and $C=40\mu F$ is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is $\_\_\_\_\Omega$ .

   Numerical answer type

   Show answer & solution

   Correct answer: 200

   In a series $LCR$ circuit, resonance occurs when the inductive reactance $\left(X_L\right)$ is equal to the capacitive reactance $\left(X_C\right)$ .
   
   $$X_L=X_C$$
   
   $$\Rightarrow {\omega }_{r}L=\frac{1}{{\omega }_{r}C}$$
   
   Where ${\omega }_r$ is the resonant angular frequency.
   
   $${\omega }_r^2=\frac{1}{LC}$$
   
   $$\Rightarrow {\omega }_r=\frac{1}{\sqrt{LC}}$$
   
   The inductance of inductor is given as $L=1.6H$ and the capacitance of capacitor is $C=40\mu F=40\times 10^{-6}F$
   
   Putting the values,
   
   $${\omega }_r=\frac{1}{\sqrt{1.6\times 40\times 10^{-6}}}$$
   
   $$\begin{array}{c}{\omega }_r=\frac{1}{\sqrt{64\times 10^{-6}}}\\ {\omega }_r=\frac{1}{8\times 10^{-3}}=\frac{1000}{8}=125rad/s\end{array}$$
   
   Using the formula for inductive reactance
   
   $$\begin{array}{c}{X}_L={\omega }_r\cdot L\\ X_L=125\times 1.6\\ X_L=125\times \frac{16}{10}=200\Omega \end{array}$$
   
   Therefore, the inductive reactance at the resonant frequency is $200\Omega$ .
3. JEE Main 2025EasyPower in AC Circuits: Wattless Current
   Q3

   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 $10V,50Hz$ . Average power dissipated by the circuit is ___________ W.

   Numerical answer type

   Show answer & solution

   Correct answer: 1

   To calculate the average power dissipated in the circuit, we follow these steps:
   
   Formula for Power:
   
   $$P=V_{\text{rms}}\times I_{\text{rms}}\times cos\varphi$$
   
   Expressing Power in Terms of Voltage and Impedance:
   
   $$P=V_{\text{rms}}\times \frac{V_{\text{rms}}}{Z}\times \frac{R}{Z}$$
   
   Simplifying the Expression:
   
   $$P=\frac{V_{\text{rms}}^2\times R}{Z^2}$$
   
   Calculate Impedance (Z):
   
   The impedance of the circuit (Z) is calculated as:
   
   $$Z=\sqrt{R^2+(X_L-X_C{)}^2}$$
   
   Given:
   
   $$X_L=100\Omega$$
   
   $$X_C=50\Omega$$
   
   $$R=50\Omega$$
   
   $$X_L-X_C=100\Omega -50\Omega =50\Omega$$
   
   Thus:
   
   $$Z=\sqrt{50^2+50^2}=50\sqrt{2}\Omega$$
   
   Substitute Values to Find Power (P):
   
   Given $V_{\text{rms}}=10\text{V}$ , we calculate $P$ as follows:
   
   $$P=\frac{10^2\times 50}{(50\sqrt{2}{)}^2}$$
   
   Simplifying further:
   
   $$P=\frac{100\times 50}{2500\times 2}=1\text{W}$$
   
   Therefore, the average power dissipated by the circuit is $1\text{W}$ .

Topic-wise question count

Where JEE Main has asked from Alternating Current, most frequent first.

• AC Circuits with Purely Resistor, Inductor, and Capacitor43
• Resonance in Series LCR Circuits36
• Series LCR Circuit: Impedance and Phase34
• AC Voltage and Current: Peak, RMS, and Average Values14
• Transformers: Principle and Types9
• Power in AC Circuits: Wattless Current8
• Alternating Current: Definition and Representation4
• AC Generators2

Year-wise Alternating Current questions in JEE Main

Number of questions from this chapter in each year's paper(s).