Alternating Current Previous Year Questions — JEE Main Physics
19 past-year Alternating Current questions from JEE Main (Physics) — page 4.
Q48 (2026)
Using a variable-frequency a.c. voltage source, the maximum current measured in the given LCR circuit is 50 mA for $V = 5\sin(100t)$. The values of L and R are shown in the figure. The capacitance of the capacitor (C) used is ____ $\mu\text{F}$.
{{IMG1}}
### Related Formula
$$\omega = \frac{1}{\sqrt{LC}}$$
### Core Logic
Current in an LCR circuit reaches a maximum value (measured as $50 \text{ mA}$) exactly when the circuit is in resonance. At resonance, the inductive reactance equals the capacitive reactance, meaning $X_L = X_C$, which gives the condition $\omega = 1/\sqrt{LC}$.
### Step 1: Identify Parameters
From the source equation $V = 5\sin(100t)$:
$\omega = 100 \text{ rad/s}$
From the circuit diagram:
$L = 2 \text{ H}$
### Step 2: Evaluate Capacitance
$$\omega^{2} = \frac{1}{LC}$$
$$C = \frac{1}{\omega^{2}L}$$
$$C = \frac{1}{(100)^{2} \times 2} = \frac{1}{10000 \times 2} = \frac{1}{2 \times 10^{4}}$$
$$C = 50 \times 10^{-6} \text{ F}$$
$$C = 50 \mu\text{F}$$
### Pattern Recognition
Sees: "maximum current" + "variable-frequency / given frequency" → Resonance! The impedance is purely resistive ($Z=R$), so $\omega L = 1/(\omega C)$. The $50 \text{ mA}$ info is a distractor/cross-check (since $V_{peak}/R = 5/100 = 50 \text{ mA}$, confirming resonance).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
Q34 (2026)
For the series LCR circuit connected with 220 V, 50 Hz a.c source as shown in the figure, the power factor is $\frac{\alpha}{10}$. The value of $\alpha$ is ____
{{IMG1}}
- $4$
- $10$
- $6$
- $8$
### Related Formula
$$\text{Impedance, } Z = \sqrt{R^2 + (X_L - X_C)^2}$$
$$\text{Power factor, } \cos \phi = \frac{R}{Z}$$
### Core Logic
Given from the circuit:
Resistance, $R = 60\, \Omega$
Inductive reactance, $X_L = 70\, \Omega$
Capacitive reactance, $X_C = 150\, \Omega$
First, calculate the impedance $Z$:
$$Z = \sqrt{60^2 + (150 - 70)^2}$$
$$Z = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100\, \Omega$$
### Step 1: Calculate Power Factor
Power factor $= \frac{R}{Z}$
$$\text{Power factor} = \frac{60}{100} = \frac{6}{10}$$
Given that the power factor is $\frac{\alpha}{10}$, comparing both sides gives $\alpha = 6$.
### Pattern Recognition
A classic 3-4-5 impedance triangle where $R=60, X_C-X_L=80$, immediately implies $Z=100$. The power factor is exactly $R/Z = 60/100 = 6/10$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
Q27 (2026)
The electric current in the circuit is given as $i = i_{o}(t/T)$ . The r.m.s current for the period $t = 0$ to $t = T$ is ____
- $\frac{i_{0}}{\sqrt{2}}$
- $i_{0}$
- $\frac{i_{0}}{\sqrt{6}}$
- $\frac{i_{0}}{\sqrt{3}}$
### Related Formula
$$I_{\mathrm{rms}} = \sqrt{ \frac{\int_{0}^{T} i^2 \, dt}{\int_{0}^{T} dt} }$$
### Core Logic
To find the root mean square (rms) value of a varying current, we integrate the square of the current over the given time period, divide by the time period, and take the square root.
### Step 1: Integration of Squared Current
$$\mathrm{i}_{\mathrm{rms}}^2 = \frac{\int_{0}^{\mathrm{T}} \left( \mathrm{i}_{0}^2 \mathrm{t}^2 / \mathrm{T}^2 \right) \mathrm{dt}}{\int_{0}^{\mathrm{T}} \mathrm{dt}}$$
$$= \frac{\mathrm{i}_{0}^2}{\mathrm{T}^3} \int_{0}^{\mathrm{T}} \mathrm{t}^2 \, \mathrm{dt}$$
### Step 2: Final Calculation
$$\mathrm{i}_{\mathrm{rms}}^2 = \frac{\mathrm{i}_{0}^2}{\mathrm{T}^3} \cdot \left[ \frac{\mathrm{t}^3}{3} \right]_{0}^{\mathrm{T}}$$
$$= \frac{\mathrm{i}_{0}^2}{\mathrm{T}^3} \cdot \frac{\mathrm{T}^3}{3} = \frac{\mathrm{i}_{0}^2}{3}$$
$$i_{\mathrm{rms}} = \frac{i_{0}}{\sqrt{3}}$$
### Pattern Recognition
For any linearly varying quantity $y = kt$ passing through origin, the RMS value over time $T$ is always $\frac{\text{Maximum Value}}{\sqrt{3}}$. A standard AC sine wave is $1/\sqrt{2}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
Q48 (2026)
An inductor stores $16 \text{ J}$ of magnetic field energy and dissipates $32 \text{ W}$ of thermal energy due to its resistance when an a.c. current of $2 \text{ A}$ (rms) and frequency $50 \text{ Hz}$ flows through it. The ratio of inductive reactance to its resistance is ____. ( $\pi = 3.14$ )
### Related Formula
$$U_B = \frac{1}{2} L i_{\text{rms}}^2$$
$$P = i_{\text{rms}}^2 R$$
$$X_L = \omega L = 2\pi f L$$
### Core Logic
{{SOL_IMG1}}
The inductor acts as a series combination of an ideal inductor $L$ and a resistor $R$.
Given: Magnetic energy stored $U = 16 \text{ J}$, Power dissipated $P = 32 \text{ W}$, $i_{\text{rms}} = 2 \text{ A}$, $f = 50 \text{ Hz}$.
### Step 1: Calculate Inductance L
$$\frac{1}{2} L i_{\text{rms}}^2 = 16$$
$$\frac{1}{2} L (2)^2 = 16$$
$$2L = 16 \Rightarrow L = 8 \text{ H}$$
### Step 2: Calculate Resistance R
$$i_{\text{rms}}^2 R = 32$$
$(2)^2 R = 32$
$$4R = 32 \Rightarrow R = 8 \, \Omega$$
### Step 3: Calculate Inductive Reactance and Ratio
$$X_L = \omega L = 2 \pi f L$$
$$X_L = 2 \times 3.14 \times 50 \times 8$$
$$X_L = 100 \times 3.14 \times 8 = 314 \times 8 \, \Omega$$
Ratio required:
$$\frac{X_L}{R} = \frac{314 \times 8}{8} = 314$$
### Pattern Recognition
The power dissipated inside a real inductor is exclusively due to its internal resistance ($I^2R$). The magnetic energy stored relates exclusively to its inductance ($\frac{1}{2}LI^2$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
