Alternating Current Previous Year Questions — JEE Main Physics
19 past-year Alternating Current questions from JEE Main (Physics).
Q15 (2025)
An electric bulb rated as 100W 220V is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is:
- 0.64 A
- 0.45 A
- 2.2 A
- 0.32 A
### Related Formula
Power consumed in a purely resistive AC device (like a lightbulb) is:
$$P = V_{\text{rms}} I_{\text{rms}}$$
The peak current $I_0$ is related to the rms current $I_{\text{rms}}$ by:
$$I_0 = \sqrt{2} I_{\text{rms}}$$
### Core Logic
Given parameters:
- Power $P = 100\mathrm{~W}$
- rms Voltage $V_{\text{rms}} = 220\mathrm{~V}$
### Step 1: Calculate RMS Current ($I_{\text{rms}}$)
$$I_{\text{rms}} = \frac{P}{V_{\text{rms}}} = \frac{100}{220} = \frac{5}{11}\mathrm{~A} \approx 0.455\mathrm{~A}$$$
### Step 2: Calculate Peak Current ($I_0$)
$$I_0 = \sqrt{2} I_{\text{rms}} = 1.414 \times \frac{5}{11} = \frac{7.07}{11} \approx 0.64\mathrm{~A}$$
### Pattern Recognition
Always remember that rated values specify the RMS limits. Since the source voltage matches the bulb's rated voltage exactly, the actual power equals the rated power ($100\mathrm{~W}$). Do not confuse $I_{\text{rms}}$ ($0.45\mathrm{~A}$) with the peak current $I_0$ ($0.64\mathrm{~A}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
Q247 (2025)
For ac circuit shown in figure, $R = 100 \, \text{k}\Omega$ and $C = 100 \, \text{pF}$ and the phase difference between $V_{\text{in}}$ and $(V_{\text{B}} - V_{\text{A}})$ is $90^\circ$ . The input signal frequency is $10^{\text{x}}$ rad/sec, where 'x' is {{IMG1}}
### Related Formula
For a series RC branch, the voltage phase angle $\theta$ is:
$$\tan\theta = \frac{X_C}{R} = \frac{1}{\omega C R}$$
If the phase difference between $V_{\text{in}}$ and $(V_B - V_A)$ is $90^\circ$:
$$\theta = 45^\circ \implies \tan\theta = 1$$
### Core Logic
{{SOL_IMG1}}{{SOL_IMG2}}
From the phasor representation of the symmetric RC divider bridge:
- The phase angle of $V_A$ is $\theta$ behind the input voltage.
- The phase angle of $V_B$ is $90^\circ - \theta$ ahead of the input.
- If their combined relative phase difference is $90^\circ$, this symmetry dictates:
{{SOL_IMG3}}{{SOL_IMG4}}
$$\theta + \theta = 90^\circ \implies \theta = 45^\circ$$
Therefore, we have the condition:
$$X_C = R \implies \frac{1}{\omega C} = R$$
### Step 1: Solve for Frequency
Rearrange to solve for $\omega$:
$$\omega = \frac{1}{R C}$$
Substitute the given values:
- $R = 100 \mathrm{~k}\Omega = 10^5 \Omega$
- $C = 100 \mathrm{~pF} = 100 \times 10^{-12} \mathrm{~F} = 10^{-10} \mathrm{~F}$
$$\omega = \frac{1}{10^5 \times 10^{-10}} = \frac{1}{10^{-5}} = 10^5 \mathrm{~rad/sec}$$
Since $\omega = 10^x$, we get $x = 5$.
### Pattern Recognition
Sees: Phase shift of $90^\circ$ across a symmetric RC bridge.
Shortcut: Clamping phase shift at $90^\circ$ implies the reactive impedance equals the resistive impedance ($X_C = R$). The frequency is simply the characteristic time constant frequency $\omega = 1/\tau = 1/(RC)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
Q9 (2025)
An ac current is represented as
$$i = 5 \sqrt {2} + 1 0 \cos \left(6 5 0 \pi t + \frac {\pi}{6}\right) \mathrm {A m p}$$
The r.m.s value of the current is
- 50 Amp
- $100\mathrm{Amp}$
- 10 Amp
- $5 \sqrt{2} \mathrm{Amp}$
### Related Formula
For a current having both DC and AC components $i = I_{\text{dc}} + I_0 \cos(\omega t + \phi)$, the mean-square value is:
$$\langle i^2 \rangle = I_{\text{dc}}^2 + \frac{I_0^2}{2}$$
The RMS current $I_{\text{rms}}$ is:
$$I_{\text{rms}} = \sqrt{\langle i^2 \rangle} = \sqrt{I_{\text{dc}}^2 + \frac{I_0^2}{2}}$$
### Core Logic
Identify the parameters from the given equation:
- $I_{\text{dc}} = 5\sqrt{2} \mathrm{~A}$
- $I_0 = 10 \mathrm{~A}$
Calculate the square of the components:
$$I_{\text{dc}}^2 = (5\sqrt{2})^2 = 50$$
$$\frac{I_0^2}{2} = \frac{100}{2} = 50$$
### Step 1: Compute Resultant RMS
Substitute back into the RMS formula:
$$I_{\text{rms}} = \sqrt{50 + 50} = \sqrt{100} = 10 \mathrm{~Amp}$$
### Pattern Recognition
Sees: Superposition of a DC current $I_{\text{dc}}$ and a pure AC cosine wave with amplitude $I_{\text{ac}}$.
Shortcut: Use the orthogonal component RMS formula: $I_{\text{rms}} = \sqrt{I_{\text{dc}}^2 + I_{\text{ac,rms}}^2}$. Here $I_{\text{dc}} = 5\sqrt{2}$ and $I_{\text{ac,rms}} = 10/\sqrt{2} = 5\sqrt{2}$. Thus, $I_{\text{rms}} = \sqrt{50 + 50} = 10 \mathrm{~A}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
Q25 (2025)
An inductor of self inductance 1 H connected in series with a resistor of $100 \pi$ ohm and an ac supply of $100 \pi$ volt, 50 Hz. Maximum current flowing in the circuit is ______ A.
### Related Formula
Inductive Reactance:
$$X_L = \omega L = 2\pi f L$$
Impedance of RL series circuit:
$$Z = \sqrt{R^2 + X_L^2}$$
Maximum current:
$$I_{\text{max}} = \sqrt{2} I_{\text{rms}} = \frac{V_{\text{max}}}{Z} \quad \text{or} \quad I_{\text{max}} = \sqrt{2} \frac{V_{\text{rms}}}{Z}$$
### Core Logic
Calculate inductive reactance $X_L$:
$$X_L = 2\pi \times 50 \times 1 = 100\pi\ \Omega$$
Given resistance $R = 100\pi\ \Omega$.
Compute total impedance $Z$:
$$Z = \sqrt{(100\pi)^2 + (100\pi)^2} = 100\pi\sqrt{2}\ \Omega$$
### Step 1: Calculate Maximum Current
Assuming the given supply voltage ($100\pi\text{ V}$) is standard RMS voltage:
$$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{100\pi}{100\pi\sqrt{2}} = \frac{1}{\sqrt{2}}\text{ A}$$
Then, peak/maximum current is:
$$I_{\text{max}} = \sqrt{2} \cdot I_{\text{rms}} = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1\text{ A}$$
Hence, the maximum current is **1**.
### Pattern Recognition
When inductive reactance equals resistance ($X_L = R$), the impedance is exactly $R\sqrt{2}$. The factor of $\sqrt{2}$ in the denominator cancels perfectly with the peak current conversion multiplier.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
Q906 (2025)
An alternating current is represented by the equation, $i=100\sqrt{2}\sin(100\pi t)$ ampere. The RMS value of current and the frequency of the given alternating current are
- $100\sqrt{2}\mathrm{~A}, 100\mathrm{~Hz}$
- $\frac{100}{\sqrt{2}}\mathrm{~A}, 100\mathrm{~Hz}$
- $100\mathrm{~A}, 50\mathrm{~Hz}$
- $50\sqrt{2}\mathrm{~A}, 50\mathrm{~Hz}$
### Related Formula
$$\beta = \frac{D\lambda}{d}$$
Therefore:
$$\beta \propto \frac{1}{d}$$
where:
* $\beta$ = fringe width
* $d$ = slit separation width
* $D$ = distance to screen
* $\lambda$ = wavelength
### Core Logic
Given data:
* Initial slit separation, $d_1 = 0.2\mathrm{~mm}$
* Final slit separation, $d_2 = 0.4\mathrm{~mm}$ ($d$ is doubled).
### Step 1: Calculate Percentage Change
Since $d_2 = 2d_1$, the new fringe width becomes:
$$\beta_2 = \frac{\beta_1}{2}$$
Percentage change formulation:
$$\text{Percentage Change} = \left| \frac{\beta_2 - \beta_1}{\beta_1} \right| \times 100 = \left| \frac{0.5\beta_1 - \beta_1}{\beta_1} \right| \times 100 = 50\%$$
### Pattern Recognition
Doubling the denominator of an inversely proportional relationship halves the primary value, yielding an absolute $50\%$ decrease.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Wave Optics
