Alternating Current Previous Year Questions — JEE Main Physics
19 past-year Alternating Current questions from JEE Main (Physics) — page 2.
Q22 (2025)
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 $10\mathrm{~V}$, $50\mathrm{~Hz}$. Average power dissipated by the circuit is ______ W. [cite: 185, 186]
### Related Formula
$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$ [cite: 786]
$$P = I_{\text{rms}}^2 R = \frac{V_{\text{rms}}^2 R}{Z^2}$$ [cite: 785, 786]
### Core Logic
First, find the total impedance $Z$ of the LCR circuit: [cite: 185, 786]
$$Z = \sqrt{50^2 + (100 - 50)^2} = \sqrt{50^2 + 50^2} = 50\sqrt{2}\ \Omega$$ [cite: 185, 786]
Now calculate the average power dissipation: [cite: 185, 786]
$$P = \frac{(10)^2 \times 50}{(50\sqrt{2})^2} = \frac{100 \times 50}{2500 \times 2} = \frac{5000}{5000} = 1\ \text{W}$$ [cite: 787]
### Pattern Recognition
Remember that only the resistive element dissipates real thermal power over a complete cycle[cite: 784, 785]. Reactive elements like ideal inductors and capacitors store and release energy alternately without net consumption.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
Q1141 (2025)
An alternating current is given by $I = I_{A}\sin\omega t + I_{B}\cos\omega t$. The r.m.s. current will be :-
- $\sqrt{I_{A}^{2}+I_{B}^{2}}$
- $\frac{\sqrt{I_{A}^{2}+I_{B}^{2}}}{2}$
- $\sqrt{\frac{I_{A}^{2}+I_{B}^{2}}{2}}$
- $\frac{|I_{A}+I_{B}|}{\sqrt{2}}$
### Related Formula
The root-mean-square current value $I_{\text{rms}}$ for a periodic function is defined as:
$$I_{\text{rms}} = \sqrt{\frac{1}{T}\int_{0}^{T} I^{2} dt}$$
For a single sinusoidal term $I = I_{0}\sin(\omega t + \phi)$, the root-mean-square value simplifies directly to:
$$I_{\text{rms}} = \frac{I_{0}}{\sqrt{2}}$$
### Core Logic
We can combine the orthogonal sine and cosine components into a single phase-shifted wave:
$$I = I_{A}\sin\omega t + I_{B}\cos\omega t = \sqrt{I_{A}^{2} + I_{B}^{2}}\sin(\omega t + \phi)$$
where peak current amplitude corresponds to:
$$I_{0} = \sqrt{I_{A}^{2} + I_{B}^{2}}$$
### Step 1: Calculating RMS Value
Using the standard peak-to-RMS conversion factor:
$$I_{\text{rms}} = \frac{I_{0}}{\sqrt{2}} = \frac{\sqrt{I_{A}^{2} + I_{B}^{2}}}{\sqrt{2}} = \sqrt{\frac{I_{A}^{2} + I_{B}^{2}}{2}}$$
### Pattern Recognition
Orthogonal sine and cosine functions are independent. Their mean squared averages add linearly, so you can think of it as a vector addition: $\frac{I_A^2}{2} + \frac{I_B^2}{2}$ under the square root.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
Q1276 (2025)
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged.
Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor $\left(\mathrm{R} / \sqrt{\mathrm{R}^2 + \omega^2\mathrm{L}^2}\right)$ , where $\omega$ is frequency of the supply across resistor R and inductor L. If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage.
In the light of the above statements, choose the most appropriate answer from the options given below:
- $\text{Both (A) and (R) are true but (R) is not the correct explanation of (A).}$
- $\text{(A) is false but (R) is true.}$
- $\text{Both (A) and (R) are true and (R) is the correct explanation of (A).}$
- $\text{(A) is true but (R) is false.}$
### Related Formula
$$Z = \sqrt{R^2 + (\omega L)^2}$$
$$\cos \phi = \frac{R}{Z}$$
### Core Logic
A choke coil has a high inductance $L$ and low resistance $R$. It reduces the current through the mercury tube without wasting electrical power as heat. Statement (A) is correct. The average power dissipation is controlled by the low power factor $\cos \phi$. Reason (R) correctly explains that without a choke coil, the direct supply voltage could cause an overflow of current, destroying the tube filament.
### Pattern Recognition
Choke coils use high $L$ and low $R$ to minimize power loss while reducing circuit current efficiently.
### Chapter Mix
Class 12 Physics: Alternating Current
Q43 (2024)
In a series LCR circuit, the capacitance is changed from $C$ to $4C$. To keep the resonance frequency unchanged, the new inductance should be:
- reduced by $ rac{1}{4} L$
- increased by $2L$
- reduced by $ rac{3}{4} L$
- increased to $4L$
### Related Formula
Resonant angular frequency formula:
$$\omega = \frac{1}{\sqrt{LC}}$$
### Core Logic
For $\omega' = \omega$:
$$\frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{LC}} \implies L'C' = LC$$
Given new capacitance $C' = 4C$:
$$L'(4C) = LC \implies L' = \frac{L}{4}$$
### Step 1: Calculate the Change Needed
The question asks for the reduction increment or statement matching the dynamic shift:
$$\Delta L = L - L' = L - \frac{L}{4} = \frac{3}{4}L$$
Therefore, the inductance must be reduced by $\frac{3}{4}L$.
### Pattern Recognition
Read options carefully. The new value is $\frac{1}{4}L$, which means it must be **reduced by** $\frac{3}{4}L$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
Q34 (2024)
In an a.c. circuit, voltage and current are given by:
$$V = 100\sin(100t)\text{ V}$$
and
$$I = 100\sin\left(100t + \frac{\pi}{3}\right)\text{ mA}$$
respectively. The average power dissipated in one cycle is:
- $5\text{ W}$
- $10\text{ W}$
- $2.5\text{ W}$
- $25\text{ W}$
### Related Formula
The average power dissipated in an AC circuit is:
$$P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos(\Delta \phi)$$
where:
* $V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$
* $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$
* $\Delta \phi$ is the phase difference between voltage and current.
### Core Logic
From the given equations:
* Peak Voltage, $V_0 = 100\text{ V}$
* Peak Current, $I_0 = 100\text{ mA} = 100 \times 10^{-3}\text{ A} = 0.1\text{ A}$
* Phase Difference, $\Delta \phi = \frac{\pi}{3}$
### Step 1: Calculate Average Power
Substitute these values into the average power formula:
$$P_{\text{avg}} = \left( \frac{100}{\sqrt{2}} \right) \times \left( \frac{100 \times 10^{-3}}{\sqrt{2}} \right) \times \cos\left( \frac{\pi}{3} \right)$$
$$P_{\text{avg}} = \frac{10^4 \times 10^{-3}}{2} \times \frac{1}{2}$$
$$P_{\text{avg}} = \frac{10}{4} = 2.5\text{ W}$$
### Pattern Recognition
Remember to look closely at units! The current is given in mA ($10^{-3}\text{ A}$). Missing this conversion leads directly to the trap answer ($2500\text{ W}$ or similar scaling errors).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
