Physics · Alternating Current

JEE Advanced 2024 — Paper 1 — Question 33

The circuit shown in the figure contains an inductor LL, a capacitor C0C_{0}, a resistor R0R_{0} and an ideal battery. The circuit also contains two keys K1\mathrm{K}_{1} and K2\mathrm{K}_{2}. Initially, both the keys are open and there is no charge on the capacitor. At an instant, key K1\mathrm{K}_{1} is closed and immediately after this the current in R0R_{0} is found to be I1I_{1}. After a long time, the current attains a steady state value I2I_{2}. Thereafter, K2K_{2} is closed and simultaneously K1\mathrm{K}_{1} is opened and the voltage across C0\mathrm{C}_{0} oscillates with amplitude V0\mathrm{V}_{0} and angular frequency ω0\omega_{0} Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

List-IList-II
(P) The value of I1I_{1} in Ampere is1.0
(Q) The value of I2I_{2} in Ampere is2. 2
(R) The value of ω0\omega_{0} in kilo-radians/s3. 4
(S) The value of V0\mathrm{V}_{0} in Volt is4. 20
5. 200
Question figure
  1. Option A:

    P→1;Q→3;R→2;S→5\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 5

    Correct
  2. Option B:

    P→1;Q→2;R→3;S→5\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 5

  3. Option C:

    P→1;Q→3;R→2 S→4\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 \mathrm{~S} \rightarrow 4

  4. Option D:

    P→2Q→5R→3 S→4\mathrm{P} \rightarrow 2 \mathrm{Q} \rightarrow 5 \mathrm{R} \rightarrow 3 \mathrm{~S} \rightarrow 4

Answer: A

Step-by-step solution

(P) At t=0,l1=0t=0, l_{1}=0

(Q) At t →∞,I2=205=4amp\rightarrow \infty, \mathrm{I}_{2}=\frac{20}{5}=4 \mathrm{amp}

(R) ω0=1 LC =125×10−3×10×10−6\omega_{0}=\frac{1}{\sqrt{\text { LC }}}=\frac{1}{\sqrt{25 \times 10^{-3} \times 10 \times 10^{-6}}} =15×10−4=2×103=2kilo−rad/sec=\frac{1}{5 \times 10^{-4}}=2 \times 10^{3}=2 \mathrm{kilo}-\mathrm{rad} / \mathrm{sec}

(S) 12LI22=12CV02\frac{1}{2} \mathrm{LI}_{2}^{2}=\frac{1}{2} \mathrm{CV}_{0}^{2} V0=(LC)I2=(25×10−310×10−6)4=5×10×4V_{0}=\left(\sqrt{\frac{L}{C}}\right) I_{2}=\left(\sqrt{\frac{25 \times 10^{-3}}{10 \times 10^{-6}}}\right) 4=5 \times 10 \times 4 V0=200\mathrm{V}_{0}=200 volts

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 1
Subject
Physics
Chapter
Alternating Current
Topic
Series LCR Circuit and Power Factor