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AIIMS 2017
Exam
AIIMS
Year
2017
Subject
Physics
Chapter
Alternating Current

AIIMS 2017 · Physics · Alternating Current

In an L C R circuit as shown in figure, both switches are open...

In an LCRL C R circuit as shown in figure, both switches are open initially. Now switch S1S_1 is closed, S2S_2 kept open. ( qq is charge on the capacitor and τ=RC\tau=R C is capacitive time constant). In an L C R circuit as shown in figure, both switches are open initially. Now switch S_1 is closed, S_2 kept open. ( q
Which of the following statement is correct?

Options

  1. AAt $t=\frac{\tau}{2}, q=C V\left(1-e^{-1}\right)
  2. BWork done by the battery is half of the energy dissipated in the resistor
  3. CAt t=τ,q=CV/2t=\tau, q=C V / 2
  4. DAt $t=2 \tau, q=C V\left(1-e^{-2}\right)

Correct answer

D. At $t=2 \tau, q=C V\left(1-e^{-2}\right)

Step-by-step solution

As switch S1S_1 is closed and switch S2S_2 is kept open. Now, capacitor is charging through a resistor RR. Charge on a capacitor at any time tt is q=q0(1−e−t/τ)=CV(1−e−t/τ)<imgsrc="https://cdn−question−pool.getmarks.app/pyq/aiims/thc9bj668CMJjbRq5sbdzZRYK2BMh02oqxG16zUBGk.original.fullsize.png"><br>q=q_0\left(1-e^{-t / \tau}\right)=C V\left(1-e^{-t / \tau}\right) <img src="https://cdn-question-pool.Quantrex.app/pyq/aiims/thc9bj668CMJjbRq5sb_dzZRYK2BMh02oqxG16zUBGk.original.fullsize.png"><br> At t=τ2,q=CV(1−e−τ/2τ)=CV(1−e−1/2) At t=τ,q=CV(1−e−τ/τ)=CV(1−e−1) At t=2τ,q=CV(1−e−2τ/τ)=CV(1−e−2)\begin{aligned} & \text{ At } t=\frac{\tau}{2}, q=C V \left(1-e^{-\tau / 2 \tau}\right)=C V \left(1-e^{-1 / 2}\right) \\ & \text{ At } t=\tau, q=C V \left(1-e^{-\tau / \tau}\right)=C V \left(1-e^{-1}\right) \\ & \text{ At } t=2 \tau, q=C V \left(1-e^{-2 \tau / \tau}\right)=C V \left(1-e^{-2}\right)\end{aligned}$
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