JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
In an \(LCR \) circuit as shown below both switches are open initially. Now switch \(S_1\) is closed, \(S_2\) kept open. ( \(q\) is charge on the capacitor and \(\tau = RC\) is Capacitive time constant). Which of the following statement is correct?

- A At \(t=\frac{\tau}{2}\),\(q=C V\) \((1-e^{-1})\)
- B Work done by the battery is half of the energy dissipated in the resistor
- C At \(t=\)\(\;\tau \) ,\(q= \) \(\frac{{{\rm{CV}}}}{2}\)
- D At \(t= \)\({2 \tau }\) ,\(q=CV\) \((1-e^{-2})\)
Answer & Solution
Correct Answer
(D) At \(t= \)\({2 \tau }\) ,\(q=CV\) \((1-e^{-2})\)
Step-by-step Solution
Detailed explanation
Charge on he capacitor at any time \(t\) is given by \(q=C V\left(1-e^{t / \tau}\right)\) at \(t=2 \tau\) \(q=C V\left(1-e^{-2}\right)\)
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