Question

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 Capacitance time constant). Which of the following statement is correct ?

• A. Work done by the battery is half of the energy dissipated in the resistor
• B. \(t=\tau,\ q=CV/2\)
• C. At \(t=2\tau,\ q=CV(1-e^{-2})\)
• D. At \(t=2\tau,\ q=CV(1-e^{-1})\)

Hint:

With only \(S_1\) closed the circuit is a series RC charging circuit: \(q=CV(1-e^{-t/\tau})\).

Answer 1

The Correct Option is C

With \(S_1\) closed and \(S_2\) open, the inductor is out of the circuit and the battery \(V\), resistor \(R\) and capacitor \(C\) are in series.
Charge on the capacitor: \(q=CV(1-e^{-t/\tau})\).
(2) At \(t=\tau\): \(q=CV(1-e^{-1})\approx0.63\,CV\ne CV/2\). False.
(3) At \(t=2\tau\): \(q=CV(1-e^{-2})\). True.
(4) At \(t=2\tau\) the exponent is \(-2\), not \(-1\). False.
(1) Total work by battery \(=qV=CV^2\); energy stored in capacitor \(=\frac12CV^2\); so energy dissipated in R \(=\frac12CV^2\). Work by battery is twice the heat in the resistor, not half. False.
Answer: (3) At \(t=2\tau,\ q=CV(1-e^{-2})\)