Electromagnetic Induction And Alternating Current Ques 66
- In a $L-C-R$ circuit as shown below, both switches are open initially. Now, switch $S _1$ and $S _2$, are closed. ( $q$ is charge on the capacitor and $\tau=R C$ is capacitance time constant). Which of the following statement is correct?
(2013 Main)
(a) Work done by the battery is half of the energy dissipated in the resistor
(b) At $t=\tau q=C V / 2$
(c) At $t=2 \tau q=C V\left(1-e^{-2}\right)$
(d) At $t=\tau / 2, q=C V\left(1-e^{-1}\right)$
Show Answer
Answer:
Correct Answer: 66.(c)
Solution:
Formula:
Decay of current in the circuit containing resistor and inductor:
- For charging of capacitor $q=C V\left(1-e^{t / \tau}\right)$
At $t=2 \tau$
$$ q=C V\left(1-e^{-2}\right) $$
BETA
