Question

A capacitor having an initial charge 0.1 C, is discharging in a simple RC circuit with the time constant 3 seconds. How much charge will remain after 6 sec?

• A. $\frac{0.1}{e^2}C$
• B. $\frac{0.01}{e^2}C$
• C. $0.1e^{2}C$
• D. $0.01e^{2}C$

Answer 1

The correct option is A $\frac{0.1}{e^2}C$

While discharging if an amount of charge q is left on a capacitor after time t,
Then,$q=Q{e}^{-\frac{t}{RC}}$, where Q is the initial charge on the capacitor at t=0.
Given, time constant=RC=3 s and t=6s,
Charge remaining after 6s, $q=0.1e^{-\frac{6}{3}}=0.1e^{-2}=\frac{0.1}{e^{2}C}$.