Q. In the capacitor shown in the circuit is changed to 5 V and left in the circuit, in 12s the charge on the capacitor will become $(e = 2.718)$
A) $\frac{10}{e}C$
B) $\frac{e}{10}C$
C) $\frac{10}{e^{2}}C$
D) $\frac{e^{2}}{10}C$
Explanation
(a) Final charge on capacitor is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ q = q_{0}e^{- t/RC}$
where $q_{0}$= charge on capacitor at $t = 0,$
$RC$=time constant of the circuit.
Putting $q_{0} = CV_{0}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ q = CV_{0}^{- 1/RC}$
Given, $C = 2F,V_{0} = 5\ volt,R = 6\ \Omega,\ t = 12\ s$
Hence, $q = (2 \times 5)\ e^{- (12/6 \times 2)}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10e^{- 1} = \frac{10}{e}C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.