A capacitor is charged to a potential difference of 100 V and is then connected across a resistor. The potential difference across the capacitor decays exponentially with respect to time. After 1 sec, the P.D. between the plates of the capacitor is 80 V. what will be the potential difference between the plates after 2 sec ?

Options

A. 40V
B. 56V
C. 60V
D. 64V

Answer

D

Explanation

During discharging of a capacitor,
q the charge after a time `t=q_(0)e^(-lamdat)`
But `q=CV` i.e. `qpropV,Cxx100ve^(-lamda1s)=80V.C`
`:.e^(-lamda)=0.8`
After `2s,V_(2)=100e^(-2lamda)=100(e^(-lamda))^(2)=100xx0.64`
`:.` V after 2s=64V.