Question

The constant of a C-R circuit is $\frac{2}{\ln \left(2\right)}$ second capacitor is discharged at time t=0 The ratio of charge on the capacitor on time t= 2s and t=6s is

• A. 3 :1
• B. 8 :1
• C. 4 :1
• D. 2 :1

Answer 1

The correct option is C 4 :1

$$\begin{array}{c}Accordingtoqueation.........\\ Timeconstant\left(\tau \right)=\frac{2}{In\left(2\right)}\\ I=Io{e}^{-t/\tau }\\ \frac{I_1}{I_2}=\frac{e^{-2/\left(2/In2\right)}}{e^{-6/\left(2/In2\right)}}\\ =\frac{e^{3In2}}{e^{In2}}=\frac{2^3}{2}=\frac{4}{1}=4:1\\ sotheoptionisC.\end{array}$$