Chapter 19, Problem 51P

To determine

The time for the capacitor to discharge the 1% of the initial voltage.

Answer to Problem 51P

The time for the capacitor to discharge the 1% of the initial voltage is $0.093\text{s}$ .

Explanation of Solution

Given info:

The resistance of the RC circuit is, $R=6.7k\Omega =6.7\times {10}^3\Omega$ .

The capacitance of the RC circuit is, $C=3\text{μF}=3\times {10}^{-6}\text{F}$ .

The initial voltage of the RC circuit is, $V_0$ .

The final voltage of the RC circuit is, $V=0.01V$ .

Formula Used:

The expression to calculate the time for the capacitor to discharge the 1% of the initial voltage is,

  $V=V_0{e}^{-\frac{t}{RC}}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}0.01V_0=V_0{e}^{-\frac{t}{\left(6.7\times {10}^3\Omega \right)\left(3\times {10}^{-6}\text{F}\right)}}\\ -\frac{t}{\left(6.7\times {10}^3\Omega \right)\left(3\times {10}^{-6}\text{F}\right)}=\ln 0.01\\ t=0.093\text{s}\end{array}$

Conclusion:

Thus, the time for the capacitor to discharge the 1% of the initial voltage is $0.093\text{s}$ .