Chapter 18, Problem 75QAP

To determine

(a)

The time required to lose 90% of its initial charge and energy

Explanation of Solution

Given:

Capacitor, $C=12.5\mu F=12.5\times {10}^{-6}F$

Resistor, $R=75.0\Omega$

Voltage of the battery, $V=50.0\text{V}$

Formula used:

Charge on the capacitor in time t is given by,
  $Q=C\times V$

Where,
Q= charge on the capacitor
C= capacitance
V = voltage
Energy stored in the capacitor is,
$E=\frac{1}{2}C{V}^2$

Calculation:

Charge on the capacitor is given by,
  $\begin{array}{l}{Q}_0=C\times V\\ Q_0=12.5\times {10}^{-6}\times 50.0\\ Q_0=625\times {10}^{-6}\text{C}\end{array}$

1. The new charge is given by.

$\begin{array}{l}Q=Q_0-0.9Q_0\\ Q=0.1Q_0\\ Q=0.1\times 625\times {10}^{-6}\\ Q=62.5\times {10}^{-6}\text{C}\end{array}$

Thus, new voltage is,
$\begin{array}{l}{V}_n=\frac{Q}{C}\\ V_n=\frac{62.5\times {10}^{-6}}{12.5\times {10}^{-6}}\\ V_n=5\text{V}\end{array}$

Voltage of the capacitor in time t is given by,
  $\begin{array}{l}{V}_n=V(1-e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$RC$}\right.})\\ 5=50.0\times (1-e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$75.0\times 12.5\times {10}^{-6}$}\right.})\\ 1-e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$75.0\times 12.5\times {10}^{-6}$}\right.}=0.1\\ e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$75.0\times 12.5\times {10}^{-6}$}\right.}=0.9\\ -0.04575=\frac{-t}{75.0\times 12.5\times {10}^{-6}}\\ t=9.8775\times {10}^{-5}\text{s}\end{array}$

1. Energy stored in the capacitor is,
$\begin{array}{l}E=\frac{1}{2}C{V}^2\\ E=0.5\times 12.5\times {10}^{-6}\times 50.0\times 50.0\\ E=0.01562\text{J}\end{array}$

Thus, new charge is,
$\begin{array}{l}{E}_n=0.1\times 0.015625\\ E_n=1.5625\times {10}^{-3}\text{J}\end{array}$

So, new voltage is,
$\begin{array}{l}{E}_n=\frac{1}{2}C{V}_n^2\\ V_n=\sqrt{\frac{2\times E_n}{C}}\\ V_n=15.81\text{V}\\ \end{array}$

Voltage of the capacitor in time t is given by,
  $\begin{array}{l}{V}_n=V(1-e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$RC$}\right.})\\ 15.81=50.0\times (1-e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$75.0\times 12.5\times {10}^{-6}$}\right.})\\ 1-e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$75.0\times 12.5\times {10}^{-6}$}\right.}=0.3162\\ e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$75.0\times 12.5\times {10}^{-6}$}\right.}=0.6838\\ -0.3800=\frac{-t}{75.0\times 12.5\times {10}^{-6}}\\ t=3.56\times {10}^{-4}\text{s}\end{array}$