Chapter 17, Problem 77E

(a)

To determine

The charge on the capacitor plates.

Answer to Problem 77E

The charge on the capacitor plates is $\underset{\_}{0.1\text{C}}$.

Explanation of Solution

Given that the capacitance is $100\mu \text{F}$, and the potential difference is $1000\text{V}$.

Write the expression for charge on the capacitor plates.

  $Q=CV$        (I)

Here, $Q$ is the charge, $C$ is the capacitance, and $V$ is the potential difference.

Conclusion:

Substitute $100\mu \text{F}$ for $C$, and $1000\text{V}$ for $V$ in equation (I) to find $Q$.

  $\begin{array}{c}Q=\left(100\mu \text{F}\right)\left(1000\text{V}\right)\\ =\left(100\mu \text{F}\times \frac{1\text{F}}{1\times {10}^6\mu \text{F}}\right)\left(1000\text{V}\right)\\ =0.1\text{C}\end{array}$

Therefore, the charge on the capacitor plates is $\underset{\_}{0.1\text{C}}$.

(b)

To determine

The resistance of the circuit.

Answer to Problem 77E

The resistance of the circuit is $\underset{\_}{10\text{ohm}}$.

Explanation of Solution

Given that the time is $0.001\text{s}$, the remaining charge is $0.37$ times the initial charge, and the capacitance is $100\mu \text{F}$.

Write the expression for the charge in the capacitor in an RC circuit under charging.

  $q\left(t\right)=q_f{e}^{-t/RC}$        (II)

Here, $q\left(t\right)$ is the charge at time $t$, $R$ is the resistance, and $C$ is the capacitance, $q_f$ is the charge when capacitor is fully charged (here the initial charge).

Since the charge remaining is $0.37$ of the initial charge, in equation (II), $q\left(t\right)$ can be replaced by $0.37q_f$.

  $\begin{array}{c}0.37q_f=q_f{e}^{-t/RC}\\ 0.37=e^{-t/RC}\end{array}$        (III)

Solve equation (III) for $R$

  $\begin{array}{c}\ln 0.37=\frac{-t}{RC}\\ R=\frac{-t}{C\ln 0.37}\end{array}$        (IV)

Conclusion:

Substitute $0.001\text{s}$ for $t$, and $100\mu \text{F}$ for $C$ in equation (IV) to find $R$.

  $\begin{array}{c}R=\frac{-\left(0.001\text{s}\right)}{\left(100\mu \text{F}\right)\ln 0.63}\\ =\frac{-\left(0.001\text{s}\right)}{\left(100\mu \text{F}\times \frac{1\text{F}}{1\times {10}^6\mu \text{F}}\right)\ln 0.37}\\ =10\text{ohm}\end{array}$

Therefore, the resistance of the circuit is $\underset{\_}{10\text{ohm}}$.

(c)

To determine

The current after $0.001\text{s}$.

Answer to Problem 77E

The current after $0.001\text{s}$ is $\underset{\_}{37\text{A}}$.

Explanation of Solution

Write the expression for the current in the series RC circuit.

  $i\left(t\right)=\frac{\epsilon }{R}{e}^{-t/R\text{}C}$        (V)

Here, $i\left(t\right)$ is the current at time $t$, $\epsilon$ is the potential difference.

Conclusion:

Substitute $10\text{ohm}$ for $R$, $100\mu \text{F}$ for $C$, $1000\text{V}$ for $\epsilon$, and $0.001\text{s}$ for $t$ in equation (V) to find $i$.

  $\begin{array}{c}i\left(t\right)=\frac{1000\text{V}}{10\text{ohm}}{e}^{-0.001\text{s}/\left(10\text{ohm}\right)\text{}\left(100\mu \text{F}\right)}\\ =\frac{1000\text{V}}{10\text{ohm}}{e}^{-0.001\text{s}/\left(10\text{ohm}\right)\text{}\left(100\mu \text{F}\times \frac{1\text{F}}{1\times {10}^6\mu \text{F}}\right)}\\ =37\text{A}\end{array}$

Therefore, the current after $0.001\text{s}$ is $\underset{\_}{37\text{A}}$.