Chapter 18, Problem 97P

(a)

To determine

The time constant of the $RC$ circuit.

Answer to Problem 97P

The time constant of the $RC$ circuit is $\underset{\_}{8.7\times {10}^{-4}\text{s}}$.

Explanation of Solution

Write the expression for the charge in the capacitor during charging.

$Q\left(t\right)=Q_0{e}^{-t/\tau }$        (I)

Here, $Q\left(t\right)$ is the charge in the capacitor during charging, $Q_0$ is the charge in the capacitor at time $t=0$, $\tau$ is the time constant.

Use equation (I) to solve for $\tau$.

$\tau =-\frac{t}{\ln \left(\frac{Q\left(t\right)}{Q_0}\right)}$        (II)

Conclusion:

Substitute $4.0\times {10}^{-3}\text{s}$ for $t$, $\left(0.010\right)Q_0$ for $Q\left(t\right)$ in equation (II) to find $\tau$.

    $\begin{array}{c}\tau =-\frac{4.0\times {10}^{-3}\text{s}}{\ln \left(\frac{\left(0.010\right)Q_0}{Q_0}\right)}\\ =-\frac{4.0\times {10}^{-3}\text{s}}{\ln \left(0.010\right)}\\ =8.7\times {10}^{-4}\text{s}\end{array}$

Therefore, the time constant of the $RC$ circuit is $\underset{\_}{8.7\times {10}^{-4}\text{s}}$.

(b)

To determine

The resistance of the flashbulb in the circuit.

Answer to Problem 97P

The resistance of the flashbulb in the circuit is $\underset{\_}{1.2\Omega }$.

Explanation of Solution

Write the expression for the time constant.

$\tau =RC$        (III)

Here, $R$ is the resistance, $C$ is the capacitance.

Use equation (III) to solve for $R$.

$R=\frac{\tau }{C}$        (IV)

Write the expression for the energy stored in the capacitor.

$U=\frac{1}{2}C{\left(\Delta V\right)}^2$        (V)

Here, $U$ is the energy stored in the capacitor, $\Delta V$ is the voltage across the capacitor.

Use equation (V) to solve for $C$.

$C=\frac{2U}{{\left(\Delta V\right)}^2}$        (VI)

Use equation (VI) and (II) in (IV) to solve for $R$.

$R=\frac{-\frac{t}{\ln \left(\frac{Q\left(t\right)}{Q_0}\right)}}{\frac{2U}{{\left(\Delta V\right)}^2}}$        (VII)

Conclusion:

Substitute $4.0\times {10}^{-3}\text{s}$ for $t$, $\left(0.010\right)Q_0$ for $Q\left(t\right)$, $32\text{J}$ for $U$, $300\text{V}$ for $\Delta V$ in equation (VII) to find $R$.

    $\begin{array}{c}R=\frac{-\frac{4.0\times {10}^{-3}\text{s}}{\ln \left(\frac{\left(0.010\right)Q_0}{Q_0}\right)}}{\frac{2\left(32\text{J}\right)}{{\left(300\text{V}\right)}^2}}\\ =-\frac{\left(4.0\times {10}^{-3}\text{s}\right){\left(300\text{V}\right)}^2}{2\left(32\text{J}\right)\ln \left(0.010\right)}\\ =1.2\Omega \end{array}$

Therefore, the resistance of the flashbulb in the circuit is $\underset{\_}{1.2\Omega }$.

(c)

To determine

The maximum power dissipated in the flashbulb.

Answer to Problem 97P

The maximum power dissipated in the flashbulb is $\underset{\_}{74\text{kW}}$.

Explanation of Solution

Write the expression for the maximum power dissipated.

$P_{\max }=\frac{V^2}{R}$        (VIII)

Here, $P_{\max }$ is the maximum power dissipated.

Conclusion:

Substitute $300\text{V}$ for $V$, $1.2\Omega$ for $R$ in equation (VIII) to find $P_{\max }$.

    $\begin{array}{c}{P}_{\max }=\frac{{\left(300\text{V}\right)}^2}{1.2\Omega }\\ =74\times {10}^3\text{W}\\ \text{=74}\text{kW}\end{array}$

Therefore, the maximum power dissipated in the flashbulb is $\underset{\_}{74\text{kW}}$.