Chapter 23, Problem 46P

(a)

To determine

The current in the circuit

Answer to Problem 46P

The current in the circuit when the switch is in position $a$ is $1.00\text{A}$.

Explanation of Solution

Write the equation for the current in the circuit.

    $I=\frac{\epsilon }{R}$

Here, $I$ is the current in the circuit when the switch is in position $a$, $e$ is the emf of the battery and $R$ is the resistance in the circuit.

Conclusion:

Substitute $12.0\text{V}$ for $\epsilon$ and $12.0\Omega$ for $R$.

    $\begin{array}{c}I=\frac{12.0\text{V}}{12.0\Omega }\\ =1.00\text{A}\end{array}$

Therefore, the current in the circuit when the switch is in position $a$ is $1.00\text{A}$.

(b)

To determine

The voltage across each resistor and the inductor

Answer to Problem 46P

The voltage across each resistor and the inductor are $12.0\text{V}$, $1200\text{V}$ and $1212\text{V}$.

Explanation of Solution

Write the equation for the voltage across the $12.0\Omega$ resistor.

    $\Delta {\text{V}}_{12}=I(12\Omega )$        (I)

Here, $\Delta {\text{V}}_{12}$ is the voltage across the $12.0\Omega$ resistor and $I$ is the current in the circuit.

Write the equation for the voltage across the $1200\Omega$ resistor.

    $\Delta {\text{V}}_{1200}=I(1200\Omega )$        (II)

Here, $\Delta {\text{V}}_{1200}$ is the voltage across the $1200\Omega$ resistor and $I$ is the current in the circuit.

The voltage across the inductor is the sum of the voltage across the $12.0\Omega$ and $1200\Omega$ resistor.

    $\Delta {\text{V}}_L=\Delta {\text{V}}_{12}+\Delta {\text{V}}_{1200}$

Here, $\Delta V_L$ is the voltage across the inductor, $\Delta {\text{V}}_{12}$ is the voltage across the $12.0\Omega$ and $\Delta {\text{V}}_{1200}$ is the voltage across the $1200\Omega$ resistor

Conclusion:

Substitute $1.00\text{A}$ for $I$ in equation (I).

    $\begin{array}{c}\Delta {\text{V}}_{12}=\left(1.00\text{A}\right)(1200\Omega )\\ =1200\text{V}\end{array}$

Therefore, the voltage across the $12.0\Omega$ resistor is $12.0\text{V}$.

Substitute $1.00\text{A}$ for $I$ in equation (II).

    $\begin{array}{c}\Delta {\text{V}}_{12}=\left(1.00\text{A}\right)(12\Omega )\\ =12.0\text{V}\end{array}$

Therefore, the voltage across the $1200\Omega$ resistor is $1200\text{V}$.

Substitute $12.0\text{V}$ for $\Delta {\text{V}}_{12}$ and $1200\text{V}$ for $\Delta {\text{V}}_{1200}$ in equation (II).

    $\begin{array}{c}\Delta {\text{V}}_L=12.0\text{V}+1200\text{V}\\ =1212\text{V}\end{array}$

Therefore, the voltage across the inductor is $1212\text{V}$.

(c)

To determine

The time elapse before the voltage drops

Answer to Problem 46P

The time elapse before the voltage drops to $12.0\text{V}$ is $7.615\times {10}^{-3}\text{s}$.

Explanation of Solution

Write the equation for the change in voltage in the circuit.

    $V\left(t\right)=V_i(1-e^{-t/\tau })$        (III)

Here, $V\left(t\right)$ is the voltage at time $t$, $V_i$ is the initial voltage across the inductor and $\tau$ is the time constant of the circuit.

Write the equation for the time constant for the circuit.

    $\tau =\frac{L}{R}$        (IV)

Here, $L$ is the inductance and $R$ is the resistance in the circuit. Substitute $2.00\text{H}$ for $L$ and $1212\Omega$ for $R$.

    $\begin{array}{c}\tau =\frac{2.00\text{H}}{1212\Omega }\\ =0.00165\text{s}\end{array}$

Rearrange equation (III) to obtain equation for the time elapse.

    $\begin{array}{c}V\left(t\right)=V_i(1-e^{-t/\tau })\\ \frac{V\left(t\right)}{V_i}=1-e^{-t/\tau }\\ t=\tau \ln \left(\frac{V\left(t\right)}{V_i}\right)\end{array}$

Conclusion:

Substitute $0.00165\text{s}$ for $\tau$, $1212\text{V}$ for $V\left(t\right)$ and $12.0\text{V}$ for $V_i$ in equation (I).

    $\begin{array}{c}t=\left(0.00165\text{s}\right)\ln \left(\frac{1212\text{V}}{12\text{V}}\right)\\ =7.615\times {10}^{-3}\text{s}\end{array}$

Therefore, the time elapse before the voltage drops to $12.0\text{V}$ is $7.615\times {10}^{-3}\text{s}$.