Chapter 32, Problem 82CP

(a)

To determine

The current in the circuit The current in the circuit a long time after the switch has been in position $a$.

Answer to Problem 82CP

The current in the circuit a long time after the switch has been in position $a$ is $1.0\text{A}$.

Explanation of Solution

Write the expression to obtain the current in the circuit The current in the circuit a long time after the switch has been in position $a$.

    $I=\frac{\epsilon }{R_1}$

Here, $I$ is the current in the circuit, $\epsilon$ is the voltage across the battery and $R_1$ is the resistance which is in series with the battery.

Conclusion:

Substitute $12.0\text{V}$ for $\epsilon$ and $12.0\text{Ω}$ for $R_1$ in the above equation to calculate $I$.

    $\begin{array}{c}I=\frac{12.0\text{V}}{12.0\text{Ω}}\\ =1.0\text{A}\end{array}$

Therefore, the current in the circuit a long time after the switch has been in position $a$ is $1.0\text{A}$.

(b)

To determine

The initial voltage across each resistor and across the inductor when switch is thrown from position $a$ to $b$.

Answer to Problem 82CP

The initial voltage across the resistor which is in series with the battery is $12.0\text{V}$ and the initial voltage across the resistor which is in parallel with the battery is $1200\text{V}$ and the initial voltage across the inductor is $\text{1212}\text{V}$ when switch is thrown from position $a$ to $b$.

Explanation of Solution

When switch is thrown from position $a$ to $b$, the current in both the resistor and inductor remain same.

Write the expression to obtain the voltage across the resistor which is in series with the battery.

    $V_{R_1}=I{R}_1$                                                                                                                (I)

Here, $V_{R_1}$ is the voltage across the resistor which is in series with battery, $I$ is the current in  the circuit and $R_1$ is the resistor which is in series with the battery.

Write the expression to obtain the voltage across the resistor which is in parallel with the battery.

    $V_{R_2}=I{R}_2$                                                                                                               (II)

Here, $V_{R_2}$ is the voltage across the resistor which is in parallel with battery, $I$ is the current in  the circuit and $R_2$ is the resistor which is in parallel with the battery.

Write the expression to obtain the voltage across the inductor.

    $V_{\text{L}}=V_{R_1}+V_{R_2}$                                                                                                         (III)

Here, $V_{\text{L}}$ is the voltage across the inductor, $V_{R_1}$ is the voltage across the resistor which is in series with battery and $V_{R_2}$ is the voltage across the resistor which is in parallel with battery.

Conclusion:

Substitute $1.0\text{A}$ for $I$ and $12.0\text{Ω}$ for $R_1$ in equation (I) to calculate $V_{R_1}$.

    $\begin{array}{c}{V}_{R_1}=\left(1.0\text{A}\right)\left(12.0\text{Ω}\right)\\ =12.0\text{V}\end{array}$

Substitute $1.0\text{A}$ for $I$ and $1200\text{Ω}$ for $R_2$ in equation (II) to calculate $V_{R_2}$.

    $\begin{array}{c}{V}_{R_2}=\left(1.0\text{A}\right)\left(1200\text{Ω}\right)\\ =1200\text{V}\end{array}$

Substitute $12.0\text{V}$ for $V_{R_1}$ and $1200\text{V}$ for $V_{R_2}$ in equation (III) to calculate $V_{\text{L}}$.

    $\begin{array}{c}{V}_{\text{L}}=12.0\text{V}+1200\text{V}\\ \text{=1212}\text{V}\end{array}$

Therefore, the initial voltage across the resistor which is in series with the battery is $12.0\text{V}$ and the initial voltage across the resistor which is in parallel with the battery is $1200\text{V}$ and the initial voltage across the inductor is $\text{1212}\text{V}$ when switch is thrown from position $a$ to $b$.

(c)

To determine

The time elapsed before the voltage across the inductor drops to $12.0\text{V}$.

Answer to Problem 82CP

The time elapsed before the voltage across the inductor drops to $12.0\text{V}$ is $7.6\text{ms}$.

Explanation of Solution

Write the expression to obtain the time elapsed before the voltage across the inductor drops to $12.0\text{V}$.

    $\epsilon =V_{\text{L}}\left(e^{\frac{-\left(R_1+R_2\right)t}{L}}\right)$

Here, $\epsilon$ is the potential across the battery, $V_{\text{L}}$ is the potential across the inductor, $R_1$ is the resistor which is in series with the battery, $R_2$ is the resistor which is in parallel with the battery, $L$ is the inductance of the inductor and $t$ is the time elapsed.

Conclusion:

Substitute $12.0\text{Ω}$ for $R_1$, $1200\text{Ω}$ for $R_2$, $12.0\text{V}$ for $\epsilon$, $2.00\text{H}$ for $L$ and $1212\text{V}$ for $V_{\text{L}}$ in the above equation to calculate $t$.

    $\begin{array}{l}12.0\text{V}=1212\text{V}\left(e^{\frac{-\left(12.0\text{Ω}+1200\text{Ω}\right)t}{2.00\text{H}}}\right)\\ \left(e^{\frac{-\left(12.0\text{Ω}+1200\text{Ω}\right)t}{2.00\text{H}}}\right)=\frac{12.0\text{V}}{1212\text{V}}\\ \left(e^{\frac{-\left(12.0\text{Ω}+1200\text{Ω}\right)t}{2.00\text{H}}}\right)=0.0099\\ \left(\frac{-\left(12.0\text{Ω}+1200\text{Ω}\right)t}{2.00\text{H}}\right)=\ln \left(0.0099\right)\end{array}$

Further solve the above equation.

    $\begin{array}{c}\left(\frac{-\left(12.0\text{Ω}+1200\text{Ω}\right)t}{2.00\text{H}}\right)=\ln \left(0.0099\right)\\ -606t=-4.61522\\ t=\frac{-4.61522}{-606}\\ =0.0076\text{s}\end{array}$

Further solve the above equation.

    $\begin{array}{c}t=0.0076\text{s}\times \frac{1000\text{ms}}{1\text{s}}\\ =7.6\text{ms}\end{array}$

Therefore, the time elapsed before the voltage across the inductor drops to $12.0\text{V}$ is $7.6\text{ms}$.