Chapter 28, Problem 66P

(a)

To determine

The current in the battery, in $100\Omega$ resistor and inductor immediately after the switch is closed.

Answer to Problem 66P

The current in the battery, in $100\Omega$ resistor and inductor immediately after the switch is closed is $\text{90}\text{.9}\text{mA}$ , $\text{90}\text{.9}\text{mA}$ and $0$ respectively.

Explanation of Solution

Given:

The inductance of the inductor is $L=\text{2}\text{H}$ .

The resistance of the resistor is $R=100\Omega$ .

The voltage of battery is ${\epsilon }_o=1\text{0}\text{V}$ .

Formula used:

The expression for the Kirchhoff’s of current at the junction given by,

  ${\displaystyle \sum I_{in}}={\displaystyle \sum I_{out}}$

Calculation:

The current through the battery and inductor is calculated as,

  $\begin{array}{l}{\displaystyle \sum I_{in}}={\displaystyle \sum I_{out}}\\ I_B=I_R{}_{\to 100}+I_L\end{array}$

As the switch is closed the value of current through the inductor,

  $I_L=0$

Then,

  $\begin{array}{c}{I}_B=I_R{}_{\to 100}+0\\ I_B=I_R{}_{\to 100}\\ I_B=\frac{{\epsilon }_o}{R_T}\\ \text{=}\frac{\text{10}\text{V}}{\left(\text{10}\text{Ω+}\text{100}\text{Ω}\right)}\end{array}$

Solve further as,

  $\begin{array}{c}{I}_B=\text{0}\text{.0909}\text{A}\left(\frac{\text{1}\text{mA}}{{\text{10}}^{-3}\text{A}}\right)\\ =\text{90}\text{.9}\text{mA}\end{array}$

Conclusion:

Therefore, the current in the battery, in $100\Omega$ resistor and inductor immediately after the switch is closed is $\text{90}\text{.9}\text{mA}$ , $\text{90}\text{.9}\text{mA}$ and $0$ respectively.

(b)

To determine

The current in the battery, in $100\Omega$ resistor and in the inductor a long time after switch is closed.

Answer to Problem 66P

The current in the battery, in $100\Omega$ resistor and in the inductor a long time after switch is closed is $\text{1}\text{A,}\text{0}$ and $\text{1}\text{A}$ respectively.

Explanation of Solution

Calculation:

The current through the resistor is calculated by using the loop rule as:

  $\left(I_{R\to 100}\right)\times \left(100\Omega \right)=2H\frac{d{I}_L}{dt}$

When the switch remains closed for long time, the current will be steady and $\frac{d{I}_L}{dt}=0$ .

  $\begin{array}{c}\left(I_{R\to 100}\right)\times \left(100\Omega \right)=2\text{H}\times \left(0\right)\\ I_{R\to 100}=0\end{array}$

The current through the battery is calculated by using the loop rule to right side:

  $\begin{array}{r}\text{10}\text{V}-{\text{I}}_{\text{B}}\times \text{10}\text{V}-{\text{I}}_{\text{R}\to \text{100}}\times \left(\text{100}\text{Ω}\right)\text{=0}\\ \text{10}\text{V}-{\text{I}}_{\text{B}}\text{×10}\text{V}-\text{0}\times \left(\text{100}\text{Ω}\right)\text{=0}\\ {\text{I}}_{\text{B}}\text{=1}\text{A}\end{array}$

The current through the inductor is calculated by using the junction rule as:

  $\begin{array}{c}{I}_B=I_R{}_{\to 100}+I_L\\ 1A=0+I_L\\ I_L=1A\end{array}$

Conclusion:

Therefore the current in the battery, in $100-\Omega$ resistor and in the inductor a long time after switch is closed is $\text{1}\text{A,}\text{0}$ and $\text{1}\text{A}$ respectively.

(c)

To determine

The current in the battery, in $100\Omega$ resistor and in the inductorimmediately after switch is opened.

Answer to Problem 66P

The current in the battery, in $100\Omega$ resistor and in the inductor immediately after switch is opened is $\text{0},-\text{1}\text{A}$ and $\text{1}\text{A}$ respectively.

Explanation of Solution

Calculation:

When the switch is reopened the current in the battery instantly becomes zero and the current in the inductor changes continuously until it takes the value of $I_L=\text{1}\text{A}$ . Now using the junction rule,

  $\begin{array}{c}{I}_B=I_R{}_{\to 100}+I_L\\ \text{0=}{I}_R{}_{\to 100}\text{+1}\text{A}\\ I_R{}_{\to 100}\text{=}-\text{1}\text{A}\end{array}$

Conclusion:

Therefore, the current in the battery, in $100\Omega$ resistor and in the inductor immediately after switch is opened is $\text{0},-\text{1}\text{A}$ and $\text{1}\text{A}$ respectively.

(d)

To determine

The current in the battery, in $100\Omega$ resistor and in the inductor after a switch is opened for a long time.

Answer to Problem 66P

The current in the battery, in $100-\Omega$ resistor and in the inductor after a switch is opened for a long time is zero.

Explanation of Solution

Calculation:

When the switch is reopened and after some time all the currents will become zero.

  $\begin{array}{c}{I}_B=I_R{}_{\to 100}\\ I_B=I_L\\ =0\end{array}$

Conclusion:

Therefore, the current in the battery, in $100-\Omega$ resistor and in the inductor after a switch is opened for a long time is zero.