Chapter 20, Problem 59P

(a)

To determine

The currents $I_1$, $I_2$, the potential difference across the resistors, power of the battery and induced emf in the inductor.

Answer to Problem 59P

The current $I_1$, $I_2$ is respectively $1.7\text{mA}$, zero, the potential difference across the resistors is respectively $45\text{V}$, zero, the power supplied by the battery is $0.075\text{W}$ and the potential difference across the inductor is $45\text{V}$.

Explanation of Solution

Write the expression to calculate $I_1$.

  $I_1=\frac{\epsilon }{R}$

Here, $\epsilon$ is the voltage of battery and R is the resistance along the path of $I_1$.

Substitute $45\text{V}$ for $\epsilon$ and $27\text{k}\Omega$ for R in the above equation to calculate $I_1$.

  $\begin{array}{c}{I}_1=\frac{45\text{V}}{27\text{k}\Omega \left(\frac{{10}^3\Omega }{1\text{kΩ}}\right)}\\ =1.7\times {10}^{-3}\text{A}\left(\frac{1\text{mA}}{{10}^{-3}\text{A}}\right)\\ =1.7\text{mA}\end{array}$

At the instance of closing the switch the current through the inductor is zero and therefore $I_2$ is zero.

The voltage across the resistance $3.0\text{kΩ}$ is zero since the current through the loop including this resistor, that is $I_2$ zero.

The whole voltage of the battery would fall across the resistor $27\text{kΩ}$. Thus, the voltage across the $27\text{kΩ}$ is $45\text{V}$.

Write the expression for the power supplied by the battery.

  $P=\frac{{\epsilon }^2}{R}$

Here, P is the power supplied by the battery.

Substitute $45\text{V}$ for $\epsilon$ and $27\text{k}\Omega$ for R in the above equation to calculate P.

  $\begin{array}{c}P=\frac{{\left(45\text{V}\right)}^2}{27\text{k}\Omega \left(\frac{{10}^3\Omega }{1\text{kΩ}}\right)}\\ =0.075\text{W}\end{array}$

Since the current through the inductor is zero ($I_2$ is zero), the voltage across the inductor is $45\text{V}$.

Conclusion:

Therefore, the current $I_1$, $I_2$ is respectively $1.7\text{mA}$, zero, the potential difference across the resistors is respectively $45\text{V}$, zero, the power supplied by the battery is $0.075\text{W}$ and the potential difference across the inductor is $45\text{V}$.

(b)

To determine

The currents $I_1$, $I_2$, the potential difference across the resistors, power of the battery and induced emf in the inductor after the switch has been closed for a long time.

Answer to Problem 59P

The current $I_1$, $I_2$ is respectively $1.7\text{mA}$, $15\text{mA}$, the potential difference across the resistors is respectively $45\text{V}$, $45\text{V}$, the power supplied by the battery is $0.75\text{W}$ and the potential difference across the inductor is $0\text{V}$.

Explanation of Solution

Write the expression to calculate $I_1$.

  $I_1=\frac{\epsilon }{R}$

Here, $\epsilon$ is the voltage of battery and R is the resistance along the path of $I_1$.

Substitute $45\text{V}$ for $\epsilon$ and $27\text{k}\Omega$ for R in the above equation to calculate $I_1$.

  $\begin{array}{c}{I}_1=\frac{45\text{V}}{27\text{k}\Omega \left(\frac{{10}^3\Omega }{1\text{kΩ}}\right)}\\ =1.7\times {10}^{-3}\text{A}\left(\frac{1\text{mA}}{{10}^{-3}\text{A}}\right)\\ =1.7\text{mA}\end{array}$

Write the expression to calculate $I_2$.

  $I_2=\frac{\epsilon }{R^{\prime }}$

Here, $\epsilon$ is the voltage of battery and $R^{\prime }$ is the resistance along the path of $I_2$.

Substitute $45\text{V}$ for $\epsilon$ and $3.0\text{k}\Omega$ for $R^{\prime }$ in the above equation to calculate $I_2$.

  $\begin{array}{c}{I}_2=\frac{45\text{V}}{3.0\text{k}\Omega \left(\frac{{10}^3\Omega }{1\text{kΩ}}\right)}\\ =15\times {10}^{-3}\text{A}\left(\frac{1\text{mA}}{{10}^{-3}\text{A}}\right)\\ =15\text{mA}\end{array}$

The resistors come into parallel and hence the voltage across the two resistors is $45\text{V}$.

Write the expression for the power supplied by the battery.

  $P=\frac{{\epsilon }^2}{R_{eq}}$

Here, P is the power supplied by the battery and $R_{eq}$ is the equivalent resistor.

Write the expression to calculate $R_{eq}$.

  $R_{eq}=\frac{R{R}^{\prime }}{R+R^{\prime }}$

Rewrite the equation for P using the above expression.

  $P=\frac{{\epsilon }^2}{\frac{R{R}^{\prime }}{R+R^{\prime }}}$

Substitute $45\text{V}$ for $\epsilon$, $27\text{k}\Omega$ for R and $3.0\text{k}\Omega$ for $R^{\prime }$ in the above equation to calculate P.

  $\begin{array}{c}P=\frac{{\left(45\text{V}\right)}^2}{\frac{\left(27\text{k}\Omega \right)\left(3.0\text{k}\Omega \right)}{27\text{k}\Omega +3.0\text{k}\Omega }}\\ =\frac{{\left(45\text{V}\right)}^2}{2.7\text{k}\Omega \left(\frac{{10}^3\Omega }{1\text{kΩ}}\right)}\\ =0.75\text{W}\end{array}$

After the switch has been closed for a long time the current reaches a constant value and the emf across the inductor is zero.

Conclusion:

Therefore, the current $I_1$, $I_2$ is respectively $1.7\text{mA}$, $15\text{mA}$, the potential difference across the resistors is respectively $45\text{V}$, $45\text{V}$, the power supplied by the battery is $0.75\text{W}$ and the potential difference across the inductor is $0\text{V}$.