Chapter 23, Problem 60P

(a)

To determine

The relation between the three currents

Answer to Problem 60P

The three currents are related by the equation $I_1=I_2+I$.

Explanation of Solution

Figure.1 shows the flow of currents in the circuit.

From figure.1, $I_1$ is the current through the resistor of resistance $R_1$, $I_2$ is the current through the resistor of resistance $R_2$ and $I$ is the current through the inductor of inductance $L$.

The current through the resistor $R_1$ is the sum of the current through the resistor $R_2$ and the inductor $L$.

Conclusion:

Therefore, the equation relating the three currents in the circuit is

    $I_1=I_2+I$        (I)

(b)

To determine

The relation using the loop in the left hand side

Answer to Problem 60P

The relation obtained from the left hand side loop is $\epsilon -I_1{R}_1-I_2{R}_2=0$.

Explanation of Solution

According to Kirchhoff’s voltage rule, the algebraic sum of all the voltages in any closed loop in a circuit is zero.

Consider the direction of current from positive terminal of battery to the negative terminal to write the Kirchhoff’s voltage rule using figure.1.

The equation for the voltage across the resistor $R_1$ is $-I_1{R}_1$ and that across $R_2$ is $-I_2{R}_2$. The voltage across the battery is given to be $\epsilon$.

Conclusion:

Write the algebraic sum of each of the voltages to obtain the required equation.

    $\epsilon -I_1{R}_1-I_2{R}_2=0$        (II)

Therefore, the relation obtained from the left hand side loop is $\epsilon -I_1{R}_1-I_2{R}_2=0$.

(c)

To determine

The relation using the outer loop

Answer to Problem 60P

The relation obtained from the outer loop is $\epsilon -I_1{R}_1-L\left(dI/dt\right)=0$.

Explanation of Solution

According to Kirchhoff’s voltage rule, the algebraic sum of all the voltages in any closed loop in a circuit is zero.

Consider the direction of current from positive terminal of battery to the negative terminal to write the Kirchhoff’s voltage rule using figure.1.

The equation for the voltage across the resistor $R_1$ is $-I_1{R}_1$ and that across $L$ is $L\left(dI/dt\right)$. The voltage across the battery is given to be $\epsilon$.

Conclusion:

Write the algebraic sum of each of the voltages to obtain the required equation.

    $\epsilon -I_1{R}_1-L\frac{dI}{dt}=0$        (II)

Therefore, the relation obtained from the outer loop is $\epsilon -I_1{R}_1-L\left(dI/dt\right)=0$.

(d)

To determine

An equation having only $I$ and no other currents

Answer to Problem 60P

The relation having only $I$ and no other currents is ${\epsilon }^{\prime }-I_1{R}^{\prime }-L\left(dI/dt\right)=0$.

Explanation of Solution

Substitute equation (I) in equation (II).

    $\begin{array}{c}\epsilon -\left(I_2+I\right)R_1-I_2{R}_2=0\\ I_2=\frac{\epsilon -I{R}_1}{R_1+R_2}\end{array}$        (III)

Substitute equation (I) in equation (III).

    $\begin{array}{c}\epsilon -\left(I_2+I\right)R_1-L\frac{dI}{dt}=0\\ I_2=\frac{\epsilon -L\frac{dI}{dt}}{R_1}-I\end{array}$        (IV)

Compare equation (III) and equation (IV).

    $\begin{array}{c}\frac{\epsilon -I{R}_1}{R_1+R_2}=\frac{\epsilon -L\frac{dI}{dt}}{R_1}-I\\ \epsilon -L\frac{dI}{dt}=\left(\frac{\epsilon -I{R}_1}{R_1+R_2}+I\right)R_1\\ =\left[\frac{\epsilon -I{R}_1+I\left(R_1+R_2\right)}{R_1+R_2}\right]R_1\\ =\left(\frac{\epsilon +I{R}_2}{R_1+R_2}\right)R_1\end{array}$

    $\begin{array}{c}L\frac{dI}{dt}=\epsilon -\left(\frac{\epsilon +I{R}_2}{R_1+R_2}\right)R_1\\ =\frac{\epsilon \left(R_1+R_2\right)-\left(\epsilon +I{R}_2\right)R_1}{R_1+R_2}\\ =\frac{\epsilon \left(R_2\right)-\left(I{R}_2\right)R_1}{R_1+R_2}\\ 0=\epsilon \frac{R_2}{R_1+R_2}-I\frac{R_1{R}_2}{R_1+R_2}-L\frac{dI}{dt}\end{array}$

Conclusion:

Substitute ${\epsilon }^{\prime }$ for $\epsilon \left(R_2/R_1+R_2\right)$ and $R^{\prime }$ for $\left(R_1{R}_2/R_1+R_2\right)$.

    ${\epsilon }^{\prime }-I{R}^{\prime }-L\frac{dI}{dt}=0$        (V)

Therefore, the relation having only $I$ and no other currents is ${\epsilon }^{\prime }-I_1{R}^{\prime }-L\left(dI/dt\right)=0$.

(e)

To determine

The equation of current with respect to time

Answer to Problem 60P

The equation of current with respect to time is $I\left(t\right)=\left(\epsilon /R_1\right)\left(1-e^{-R^{\prime }t/L}\right)$.

Explanation of Solution

The Kirchhoff’s loop rule for the reference equation is

    $\epsilon -IR-L\frac{dI}{dt}=0$        (VI)

Write the equation for the solution of equation (VI).

    $I\left(t\right)=\frac{{\epsilon }^{\prime }}{R^{\prime }}\left(1-e^{-R^{\prime }t/L}\right)$        (VII)

Here, $\begin{array}{c}\frac{{\epsilon }^{\prime }}{R^{\prime }}=\frac{\epsilon R_2/\left(R_1+R_2\right)}{R_1{R}_2/\left(R_1+R_2\right)}\\ =\frac{\epsilon }{R_1}\end{array}$        (VIII)

Conclusion:

Substitute equation (VIII) in equation (VII0.

    $I\left(t\right)=\frac{\epsilon }{R_1}\left(1-e^{-R^{\prime }t/L}\right)$

Here, $R^{\prime }=\frac{R_1{R}_2}{R_1+R_2}$

Therefore, equation of current with respect to time is $I\left(t\right)=\left(\epsilon /R_1\right)\left(1-e^{-R^{\prime }t/L}\right)$.