Chapter 32, Problem 70AP

(a)

To determine

The relation among the currents using Kirchhoff’s law.

Answer to Problem 70AP

The relation among the currents using Kirchhoff’s law is $I_1=I_2+I$.

Explanation of Solution

Apply Kirchhoff’s junction rule which states that at any junction the sum of the cuurent must be equal to zero.

    $I_1=I_2+I$                                                                                                                    (I)

Here, $I$ is the equivalent current in the circuit, $I_1$ is the current through resistor $R_1$ and $I_2$ is the current through resistor $R_2$.

Therefore, the relation among the currents using Kirchhoff’s law is $I_1=I_2+I$.

(b)

To determine

The relation using Kirchhoff’s rule in the left loop.

Answer to Problem 70AP

The relation using Kirchhoff’s rule in the left loop is $I_1=\frac{\epsilon +I{R}_2}{R_1+R_2}$.

Explanation of Solution

Apply Kirchhoff’s loop rule in the left loop.

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

Here, $\epsilon$ is the emf across the left loop.

Rearrange equation (I) to get the expression for $I_2$.

    $I_2=I_1-I$

Substitute $I_1-I$ for $I_2$ in equation (II).

    $\begin{array}{c}\epsilon =I_1{R}_1+\left(I_1-I\right)R_2\\ I_1=\frac{\epsilon +I{R}_2}{R_1+R_2}\end{array}$

Therefore, the relation using Kirchhoff’s rule in the left loop is $I_1=\frac{\epsilon +I{R}_2}{R_1+R_2}$.

(c)

To determine

The relation using Kirchhoff’s rule in the outer loop.

Answer to Problem 70AP

The relation using Kirchhoff’s rule in the outer loop is $\epsilon =\left(\frac{\epsilon +I{R}_2}{R_1+R_2}\right)R_1+L\frac{dI}{dt}$.

Explanation of Solution

Apply Kirchhoff’s rule in the outer loop.

    $\epsilon =I_1{R}_1+L\frac{dI}{dt}$                                                                                                        (III)

Substitute $\frac{\epsilon +I{R}_2}{R_1+R_2}$ for $I_1$ in equation (III).

    $\epsilon =\left(\frac{\epsilon +I{R}_2}{R_1+R_2}\right)R_1+L\frac{dI}{dt}$                                                                                          (IV)

Therefore, the relation using Kirchhoff’s rule in the outer loop is $\epsilon =\left(\frac{\epsilon +I{R}_2}{R_1+R_2}\right)R_1+L\frac{dI}{dt}$.

(d)

To determine

The equation involving only current.

Answer to Problem 70AP

The equation involving only current is $\frac{\epsilon R_2}{\left(R_1+R_2\right)}-\frac{I{R}_1{R}_2}{\left(R_1+R_2\right)}-L\frac{dI}{dt}=0$.

Explanation of Solution

Multiply and divide left hand side of equation (IV).

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

Therefore, the equation involving only current is $\frac{\epsilon R_2}{\left(R_1+R_2\right)}-\frac{I{R}_1{R}_2}{\left(R_1+R_2\right)}-L\frac{dI}{dt}=0$.

(e)

To determine

The given expression.

Answer to Problem 70AP

The given equation $I=\frac{\epsilon }{R_1}\left(\left(1-e^{\left(\frac{R^{\prime }}{\left(L\right)}\right)\left(t\right)}\right)\right)$ has been derived.

Explanation of Solution

Equation (V) is of the form.

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

The solution to equation (V) is

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

Compare equation (V), (VI) and (VII) for the value of $I$.

    $\begin{array}{c}I=\frac{\frac{\epsilon R_2}{\left(R_1+R_2\right)}}{\frac{R_1{R}_2}{\left(R_1+R_2\right)}}\left(1-e^{\left(\frac{R_1{R}_2}{\left(R_1+R_2\right)}\right)\left(\frac{t}{L}\right)}\right)\\ =\frac{\epsilon }{R_1}\left(\left(1-e^{\left(\frac{R_1{R}_2}{\left(R_1+R_2\right)}\right)\left(\frac{t}{L}\right)}\right)\right)\end{array}$                                                                             (VIII)

Substitute $R^{\prime }$ for $\frac{R_1{R}_2}{R_1+R_2}$ in equation (VIII)

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

Therefore, the given equation $I=\frac{\epsilon }{R_1}\left(\left(1-e^{\left(\frac{R^{\prime }}{\left(L\right)}\right)\left(t\right)}\right)\right)$ has been derived.