Chapter 28, Problem 12P

(a)

To determine

The resistance $R_1$ for the given circuit.

Answer to Problem 12P

The resistance $R_1$ for the given circuit is $\epsilon \left(\frac{1}{I_{\text{b}}}+\frac{2}{I_{\text{a}}}-\frac{2}{I_0}\right)$.

Explanation of Solution

At the start when the switch is open.

The resistors $R_1$, $R_2$ and $R_3$ are connected in series.

Write the expression for the equivalent resistance for the resistors connected in series.

    $R_{\text{S}}=R_1+R_2+R_3$                          (I)

Here, $R_{\text{S}}$ is the equivalent resistance for the resistors connected in series and $R_1$, $R_2$ and $R_3$ are the resistors.

Write the expression for the Ohm’s law when the switch is open.

    $\epsilon =I_{\text{0}}{R}_{\text{S}}$                    (II)

Here, $\epsilon$ is the potential difference and $I_0$ is the current when the switch is open.

Substitute $R_1+R_2+R_3$ for $R_{\text{S}}$ in equation (I).

    $\begin{array}{c}\epsilon =I_{\text{0}}\left(R_1+R_2+R_3\right)\\ \left(R_1+R_2+R_3\right)=\frac{\epsilon }{I_{\text{0}}}\end{array}$                                (III)

When the switch is closed.

Resistors $R_2$ and $R_2$ are parallel to each other.

Write the expression for the equivalent resistance for the resistors connected in parallel.

    $\begin{array}{c}\frac{1}{R_{\text{P}}}=\frac{1}{R_2}+\frac{1}{R_2}\\ R_{\text{P}}=\frac{R_2}{2}\end{array}$                         (IV)

Here, $R_{\text{P}}$ is the equivalent resistance for the resistors connected in parallel.

Write the expression for the Ohm’s law when the switch is closed at position $a$.

    $\epsilon =I_{\text{a}}{R}_{\text{P}}$                    (V)

Here, $\epsilon$ is the potential difference and $I_{\text{a}}$ is the current when the switch is closed at position $a$.

Substitute $\frac{R_2}{2}$ for $R_{\text{P}}$ in equation (V).

    $\begin{array}{c}\epsilon =\left(\frac{R_2}{2}\right)\\ R_2=\frac{2\epsilon }{I_{\text{a}}}\end{array}$                                           (VI)

Now, $R_1$, $R_{\text{P}}$ and $R_3$ are connected in series.

Write the expression for equivalent resistance of the resistor in the circuit.

    $R_{\text{e}}=R_1+R_{\text{P}}+R_3$                                     (VII)

Substitute $\frac{R_2}{2}$ for $R_{\text{P}}$ in equation (VII).

    $R_{\text{eq}}=R_1+\frac{R_2}{2}+R_3$

Here, $R_{\text{eq}}$ is the equivalent resistance of  the resistors in the circuit

Write the expression for the Ohm’s law for equivalent circuit.

    $\epsilon =I_{\text{a}}{R}_{\text{eq}}$                     (VIII)

Here, $\epsilon$ is the potential difference and $I$ is the current.

Substitute $R_1+\frac{R_2}{2}+R_3$ for $R_{\text{eq}}$ in equation (VIII).

    $\begin{array}{c}\epsilon =I_{\text{a}}\left(R_1+\frac{R_2}{2}+R_3\right)\\ \left(R_1+\frac{R_2}{2}+R_3\right)=\frac{\epsilon }{I_{\text{a}}}\end{array}$                           (IX)

When the switch is at position $b$.

The resistors $R_1$ and $R_2$ are in series.

Write the expression for the equivalent resistance for the resistors in series

    $R_{\text{S}}{}^{\prime }=R_1+R_2$

Here, $R_{\text{S}}{}^{\prime }$ is the equivalent resistance for the resistors in series.

Write the expression for the Ohm’s law.

    $\epsilon =I_{\text{b}}{R}_{\text{S}}{}^{\prime }$                    (X)

Here, $\epsilon$ is the potential difference and $I$ is the current when the witch is at $b$.

Conclusion:

Substitute $R_1+R_2$ for $R_{\text{S}}{}^{\prime }$ equation (X).

    $\begin{array}{c}\epsilon =I_{\text{b}}\left(R_1+R_2\right)\\ \left(R_1+R_2\right)=\frac{\epsilon }{I_{\text{b}}}\end{array}$                               (XI)

Substitute $\frac{\epsilon }{I_{\text{b}}}$ for $\left(R_1+R_2\right)$ in equation (III).

    $\begin{array}{l}\left(\frac{\epsilon }{I_{\text{b}}}+R_3\right)=\frac{\epsilon }{I_{\text{0}}}\\ R_3=\epsilon \left(\frac{1}{I_{\text{0}}}-\frac{1}{I_{\text{b}}}\right)\end{array}$                                  (XII)

Substitute $\epsilon \left(\frac{1}{I_{\text{0}}}-\frac{1}{I_{\text{b}}}\right)$ for $R_3$ in equation (IX).

    $\begin{array}{c}\left(R_1+\frac{R_2}{2}+\epsilon \left(\frac{1}{I_{\text{0}}}-\frac{1}{I_{\text{b}}}\right)\right)=\frac{\epsilon }{I_{\text{a}}}\\ R_1+\frac{R_2}{2}=\frac{\epsilon }{I_{\text{a}}}+\frac{\epsilon }{I_{\text{b}}}-\frac{\epsilon }{I_0}\end{array}$                         (XIII)

Substitute $\left(R_1+R_2\right)$ for $\frac{\epsilon }{I_{\text{b}}}$ in equation (XIII).

    $\begin{array}{c}{R}_1+\frac{R_2}{2}=\frac{\epsilon }{I_{\text{a}}}+\left(R_1+R_2\right)-\frac{\epsilon }{I_0}\\ 2R_2-R_2=2\epsilon \left(\frac{1}{I_0}-\frac{1}{I_{\text{a}}}\right)\\ R_2=2\epsilon \left(\frac{1}{I_0}-\frac{1}{I_{\text{a}}}\right)\end{array}$                        (XIV)

Substitute $2\epsilon \left(\frac{1}{I_0}-\frac{1}{I_{\text{a}}}\right)$ for $R_2$ in equation (XI).

    $\begin{array}{c}\left(R_1+2\epsilon \left(\frac{1}{I_0}-\frac{1}{I_{\text{a}}}\right)\right)=\frac{\epsilon }{I_{\text{b}}}\\ R_1=\frac{\epsilon }{I_{\text{b}}}-2\epsilon \left(\frac{1}{I_0}-\frac{1}{I_{\text{a}}}\right)\\ =\epsilon \left(\frac{1}{I_{\text{b}}}+\frac{2}{I_{\text{a}}}-\frac{2}{I_0}\right)\end{array}$

Therefore, the resistance $R_1$ for the given circuit is $\epsilon \left(\frac{1}{I_{\text{b}}}+\frac{2}{I_{\text{a}}}-\frac{2}{I_0}\right)$.

(b)

To determine

The resistance $R_2$ for the given circuit

Answer to Problem 12P

The resistance $R_2$ for the given circuit is $2\epsilon \left(\frac{1}{I_0}-\frac{1}{I_{\text{a}}}\right)$.

Explanation of Solution

Conclusion:

Substitute $R_1+R_2$ for $R_{\text{S}}{}^{\prime }$ equation (X).

    $\begin{array}{c}\epsilon =I_{\text{b}}\left(R_1+R_2\right)\\ \left(R_1+R_2\right)=\frac{\epsilon }{I_{\text{b}}}\end{array}$                               (XI)

Substitute $\frac{\epsilon }{I_{\text{b}}}$ for $\left(R_1+R_2\right)$ in equation (III).

    $\begin{array}{l}\left(\frac{\epsilon }{I_{\text{b}}}+R_3\right)=\frac{\epsilon }{I_{\text{0}}}\\ R_3=\epsilon \left(\frac{1}{I_{\text{0}}}-\frac{1}{I_{\text{b}}}\right)\end{array}$                                  (XII)

Substitute $\epsilon \left(\frac{1}{I_{\text{0}}}-\frac{1}{I_{\text{b}}}\right)$ for $R_3$ in equation (IX).

    $\begin{array}{c}\left(R_1+\frac{R_2}{2}+\epsilon \left(\frac{1}{I_{\text{0}}}-\frac{1}{I_{\text{b}}}\right)\right)=\frac{\epsilon }{I_{\text{a}}}\\ R_1+\frac{R_2}{2}=\frac{\epsilon }{I_{\text{a}}}+\frac{\epsilon }{I_{\text{b}}}-\frac{\epsilon }{I_0}\end{array}$                         (XIII)

Substitute $\left(R_1+R_2\right)$ for $\frac{\epsilon }{I_{\text{b}}}$ in equation (XIII).

    $\begin{array}{c}{R}_1+\frac{R_2}{2}=\frac{\epsilon }{I_{\text{a}}}+\left(R_1+R_2\right)-\frac{\epsilon }{I_0}\\ 2R_2-R_2=2\epsilon \left(\frac{1}{I_0}-\frac{1}{I_{\text{a}}}\right)\\ R_2=2\epsilon \left(\frac{1}{I_0}-\frac{1}{I_{\text{a}}}\right)\end{array}$                         (XIV)

Therefore, the resistance $R_2$ for the given circuit is $2\epsilon \left(\frac{1}{I_0}-\frac{1}{I_{\text{a}}}\right)$.

(c)

To determine

The resistance $R_3$ for the given circuit

Answer to Problem 12P

The resistance $R_3$ for the given circuit is $\epsilon \left(\frac{1}{I_{\text{0}}}-\frac{1}{I_{\text{b}}}\right)$.

Explanation of Solution

Conclusion:

Substitute $R_1+R_2$ for $R_{\text{S}}{}^{\prime }$ equation (X).

    $\begin{array}{c}\epsilon =I_{\text{b}}\left(R_1+R_2\right)\\ \left(R_1+R_2\right)=\frac{\epsilon }{I_{\text{b}}}\end{array}$                               (XI)

Substitute $\frac{\epsilon }{I_{\text{b}}}$ for $\left(R_1+R_2\right)$ in equation (III).

    $\begin{array}{l}\left(\frac{\epsilon }{I_{\text{b}}}+R_3\right)=\frac{\epsilon }{I_{\text{0}}}\\ R_3=\epsilon \left(\frac{1}{I_{\text{0}}}-\frac{1}{I_{\text{b}}}\right)\end{array}$                                  (XII)

Therefore, the resistance $R_3$ for the given circuit is $\epsilon \left(\frac{1}{I_{\text{0}}}-\frac{1}{I_{\text{b}}}\right)$.