Chapter 28, Problem 17P

(a)

To determine

The equivalent resistance between points $a$ and $b$.

Answer to Problem 17P

The equivalent resistance between points $a$ and $b$ is $11.7\Omega$.

Explanation of Solution

Consider the circuit diagram as shown below.

Figure-(1)

Resistors $R_1$ and $R_2$ are connected in parallel.

Write the expression to calculate the equivalent resistance for $R_1$ and $R_2$ resistors.

    $\frac{1}{R_{\text{a}}}=\frac{1}{R_1}+\frac{1}{R_2}$                                                                                                             (I)

Here, $R_{\text{a}}$ is the equivalent resistance for $R_1$ and $R_2$ resistors.

Resistors $R_4$ and $R_5$ are also connected in parallel.

Write the expression to calculate the equivalent resistance for $R_4$ and $R_5$ resistors.

    $\frac{1}{R_{\text{b}}}=\frac{1}{R_4}+\frac{1}{R_5}$                                                                                                           (II)

Here, $R_{\text{a}}$ is the equivalent resistance for $R_4$ and $R_5$ resistors.

Write the expression to calculate the equivalent resistance of the circuit.

    $R_{\text{eq}}=R_{\text{a}}+R_3+R_{\text{b}}$                                                                                                    (III)

Here, $R_{\text{eq}}$ is the equivalent resistance for the circuit.

Conclusion:

Substitute $6.00\Omega$ for $R_1$ and $12.0\Omega$ for $R_2$ in equation (I) to calculate $R_{\text{a}}$.

    $\begin{array}{c}\frac{1}{R_{\text{a}}}=\frac{1}{6.00\Omega }+\frac{1}{12.0\Omega }\\ R_{\text{a}}=\frac{\left(6.00\Omega \right)\left(12.0\Omega \right)}{6.00\Omega +12.0\Omega }\\ =\frac{72.0{\Omega }^2}{18\Omega }\\ =4\Omega \end{array}$

Substitute $4.00\Omega$ for $R_4$ and $8.00\Omega$ for $R_5$ in equation (II) to calculate $R_{\text{b}}$.

    $\begin{array}{c}\frac{1}{R_{\text{b}}}=\frac{1}{4.00\Omega }+\frac{1}{8.00\Omega }\\ R_{\text{b}}=\frac{\left(4.00\Omega \right)\left(8.00\Omega \right)}{4.00\Omega +8.00\Omega }\\ =\frac{32.0{\Omega }^2}{12\Omega }\\ =2.7\Omega \end{array}$

Substitute $4\Omega$ for $R_{\text{a}}$, $5.00\Omega$ for $R_3$ and $2.7\Omega$ for $R_{\text{b}}$ in equation (III) to calculate $R_{\text{eq}}$.

    $\begin{array}{c}{R}_{\text{eq}}=4\Omega +5.00\Omega +2.7\Omega \\ =11.7\Omega \end{array}$

Therefore, the equivalent resistance between points $a$ and $b$ is $11.7\Omega$.

(b)

To determine

The current in each resistor.

Answer to Problem 17P

The current across the resistor of $12.0\Omega$ is $1.00\text{A}$, $6.00\Omega$ is $2.00\text{A}$, $5.00\Omega$ is $3.00\text{A}$, $4.00\Omega$ is $2.00\text{A}$ and $8.00\Omega$ is $1.00\text{A}$.

Explanation of Solution

Write the expression to calculate the total current.

    $I=\frac{V}{R_{\text{eq}}}$                                                                                                                    (IV)

Here, $I$ is the current and $V$ is the voltage.

Since the resistances $R_{\text{a}}$, $R_{\text{b}}$ and $R_3$ are in series connection, the current across each resistance remains the same and equal to $I$.

Write the expression to calculate the voltage for resistance $R_{\text{a}}$.

    $V_{\text{a}}=I{R}_{\text{a}}$                                                                                                                    (V)

Here, $V_{\text{a}}$ is the voltage for resistance $R_{\text{a}}$.

Since resistors $R_1$ and $R_2$ are connected in parallel, therefore the voltage is same.

Write the expression to calculate the voltage for resistance $R_{\text{b}}$.

    $V_{\text{b}}=I{R}_{\text{b}}$                                                                                                                   (VI)

Here, $V_{\text{b}}$ is the voltage for resistance $R_{\text{b}}$.

Since resistors $R_4$ and $R_4$ are connected in parallel, therefore the voltage is same.

Write the expression to calculate the current across resistor $R_2$.

    $I=\frac{V}{R}$                                                                                                                     (VII)

Conclusion:

Substitute $35.0\text{V}$ for $V$ and $11.67\Omega$ for $R_{\text{eq}}$ in equation (IV) to calculate $I$.

    $\begin{array}{c}I=\frac{35\text{V}}{11.67\Omega }\\ =3.00\text{A}\end{array}$

Substitute $3.00\text{A}$ for $I$ and $4\Omega$ for $R_{\text{a}}$ to calculate $V_{\text{a}}$ in equation (V) to calculate $V_{\text{a}}$.

    $\begin{array}{c}{V}_{\text{a}}=\left(3.0\text{A}\right)\left(4\Omega \right)\\ =12.0\text{V}\end{array}$

Substitute $3.00\text{A}$ for $I$ and $2.67\Omega$ for $R_{\text{b}}$ to calculate $V_{\text{b}}$ in equation (VI) to calculate $V_{\text{b}}$.

    $\begin{array}{c}{V}_{\text{b}}=\left(3.0\text{A}\right)\left(2.67\Omega \right)\\ =8.0\text{V}\end{array}$

Substitute $12.0\text{V}$ for $V$ and $12.0\Omega$ for $R$ in equation (VII) to calculate $I_1$.

    $\begin{array}{c}{I}_1=\frac{12.0\text{V}}{12.0\Omega }\\ =1.00\text{A}\end{array}$

Substitute $12.0\text{V}$ for $V$ and $6.00\Omega$ for $R$ in equation (VII) to calculate $I_2$.

    $\begin{array}{c}{I}_2=\frac{12.0\text{V}}{6.00\Omega }\\ =2.00\text{A}\end{array}$

Substitute $8.0\text{V}$ for $V$ and $4.00\Omega$ for $R$ in equation (VII) to calculate $I_4$.

    $\begin{array}{c}{I}_4=\frac{8.0\text{V}}{4.00\Omega }\\ =2.00\text{A}\end{array}$

Substitute $8.0\text{V}$ for $V$ and $8.00\Omega$ for $R$ in equation (VII) to calculate $I_5$.

    $\begin{array}{c}{I}_5=\frac{8.0\text{V}}{8.00\Omega }\\ =1.00\text{A}\end{array}$

Current across resistor $R_3$.

    $I_3=3.00\text{A}$

Therefore, the current across resistor $12.0\Omega$ is $1.00\text{A}$, $6.00\Omega$ is $2.00\text{A}$, $5.00\Omega$ is $3.00\text{A}$, $4.00\Omega$ is $2.00\text{A}$ and $8.00\Omega$ is $1.00\text{A}$.