Chapter 29, Problem 45PQ

Figure P29.45 shows five resistors connected between terminals a and b.

1. a. What is the equivalent resistance of this combination of resistors?
2. b. What is the current through each resistor if a 24.0-V battery is connected across the terminals?

To determine

Find the equivalent resistance of the combination of the resistors.

Answer to Problem 45PQ

The equivalent resistance of the combination is $\underset{\_}{14.7\text{Ω}}$.

Explanation of Solution

Consider $R_1$ as the $20\Omega$ resistance, $R_2$ as the $\text{5}\Omega$ resistance, $R_3$ as the $\text{9}\Omega$ resistance, $R_4$ as the $2\Omega$ resistance and $R_5$ as the $12\Omega$ resistance.

Refer to figure above, the resistance $R_1$ and $R_2$ are in parallel.

Write the equivalent resistance across the two resistances as.

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

Here, $R_{eq}$ is the equivalent parallel resistance of $R_1$ and $R_2$.

Further, the resistances $R_4$ and $R_5$ are in parallel.

Write the equivalent resistance across the two resistances as.

  $\frac{1}{{R^{\prime }}_{\text{eq}}}=\frac{1}{R_4}+\frac{1}{R_5}$                                                                                     (II)

Here, ${R^{\prime }}_{\text{eq}}$ is the equivalent parallel resistance of $R_4$ and $R_5$.

The circuit has $R_{eq}$ , $R_3$ and ${R^{\prime }}_{\text{eq}}$ are three resistances connected in series.

Write the equivalent resistance of the circuit as.

  $R_{total}=R_{eq}+R_3+{R^{\prime }}_{\text{eq}}$                                                                                   (III)

Here, $R_{total}$ is the equivalent series resistance of circuit.

Conclusion:

Substitute $20\Omega$ for $R_1$ and $5\Omega$ for $R_2$ in equation (I).

  $\begin{array}{l}\frac{1}{R_{eq}}=\frac{1}{20\Omega }+\frac{1}{5\Omega }\\ \frac{1}{R_{eq}}=\frac{25}{100}\Omega \end{array}$

Rearrange the terms in above equation

  $R_{eq}=4\Omega$

Substitute $2\Omega$ as $R_4$ and $12\Omega$ as $R_5$ in equation (II).

  $\begin{array}{l}\frac{1}{{R^{\prime }}_{\text{eq}}}=\frac{1}{2\Omega }+\frac{1}{12\Omega }\\ \frac{1}{{R^{\prime }}_{\text{eq}}}=\frac{14}{24}\Omega \end{array}$

Rearrange the terms in above equation

  ${R^{\prime }}_{\text{eq}}=1.7\Omega$

Substitute $4\Omega$ as $R_{eq}$, $9\Omega$ as $R_3$ and $1.7\Omega$ as ${R^{\prime }}_{\text{eq}}$ in equation (III).

  $\begin{array}{c}{R}_{total}=4\Omega +9\Omega +1.7\Omega \\ =14.7\Omega \end{array}$

Thus, the equivalent resistance of the combination is $\underset{\_}{14.7\text{Ω}}$.

To determine

Find the current in each resistor in the circuit.

Answer to Problem 45PQ

The current flowing through $20\Omega$ resistor is $\underset{\_}{0.326\text{A}}$, the current flowing through $5\Omega$ resistor is $\underset{\_}{1.30\text{A}}$, the current flowing through $9\Omega$ resistor is $\underset{\_}{1.63\text{A}}$, the current flowing through $2\Omega$ resistor is $\underset{\_}{1.40\text{A}}$ and the current flowing through $12\Omega$ resistor is $\underset{\_}{0.233\text{A}}$.

Explanation of Solution

Write the expression for the current drawn by the circuit from the battery as.

  $I=\frac{\epsilon }{R_{total}}$                                                                                                      (IV)

Here, $I$ is the current drawn, $\epsilon$ is the Emf of the battery and $R_{total}$ is the equivalent resistance of circuit.

Write the expression for the voltage drop across $R_1$ and $R_2$ as.

  $V_{R_{eq}}=I{R}_{eq}$                                                                                                     (V)

Here $V_{R_{eq}}$ is the voltage drop across $R_{eq}$.

Write the expression for the voltage drop across $R_4$ and $R_5$ as.

  $V_{R_{e{q}^{\text{'}}}}=I{R^{\prime }}_{\text{eq}}$                                                                                                   (VI)

Here $V_{R_{e{q}^{\text{'}}}}$ is the voltage drop across ${R^{\prime }}_{\text{eq}}$.

In a parallel connection, the voltage drop across all the elements remains same. Therefore the voltage drop across $R_1$ and $R_2$ is same as $V_{R_{eq}}$ and the voltage drop across $R_4$ and $R_5$ is same as $V_{{R^{\prime }}_{\text{eq}}}$.

Write the expression for the current through $R_1$ and $R_2$ as.

  $I^{\text{'}}=\frac{V_{R_{eq}}}{R}$                                                                                                     (VII)

Here $I^{\text{'}}$ is the current through respective resistor.

Write the expression for the current through $R_4$ and $R_5$ as.

  $I\text{'}\text{'}=\frac{V_{{R^{\prime }}_{\text{eq}}}}{R}$                                                                                                   (VIII)

Here $I^{\text{'}\text{'}}$ is the current through respective resistor and is the value of respective resistance.

The resistance $R_3$ is connected in series to the circuit so the current flowing through the circuit will be the same as the current through $R_{eq}$.

Therefore, the current through the resistance $R_3$ is same as $I$ (current drawn from battery).

Conclusion:

Substitute $24\text{V}$ for $\epsilon$ and $14.7\Omega$ for $R_{total}$ in equation (IV).

  $\begin{array}{c}I=\frac{24\text{V}}{14.7\Omega }\\ =1.63\text{A}\end{array}$

Substitute $1.63\text{A}$ for $I$ and $4\Omega$ for $R_{eq}$ in the equation (V).

  $\begin{array}{c}{V}_{R_{eq}}=\left(1.63\text{A}\right)\left(4\Omega \right)\\ =6.52\text{V}\end{array}$

Substitute $1.63\text{A}$ for $I$ and $1.7\Omega$ for ${R^{\prime }}_{\text{eq}}$ in equation (VI).

    $\begin{array}{c}{V}_{R_{eq}{}^{\text{'}}}=\left(1.63\text{A}\right)\left(1.7\Omega \right)\\ =2.77\Omega \\ \approx 2.8\Omega \end{array}$

Substitute $I_1$ for $I^{\text{'}}$, $6.52\text{V}$ for $V_{R_{eq}}$ and $20\Omega$ for $R$ in equation (VII).

  $\begin{array}{c}{I}_1=\frac{6.52\text{V}}{20\Omega }\\ =0.326\text{A}\end{array}$

Substitute $I_2$ for $I^{\text{'}}$, $6.52\text{V}$ for $V_{R_{eq}}$ and $5\Omega$ for $R_{}$ in equation (IX).

  $\begin{array}{c}{I}_2=\frac{6.52\text{V}}{5\Omega }\\ =1.30\text{A}\end{array}$

The current through $R_3$ is same as $I$ .

  $I_3=1.63\text{A}$

Substitute $I_4$ for $I^{\text{'}\text{'}}$, $2.78\text{V}$ for $V_{{R^{\prime }}_{\text{eq}}}$ and $2\Omega$ for $R_{}$ in equation (VIII).

  $\begin{array}{c}{I}_4=\frac{2.8\text{V}}{2\Omega }\\ =1.40\text{A}\end{array}$

Substitute $I_5$ for $I^{\text{'}\text{'}}$, $2.78\text{V}$ for $V_{{R^{\prime }}_{\text{eq}}}$ and $12\Omega$ for $R_{}$ in equation (VIII).

  $\begin{array}{c}{I}_5=\frac{2.8\text{V}}{12\Omega }\\ =0.233\text{A}\end{array}$

Thus, the current flowing through $20\Omega$ resistor is $\underset{\_}{0.326\text{A}}$, the current flowing through $5\Omega$ resistor is $\underset{\_}{1.30\text{A}}$, the current flowing through $9\Omega$ resistor is $\underset{\_}{1.63\text{A}}$, the current flowing through $2\Omega$ resistor is $\underset{\_}{1.40\text{A}}$ and the current flowing through $12\Omega$ resistor is $\underset{\_}{0.233\text{A}}$.