Chapter 29, Problem 40PQ

The emf in Figure P29.40 is 4.54 V. The resistances are $R_1=13.0\Omega$, $R_2=26.0\Omega$, and $R_3=39.0\Omega$ Find

1. a. the current in each resistor,
2. b. the power consumed by each resistor, and
3. c. the power supplied by the emf device.

To determine

Determine the current in each resistor.

Answer to Problem 40PQ

The current in resistor $R_1$ is $0.349\text{ A}$, The current in resistor $R_2$ is $0.069\text{A}$ and The current in resistor $R_3$ is $0.069\text{A}$.

Explanation of Solution

Refer to figure P29.40, in the given circuit, $R_1$ is parallel to the $R_2$ and $R_3$ where $R_2$ and $R_3$ is in series.

Write the expression for current in resistor $R_1$ as.

  $I_1=\frac{\epsilon }{R_1}$                                                                                                          (I)

Here, $I_1$ is current passes through $R_1$, $R_1$ is resistor and $\epsilon$ is emf of device.

Write the expression for current in $R_2$ as.

  $I_2=\frac{\epsilon }{R_2+R_3}$                                                                                                (II)

Here, $I_2$ is current passes through $R_2$, $R_2$ is the resistance and $\epsilon$ is the emf of the device.

The current through resistance $R_2$ and $R_3$ will be same.

  $I_2=I_3$

Here, $I_3$ is the current through resistance $R_3$ .

Conclusion:

Substitute $4.54\text{ V}$ for $\epsilon$ and $13.0\Omega$ for $R_1$ in equation (I).

  $\begin{array}{c}{I}_1=\frac{4.54\text{ V}}{13.0\Omega }\\ =0.349\text{ A}\end{array}$

Substitute $4.54\text{ V}$ for $\epsilon$ and $26.0\Omega$ for $R_2$ and $39.0\Omega$ for $R_3$ in equation (II).

    $\begin{array}{c}{I}_2=\frac{4.54\text{ V}}{26.0\Omega +39.0\Omega }\\ =0.069\text{A}\end{array}$

As the current $I_3$ is equal to $I_2$ therefore, the current passing through resistor $R_3$ is $0.069\text{A}$.

Therefore, the current in resistor $R_1$ is $0.349\text{ A}$, The current in resistor $R_2$ is $0.069\text{A}$ and The current in resistor $R_3$ is $0.069\text{A}$.

To determine

Power consumed by each resistor.

Answer to Problem 40PQ

Power consumed by resistor $R_1$ is $\underset{\_}{1.5925\text{ W}}$, $R_2$ is $\underset{\_}{0.1274\text{ W}}$ and $R_3$ is $\underset{\_}{0.1911\text{ W}}$.

Explanation of Solution

Write the expression for power consumed by each resistor as.

  $P=I^{2}R$                                                                                                       (III)

Here, $P$ is power, $I$ is current and $R$ is resistance.

Substitute $P_1$ for $P$ and $R_1$ for $R$ in equation (III).

    $P_1=I^2{R}_1$                                                                                                      (IV)

Here, $P_1$ is power absorbed by resistor $R_1$ and $R_1$ is resistance of first resistor.

Substitute $P_2$ for $P$ and $R_2$ for $R$ in equation (III)

    $P_2=I^2{R}_2$                                                                                                     (V)

Here, $P_2$ is power absorbed by resistor $R_2$ and $R_2$ is resistance of second resistor.

Substitute $P_3$ for $P$ and $R_3$ for $R$ in equation (III)

    $P_3=I^2{R}_3$                                                                                                     (VI)

Here, $P_3$ is power absorbed by resistor $R_3$ and $R_3$ is the resistance of third resistor.

Conclusion:

Substitute $0.349\text{ A}$ for $I_1$ and $13.0 \Omega$ for $R_1$ in equation (IV).

  $\begin{array}{c}{P}_1=I_1{}^2{R}_1\\ ={(0.349\text{ A})}^2\left(13.0\Omega \right)\\ =1.58\text{ W}\end{array}$

Substitute $0.0698\text{ A}$ for $I_2$ and $26.0 \Omega$ for $R_2$ in equation (V).

  $\begin{array}{c}{P}_2=I_2{}^2{R}_2\\ ={(0.0698\text{ A})}^2\left(26.0\Omega \right)\\ =0.127\text{ W}\end{array}$

Substitute $0.0698\text{ A}$ for $I_3$ and $39.0 \Omega$ for $R_3$ in equation (VI).

  $\begin{array}{c}{P}_3=I_3{}^2{R}_3\\ ={(0.0698\text{ A})}^2\left(39.0\Omega \right)\\ =0.190\text{ W}\end{array}$

Thus, the power consumed by resistor $R_1$ is $\underset{\_}{1.58\text{ W}}$, $R_2$ is $\underset{\_}{0.127\text{ W}}$ and $R_3$ is $\underset{\_}{0.190\text{ W}}$.

To determine

Power supplied by the Emf device.

Answer to Problem 40PQ

Power supplied by the emf device is $\underset{\_}{1.90\text{ W}}$.

Explanation of Solution

Write the expression for power drawn by the circuit as.

  $P=\frac{{\epsilon }^2}{R_{eq}}$                                                                                                      (VII)

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

Write the expression for equivalent resistance of the given circuit as.

  $R_{eq}=\frac{R_1(R_2+R_3)}{R_1+(R_2+R_3)}$

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

    $P=\frac{{\epsilon }^2}{\frac{R_1(R_2+R_3)}{R_1+(R_2+R_3)}}$

Rearrange the above expression as.

  $P=\frac{{\epsilon }^2\left(R_1+(R_2+R_3)\right)}{R_1(R_2+R_3)}$                                                                            (VIII)

Here, $R_1$, $R_2$ and $R_3$ is circuit elements of the circuit.

Conclusion:

Substitute $4.54\text{ V}$ for $\epsilon$, $13.0 \Omega$ for $R_1$, $26.0 \Omega$ for $R_2$ and $39.0 \Omega$ for $R_3$ in equation (VIII).

  $\begin{array}{c}P=\frac{{\left(4.54\text{ V}\right)}^2\left(13.0\Omega +\left(26.0\Omega +39.0\Omega \right)\right)}{13.0\Omega \left(26.0\Omega +39.0\Omega \right)}\\ =\frac{20.6116{\text{ V}}^2\left(13.0\Omega +65.0\Omega \right)}{\left(13.0\Omega \right)\left(65.0\Omega \right)}\\ =\frac{\left(20.6116{\text{ V}}^2\right)\left(78.0\Omega \right)}{845.0\Omega }\\ =1.9026\text{ W}\\ \approx 1.90\text{ W}\end{array}$

Thus, the power supplied by the emf device is $\underset{\_}{1.90\mathrm{W}}$.