Chapter 29, Problem 44PQ

(a)

To determine

The power absorbed by each resistor when switch is open.

Answer to Problem 44PQ

The power absorbed by resistor $R_1$ is $\underset{\_}{0.265\text{W}}$, $R_2$ is $\underset{\_}{0.531\text{W}}$, $R_3$ is $\underset{\_}{0.796\text{W}}$, and $R_4$ is $\underset{\_}{0\text{W}}$.

Explanation of Solution

When switch is open, battery ${\epsilon }_1$ does not supply voltage to the circuit. Only battery ${\epsilon }_2$ will supply power to the resistors $R_1$, $R_2$ and $R_3$.

Write the expression for total current passes through the loop 2 as.

    $I=\frac{{\epsilon }_2}{R_{eq}}$                                                                                                          (I)

Here, $I$ is the current through loop 2, ${\epsilon }_2$ is Emf of the device in loop 2 and $R_{eq}$ is the equivalent resistance of the loop 2.

Write the expression for equivalent resistance of the loop 2 as.

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

Here, $R_1$, $R_2$ and $R_3$ are the resistance of the three elements.

Write the expression for the power-absorbed in resistor as.

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

Here, $P$ is the power absorbed by each resistor and $R$ is the circuit element.

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 $R_1$.

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 $R_2$.

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 $R_3$.

Substitute $P_4$ for $P$ and $R_4$ for $R$ in equation (III)

    $P_4=I^2{R}_4$                                                                                                  (VII)

Here, $P_2$ is power absorbed by resistor $R_2$.

Conclusion:

Substitute $15.0\Omega$ for $R_1$, $30.0\Omega$ for $R_2$ and $45.0\Omega$ for $R_3$ in equation (II).

  $\begin{array}{c}{R}_{eq}=15.0\Omega \text{+30}\text{.0 }\Omega \text{+45}\text{.0 }\Omega \\ =90.0\Omega \end{array}$

Substitute $90.0\Omega$ for $R_{eq}$ and $12.0\text{ V}$ for ${\epsilon }_2$ in equation (I).

    $\begin{array}{c}I=\frac{12.0\text{ V}}{90.0\Omega }\\ =0.133\text{ A}\end{array}$

Substitute $0.133\text{ A}$ for $I$ and $15.0\Omega$ for $R_1$ in equation (IV).

    $\begin{array}{c}{P}_1={\left(0.133\text{ A}\right)}^2\times 15.0\Omega \\ =0.265\text{ W}\end{array}$

Substitute $0.133\text{ A}$ for $I$ and $30.0\Omega$ for $R_2$ in equation (V).

    $\begin{array}{c}{P}_2={\left(0.133\text{ A}\right)}^2\times 30.0\Omega \\ =0.5306\text{ W}\\ \approx \text{0}\text{.531}\text{W}\end{array}$

Substitute $0.133\text{ A}$ for $I$ and $45.0\Omega$ for $R_3$ in equation (VI).

    $\begin{array}{c}{P}_3={\left(0.133\text{ A}\right)}^2\times 45.0\Omega \\ =0.7960\text{ W}\\ \approx \text{0}\text{.796}\text{W}\end{array}$

Substitute $0\text{ A}$ for $I$ and $\text{60}\text{.0 }\Omega$ for $R_4$ in equation (VII).

    $\begin{array}{c}{P}_4={\left(0\text{A}\right)}^2\times 60.0\Omega \\ =0\text{W}\end{array}$

Thus, the power absorbed by resistor $R_1$ is $\underset{\_}{0.265\text{ W}}$, $R_2$ is $\underset{\_}{0.5306\text{ W}}$, $R_3$ is $\underset{\_}{0.7960\text{ W}}$ and $R_4$ is $\underset{\_}{0\text{W}}$.

(b)

To determine

The power absorbed by each resistor when switch is closed.

Answer to Problem 44PQ

When switch is closed, the power absorbed by resistor $R_1$ is $\underset{\_}{0.0710\text{ W}}$, $R_2$ is $\underset{\_}{\text{0}\text{.461 W}}$, $R_3$ is $\underset{\_}{0.692\text{ W}}$ and $R_4$ is $\underset{\_}{\text{0}\text{.183 W}}$.

Explanation of Solution

When switch is closed, battery ${\epsilon }_1$ and ${\epsilon }_2$ start to supply voltage to the circuit, resistor $R_2$ and $R_3$ are in series. So, equivalent circuit can be redrawn as.

Refer to above figure, apply Kirchhoff’s voltage law in the first loop.

Write the expression for current as.

    $-{\epsilon }_1+R_1(I_1+I_2)+R_4{I}_1=0$

Simplify the above equation as.

  $I_1\left(R_1+R_4\right)+I_2{R}_1={\epsilon }_1$                                                                                (VII)

Here, $I_1$ is current flowing in loop 1, ${\epsilon }_1$ is device emf of battery in the loop 1, and $I_2$ is current flowing in loop 2.

Refer to above figure, apply Kirchhoff’s voltage law in the loop 2.

Write the expression for current as.

    $-{\epsilon }_2+R_3{I}_2+R_1(I_1+I_2)+R_2{I}_2=0$

Simplify the above equation as.

  $I_2\left(R_1+R_2+R_3\right)+I_1{R}_1={\epsilon }_2$                                                                      (VIII)

Here, ${\epsilon }_2$ is device emf of battery in the loop 2.

Write the expression for power absorbed by resistor $R_4$ as.

    $P_4=I_4^2{R}_4$                                                                                                    (IX)

Here, $P_4$ is power absorbed in resistor $R_4$.

Write the expression for power absorbed in resistor $R_1$ as.

    $P_1^{\text{'}}={\left(I_1-I_2\right)}^2{R}_1$                                                                                         (X)

Here, $P_1^{\text{'}}$ is power absorbed by resistor $R_1$ when switch is closed.

Write the expression for power absorbed by resistor $R_2$ as.

    $P_2^{\text{'}}=I_2^2{R}_2$                                                                                                     (XI)

Here, $P_2^{\text{'}}$ is power absorbed by resistor $R_2$ when switch is closed.

Write the expression for power absorbed by resistor $R_3$ as.

    $P_3^{\text{'}}=I_3^2{R}_3$                                                                                                    (XII)

Here, $P_3^{\text{'}}$ is power absorbed by resistor $R_3$ when switch is closed.

Conclusion:

Substitute $15.0\Omega$ for $R_1$, $60.0\Omega$ for $R_4$ and $6.00\text{ V}$ for ${\epsilon }_1$ in equation (VII).

    $\begin{array}{c}{I}_1\left(15.0\Omega +60.0\Omega \right)+I_2\times 15.0\Omega =6.00\text{ V}\\ I_1\times 75.0\Omega +I_{2}15.0\Omega \text{=6}\text{.00 V}\end{array}$

Rearrange the above equation as.

    $75I_1+15I_2=6\text{ V}$                                                                                     (XIII)

Substitute $15.0\Omega$ for $R_1$, $30.0\Omega$ for $R_2$, $45.0\Omega$ for $R_3$ and $12.0\text{ V}$ in equation (VIII)

    $\begin{array}{c}{I}_2\left(15.0\Omega +30.0\Omega +45.0\Omega \right)+I_1\times 15.0\Omega =12.0\text{V}\\ I_2\times 90\Omega \text{+}{I}_1\times 15.0\Omega =12.0\text{ V}\end{array}$

Rearrange the above equation as.

    $15I_1+90I_2=12$                                                                                        (XIV)

Simplify and solve equation (XIII) and equation (XIV) as.

    $I_4=0.0552\text{ A}$

The current in the loop 2 is given as.

  $I_2=I_3=0.124\text{ A}$

Substitute $60.0\Omega$ for $R_4$ and $0.0552\text{ A}$ for $I_2$ in equation (IX).

    $\begin{array}{c}{P}_4={\left(0.0552\text{A}\right)}^2\times 60.0\Omega \\ =0.183\text{W}\end{array}$

Substitute $15.0\Omega$ for $R_1$, $0.0552\text{ A}$ for $I_1$ and $0.1241\text{ A}$ for $I_2$ in equation (X)

    $\begin{array}{c}{P}_1^{\text{'}}={\left(0.1241\text{ A}-0.0552\text{ A}\right)}^2\times 15.0\Omega \\ ={\left(0.0688\text{ A}\right)}^2\times 15.0\Omega \\ =0.0710\text{ W}\end{array}$

Substitute $0.1241\text{ A}$ for $I_2$ and $30.0\Omega$ for $R_2$ in equation (XI)

    $\begin{array}{c}{P}_2^{\text{'}}={\left(0.1241\text{ A}\right)}^2\times 30.0\Omega \\ =0.461\text{W}\end{array}$

Substitute $0.1241\text{ A}$ for $I_2$ and $45.0\Omega$ for $R_3$ in equation (XII)

    $\begin{array}{c}{P}_3^{\text{'}}={\left(0.1241\text{ A}\right)}^2\times 45.0\Omega \\ =0.692\text{ W}\end{array}$

Thus, when switch is closed, the power absorbed by resistor $R_1$ is $\underset{\_}{0.0710\text{ W}}$, $R_2$ is $\underset{\_}{\text{0}\text{.461 W}}$, $R_3$ is $\underset{\_}{0.692\text{ W}}$ and $R_4$ is $\underset{\_}{\text{0}\text{.183 W}}$.