Chapter 17, Problem 31E

(a)

To determine

The current in the circuit.

Answer to Problem 31E

The current in the circuit is $\underset{\_}{0.4\text{A}}$.

Explanation of Solution

The circuit is given in Figure 17.24. The resistances are in series and the battery are connected in such a way that the polarities are opposite to each other.

The net potential drop in the circuit can be computed as,

  $\epsilon ={\epsilon }_1-{\epsilon }_2$        (I)

Here, $\epsilon$ is the net potential drop, ${\epsilon }_1$ is the potential difference of the battery in the upper portion of the loop, and ${\epsilon }_2$ is the potential difference of the battery in the bottom portion of the loop.

Since the resistors are connected in series, the net resistance of the circuit is obtained as,

  $R=R_1+R_2$        (II)

Here, $R$ is the net resistance, and $R_1$, $R_2$ are the individual resistances.

Write the expression for the current according to Ohm’s law.

  $I=\frac{\epsilon }{R}$        (III)

Conclusion:

Substitute $12\text{V}$ for ${\epsilon }_1$, and $8\text{V}$ for ${\epsilon }_2$ in equation (I) to find $\epsilon$.

  $\begin{array}{c}\epsilon =12\text{V}-8\text{V}\\ =4\text{V}\end{array}$

Substitute $6\text{ohm}$ for $R_1$, and $4\text{ohm}$ for $R_2$ in equation (II) to find $R$.

  $\begin{array}{c}R=6\text{ohm}+4\text{ohm}\\ =10\text{ohm}\end{array}$

Substitute $4\text{V}$ for $\epsilon$, and $10\text{ohm}$ for $R$ in equation (III) to find $I$.

  $\begin{array}{c}I=\frac{4\text{V}}{10\text{ohm}}\\ =0.4\text{A}\end{array}$

Therefore, the current in the circuit is $\underset{\_}{0.4\text{A}}$.

(b)

To determine

The power supplied by each battery.

Answer to Problem 31E

The power supplied by the $12\text{V}$ battery is $\underset{\_}{4.8\text{W}}$ and that by the $8\text{V}$ battery is $\underset{\_}{-3.2\text{W}}$.

Explanation of Solution

Write the expression for power supplied by the first battery.

  $P_1=I{\epsilon }_1$        (IV)

Here, $P_1$ is the power supplied by the first battery.

Similarly, the power supplied by the second battery can be computed as,

  $P_2=I{\epsilon }_2$        (V)

Here, $P_2$ is the power supplied by the first battery.

Conclusion:

Substitute $0.4\text{A}$ for $I$, and $12\text{V}$ for ${\epsilon }_1$ in equation (IV) to find $P_1$.

  $\begin{array}{c}{P}_1=\left(0.4\text{A}\right)\left(12\text{V}\right)\\ =4.8\text{W}\end{array}$

Substitute $0.4\text{A}$ for $I$, and $-8\text{V}$ for ${\epsilon }_2$ in equation (V) to find $P_2$ (the negative sign for ${\epsilon }_2$ is because the polarity of ${\epsilon }_2$ is opposite to that of ${\epsilon }_1$.

  $\begin{array}{c}{P}_2=\left(0.4\text{A}\right)\left(-8\text{V}\right)\\ =-3.2\text{W}\end{array}$

Therefore, the power supplied by the $12\text{V}$ battery is $\underset{\_}{4.8\text{W}}$ and that by the $8\text{V}$ battery is $\underset{\_}{-3.2\text{W}}$.

(c)

To determine

The power dissipated in each resistor.

Answer to Problem 31E

The power dissipated in $6\text{ohm}$ resistor is $\underset{\_}{0.96\text{W}}$, and that in $4\text{ohm}$ resistor is $\underset{\_}{0.64\text{W}}$.

Explanation of Solution

Write the expression for power dissipated in the first resistor.

  $P_{R1}=I^2{R}_1$        (VI)

Here, $P_{R1}$ is the power dissipated in the first resistor.

Write the expression for power dissipated in the second resistor.

  $P_{R2}=I^2{R}_2$        (VII)

Here, $P_{R2}$ is the power dissipated in the second resistor.

Conclusion:

Substitute $0.4\text{A}$ for $I$, and $6\text{ohm}$ for $R_1$ in equation (VI) to find $P_{R1}$.

  $\begin{array}{c}{P}_{R1}={\left(0.4\text{A}\right)}^2\left(6\text{ohm}\right)\\ =0.96\text{W}\end{array}$

Substitute $0.4\text{A}$ for $I$, and $4\text{ohm}$ for $R_2$ in equation (VII) to find $P_{R2}$.

  $\begin{array}{c}{P}_{R2}={\left(0.4\text{A}\right)}^2\left(4\text{ohm}\right)\\ =0.64\text{W}\end{array}$

Therefore, the power dissipated in $6\text{ohm}$ resistor is $\underset{\_}{0.96\text{W}}$, and that in $4\text{ohm}$ resistor is $\underset{\_}{0.64\text{W}}$.