Chapter 13, Problem 3SP

In the circuit shown, the 8-V battery is opposing the 12-V battery as they are positioned. The total voltage of the two batteries will be found by subtracting.

a.    What is the current flowing around the circuit?

b.    What is the voltage difference across the 20 Ω resistor?

c.    What is the power delivered by the 12-V battery?

d.    Is the 12-V battery discharging or charging in this arrangement?

To determine

The current flowing in the circuit.

Answer to Problem 3SP

The current through the circuit is $0.057\text{A}$.

Explanation of Solution

Given info: The given circuit is shown below.

The two batteries in the given circuit have opposite polarity in the circuit. So the net potential difference of the circuit will be the difference of the two potential of the two batteries.

Write the formula for the net potential difference of the circuit.

$V=V_1-V_2$

Here,

$V$ is the net potential difference of the circuit

$V_1$ is the potential of first battery

$V_2$ is the potential of the second battery

The two resistance in the circuit are in series.

Write the formula for the equivalent resistance of the two resistances.

$R=R_1+R_2$

Here,

$R$ is the equivalent resistance

$R_1$ is the first resistance

$R_2$ is the second resistance

Write the formula for current.

$I=\frac{V}{R}$

Here,

$I$ is the current in the circuit

Substitute the expression for $R$ and $I$ in the above equation.

$I=\frac{V_1-V_2}{R_1+R_2}$

Substitute $12\text{V}$ for $V_1$, $8\text{V}$ for $V_2$, $50\text{Ω}$ for $R_1$, $20\text{Ω}$ for $R_2$ to determine $I$.

$\begin{array}{c}I=\frac{12\text{V}-8\text{V}}{50\text{Ω}+20\text{Ω}}\\ =0.057\text{A}\end{array}$

Conclusion:

The current through the circuit is $0.057\text{A}$.

To determine

The voltage difference across the $20\text{Ω}$ resistor.

Answer to Problem 3SP

The voltage difference across the $20\text{Ω}$ resistor is $1.14\text{V}$.

Explanation of Solution

Given info: The given circuit is shown below.

The two batteries in the given circuit have opposite polarity in the circuit. So the net potential difference of the circuit will be the difference of the two potential of the two batteries.

Write the formula for the voltage difference across a resistor when two resistors are in series.

$V_1=\left(\frac{R_1}{R_1+R_2}\right)V$

Here,

$V$ is the net potential difference of the circuit

$R_1$ is the first resistance

$R_2$ is the second resistance

$V_1$ is the potential across the first resistance

$V_2$ is the potential across the second resistance

Substitute $4\text{V}$ for $V$, $20\text{Ω}$ for $R_1$, $50\text{Ω}$ for $R_2$ to determine $V_1$.

$\begin{array}{c}I=\left(\frac{20\text{Ω}}{50\text{Ω}+20\text{Ω}}\right)\left(4.0\text{V}\right)\\ =1.14\text{V}\end{array}$

Conclusion:

The voltage difference across the $20\text{Ω}$ resistor is $1.14\text{V}$.

To determine

The power delivered by the battery with voltage $12\text{V}$

Answer to Problem 3SP

The power delivered by the battery with voltage $12\text{V}$ is $0.686\text{W}$.

Explanation of Solution

Given info: The given circuit is shown below.

Write the formula for the power delivered.

$P=VI$

Here,

$P$ is the power delivered

$V$ is the voltage

$I$ is the current

From section (a) the current in the circuit is $0.057\text{A}$.

Substitute $0.057\text{A}$ for $I$, $12\text{V}$ for $V$ to determine $P$.

$\begin{array}{c}P=\left(12\text{V}\right)\left(0.057\text{A}\right)\\ =0.686\text{W}\end{array}$

Conclusion:

The power delivered by the battery with voltage $12\text{V}$ is $0.686\text{W}$.

To determine

Whether the battery with voltage $12\text{V}$ is discharging or charging.

Answer to Problem 3SP

The battery with voltage $12\text{V}$ is charging.

Explanation of Solution

Given info: The given circuit is shown below.

The overall voltage of the circuit is $4\text{V}$. Current flows from the positive terminal to the negative terminal of the $12\text{V}$ battery.

Thus the battery with voltage $12\text{V}$ is charging.

Conclusion:

The battery with voltage $12\text{V}$ is charging.