Chapter 13, Problem 9E

Three resistors are connected to a 12-V battery as shown. The internal resistance of the battery is negligible.

a.    What is the total resistance in the circuit?

b.    What is the current coming from the battery?

c.    What is the current through the 15 Ω. resistance?

d.    Does this same current flow through the 25 Ω resistance?

e.    What is the voltage difference across the 20 Ω resistance?

To determine

The total resistance of the circuit.

Answer to Problem 9E

The equivalent resistance of the circuit is $\text{60}\text{Ω}$.

Explanation of Solution

Given info: The voltage of the battery is $12\text{V}$. The three resistances connected in the circuit is $25\text{Ω}$, $20\text{Ω}$, $15\text{Ω}$.

Write the formula for the equivalent resistance when three resistances are connected in series.

$R_e=R_1+R_2+R_3$

Here,

$R_e$ is the equivalent resistance

$R_1$ is the first charge

$R_2$ is the second charge

$R_3$ is the third charge

Substitute $25\text{Ω}$ for $R_1$, $20\text{Ω}$ for $R_2$, $15\text{Ω}$ for $R_3$ to determine $R_e$.

$\begin{array}{c}{R}_e=25\text{Ω}+20\text{Ω}+15\text{Ω}\\ \text{=60}\text{Ω}\end{array}$

Conclusion:

The equivalent resistance of the circuit is $\text{60}\text{Ω}$.

To determine

The current in the circuit.

Answer to Problem 9E

The current in the circuit is $0.2\text{A}$.

Explanation of Solution

Given info: The voltage of the battery is $12\text{V}$. The three resistances connected in the circuit is $25\text{Ω}$, $20\text{Ω}$, $15\text{Ω}$.

Write the formula for the equivalent resistance when three resistances are connected in series.

$R_e=R_1+R_2+R_3$

Here,

$R_e$ is the equivalent resistance

$R_1$ is the first charge

$R_2$ is the second charge

$R_3$ is the third charge

Write the formula for the current in the circuit.

$I=\frac{V}{R_1+R_2+R_3}$

Here,

$V$ is the voltage of the circuit.

Substitute $25\text{Ω}$ for $R_1$, $20\text{Ω}$ for $R_2$, $15\text{Ω}$ for $R_3$, $12\text{V}$ for $V$ to determine $I$.

$\begin{array}{c}I=\frac{\text{12}\text{V}}{25\text{Ω}+20\text{Ω}+15\text{Ω}}\\ =0.2\text{A}\end{array}$

Conclusion:

The current in the circuit is $0.2\text{A}$.

To determine

The current through the resistor $15\text{Ω}$.

Answer to Problem 9E

The current through the $15\text{Ω}$ resistor is $0.2\text{A}$.

Explanation of Solution

Given info: The voltage of the battery is $12\text{V}$. The three resistances connected in the circuit is $25\text{Ω}$, $20\text{Ω}$, $15\text{Ω}$.

When resistors are connected in series, the path for current is same. So the current flowing through all the elements which are in series with each other is same.

In the given circuit, the three resistances are connected in series. So the current through each resistor will be same which will be equal to the current coming from the battery.

In the section (b), the current from the battery was found out to be $0.2\text{A}$. Thus the current through the $15\text{Ω}$ resistor will also be $0.2\text{A}$.

Conclusion:

The current through the $15\text{Ω}$ resistor is $0.2\text{A}$.

To determine

Whether the current flowing through $15\text{Ω}$ resistor is same as $25\text{Ω}$ resistor.

Answer to Problem 9E

The current flowing through $15\text{Ω}$ resistor is same as $25\text{Ω}$ resistor.

Explanation of Solution

Given info: The voltage of the battery is $12\text{V}$. The three resistances connected in the circuit is $25\text{Ω}$, $20\text{Ω}$, $15\text{Ω}$.

When resistors are connected in series, the path for current is same. So the current flowing through all the elements which are in series with each other is same.

In the given circuit, the three resistances are connected in series. So the current through each resistor will be same which will be equal to the current coming from the battery.

Thus the current flowing through $15\text{Ω}$ resistor is same as $25\text{Ω}$ resistor.

Conclusion:

The current flowing through $15\text{Ω}$ resistor is same as $25\text{Ω}$ resistor.

To determine

The voltage across $20\text{Ω}$.

Answer to Problem 9E

The voltage across $20\text{Ω}$ is $4\text{V}$.

Explanation of Solution

Given info: The voltage of the battery is $12\text{V}$. The three resistances connected in the circuit is $25\text{Ω}$, $20\text{Ω}$, $15\text{Ω}$.

Write the formula for the voltage across a resistor in series with two other resistors.

$V_1=\left(\frac{R_1}{R_1+R_2+R_3}\right)V$

Here,

$R_1$ is the first resistor

$R_2$ is the second resistor

$R_3$ is the third resistor

$V_1$ is the voltage across $R_1$

Substitute $20\text{Ω}$ for $R_1$, $25\text{Ω}$ for $R_2$, $15\text{Ω}$ for $R_3$, $12\text{V}$ for $V$, to determine $V_1$.

$\begin{array}{c}{V}_1=\left(\frac{20\text{Ω}}{20\text{Ω}+25\text{Ω}+15\text{Ω}}\right)\left(12\text{V}\right)\\ =4\text{V}\end{array}$

Conclusion:

The voltage across $20\text{Ω}$ is $4\text{V}$.