Chapter 23, Problem 60A

(a)

To determine

The potential difference across the battery.

Answer to Problem 60A

The potential difference across the battery is $20\text{ V}$ .

Explanation of Solution

Given:

The given circuit is shown below.

The current in ammeter 3 is $I_3=0.40\text{ A}$ .

The resistance is $R_1=20.0\text{Ω}$ .

The resistance is $R_2=50.0\text{Ω}$ .

The resistance is $R_3=10.0\text{Ω}$ .

Formula used:

The expression for the resistance is,

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

Here, $V$ is the voltage and $I$ is current.

Calculation:

The potential difference across the battery is,

$\begin{array}{l}{R}_2=\frac{V}{I_3}\\ V=I_3{R}_2\\ V=\left(0.40\text{ A}\right)\left(50.0\text{ Ω}\right)\\ V=20\text{ V}\end{array}$

Conclusion:

Thus, the potential difference across the battery is $20\text{ V}$ .

(b)

To determine

The current in ammeter 1.

Answer to Problem 60A

The current in ammeter 1 is $3.4\text{ A}$ .

Explanation of Solution

Given:

The given circuit is shown below.

The current in ammeter 3 is $I_3=0.40\text{ A}$ .

The resistance is $R_1=20.0\text{Ω}$ .

The resistance is $R_2=50.0\text{Ω}$ .

The resistance is $R_3=10.0\text{Ω}$ .

Formula used:

The expression for the resistance is,

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

Here, $V$ is the voltage and $I$ is current.

Calculation:

From part (a), the potential difference across the battery is $V=20\text{V}$ .

The equivalent resistance for ammeter 1 is,

$\begin{array}{l}\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\ R=\frac{1}{\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)}\\ R=\frac{1}{\left(\frac{1}{20.0\Omega }+\frac{1}{50.0\Omega }+\frac{1}{30.0\Omega }\right)}\\ R=\frac{1}{\text{0}\text{.17 }\Omega }\\ R=5.88\Omega \end{array}$

The current in ammeter 1 is

$\begin{array}{l}R=\frac{V}{I_1}\\ I_1=\frac{2.0\times {10}^1\text{ V}}{5.88\Omega }\\ I_1=3.4\text{ A}\end{array}$

Conclusion:

Thus, the current in ammeter 1 is $3.4\text{ A}$ .

(c)

To determine

The current in ammeter 2.

Answer to Problem 60A

The current in ammeter 2 is $1.0\text{ A}$ .

Explanation of Solution

Given info:

The given circuit is shown below.

The current in ammeter 3 is $I_3=0.40\text{ A}$ .

The resistance is $R_1=20.0\text{Ω}$ .

The resistance is $R_2=50.0\text{Ω}$ .

The resistance is $R_3=10.0\text{Ω}$ .

Formula used:

The expression for the resistance is,

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

Here, $V$ is the voltage and $I$ is current.

Calculation:

From part (a), the potential difference across the battery is $V=20\text{ V}$ .

The current in ammeter 2 is,

$\begin{array}{l}{R}_1=\frac{V}{I_2}\\ I_2=\frac{2.0\times {10}^1\text{ V}}{\text{20}\text{.0 }\Omega }\\ I_2=1.0\text{ A}\end{array}$

Conclusion:

Thus, the current in ammeter 2 is $1.0\text{ A}$ .

(c)

To determine

The current in ammeter 4.

Answer to Problem 60A

The current in ammeter 4 is $2.0\text{ A}$ .

Explanation of Solution

Given info:

The given circuit is shown below.

The current in ammeter 3 is $I_3=0.40\text{ A}$ .

The resistance is $R_1=20.0\text{Ω}$ .

The resistance is $R_2=50.0\text{Ω}$ .

The resistance is $R_3=10.0\text{Ω}$ .

Formula used:

The expression for the resistance is,

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

Here, $V$ is the voltage and $I$ is current.

Calculation:

From part (a), the potential difference across the battery is $V=20\text{ V}$ .

The current in ammeter 4 is,

$\begin{array}{l}{R}_3=\frac{V}{I_4}\\ I_4=\frac{2.0\times {10}^1\text{ V}}{\text{10}\text{.0 }\Omega }\\ I_4=2.0\text{ A}\end{array}$

Conclusion:

Thus, the current in ammeter 4 is $2.0\text{ A}$ .