Chapter 23, Problem 59A

(a)

To determine

The hottest resistor in the circuit.

Answer to Problem 59A

The hottest resistor in the circuit having the resistance of $10.0\Omega$ .

Explanation of Solution

Given:

The given circuit is shown below.

The voltage is $V=110\text{ V}$ .

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 power is,

$P=\frac{V^2}{R}$

Here, $V$ is the voltage and $R$ is the resistance.

Calculation:

The voltage $V$ is constant in parallel circuit.

The power of $20.0\Omega$ is,

$\begin{array}{l}P=\frac{V^2}{R}\\ P_{20.0\Omega }=\frac{{\left(110\text{ V}\right)}^2}{20.0\Omega }\\ P_{20.0\Omega }=6.05\times {10}^2\text{ W}\end{array}$

The power of $50.0\Omega$ is,

$\begin{array}{l}P=\frac{V^2}{R}\\ P_{50.0\Omega }=\frac{{\left(110\text{ V}\right)}^2}{50.0\Omega }\\ P_{50.0\Omega }=2.42\times {10}^2\text{ W}\end{array}$

The power of $10.0\Omega$ is,

$\begin{array}{l}P=\frac{V^2}{R}\\ P_{10.0\Omega }=\frac{{\left(110\text{ V}\right)}^2}{10.0\Omega }\\ P_{10.0\Omega }=1.21\times {10}^3\text{ W}\end{array}$

It is observed that maximum power will developed by the smallest resistor.

Conclusion:

Thus, the hottest resistor in the circuit having the resistance of $10.0\Omega$ .

(b)

To determine

The coolest resistor in the circuit.

Answer to Problem 59A

The coolest resistor in the circuit having the resistance of $50.0\Omega$ .

Explanation of Solution

Given:

The given circuit is shown below.

The voltage is $V=110\text{ V}$ .

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 power is,

$P=\frac{V^2}{R}$

Here, $V$ is the voltage and $R$ is the resistance.

Calculation:

From part (a),

The power of $20.0\Omega$ is $P_{20.0\Omega }=6.05\times {10}^2\text{ W}$ .

The power of $50.0\Omega$ is $P_{50.0\Omega }=2.42\times {10}^2\text{ W}$ .

The power of $10.0\Omega$ is $P_{10.0\Omega }=1.21\times {10}^3\text{ W}$ .

The least power will developed by the largest resistor.

Conclusion:

Thus, the coolest resistor in the circuit having the resistance of $50.0\Omega$ .

(c)

To determine

The current in ammeter 1.

Answer to Problem 59A

The current in ammeter 1 is $19\text{ A}$ .

Explanation of Solution

Given:

The given circuit is shown below.

The voltage is $V=110\text{ V}$ .

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 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,

$\begin{array}{l}R=\frac{V}{I_1}\\ I_1=\frac{110\text{ V}}{5.88\Omega }\\ I_1=19\text{ A}\end{array}$

Conclusion:

Thus, the current in ammeter 1 is $19\text{ A}$ .

(d)

To determine

The current in ammeter 2.

Answer to Problem 59A

The current in ammeter 2 is $5.5\text{ A}$ .

Explanation of Solution

Given:

The given circuit is shown below.

The voltage is $V=110\text{ V}$ .

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 current in ammeter 2 is,

$\begin{array}{l}{R}_1=\frac{V}{I_2}\\ I_2=\frac{110\text{ V}}{\text{20}\text{.0 }\Omega }\\ I_2=5.5\text{ A}\end{array}$

Conclusion:

Thus, the current in ammeter 2 is $5.5\text{ A}$ .

(e)

To determine

The current in ammeter 3.

Answer to Problem 59A

The current in ammeter 3 is $2.2\text{ A}$ .

Explanation of Solution

Given:

The given circuit is shown below.

The voltage is $V=110\text{ V}$ .

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 current in ammeter 3 is,

$\begin{array}{l}{R}_2=\frac{V}{I_3}\\ I_3=\frac{110\text{ V}}{\text{50}\text{.0 }\Omega }\\ I_3=2.2\text{ A}\end{array}$

Conclusion:

Thus, the current in ammeter 3 is $2.2\text{ A}$ .

(f)

To determine

The current in ammeter 4.

Answer to Problem 59A

The current in ammeter 4 is $11\text{ A}$ .

Explanation of Solution

Given:

The given circuit is shown below.

The voltage is $V=110\text{ V}$ .

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 current in ammeter 4 is,

$\begin{array}{l}{R}_3=\frac{V}{I_4}\\ I_4=\frac{110\text{ V}}{\text{10}\text{.0 }\Omega }\\ I_4=11\text{ A}\end{array}$

Conclusion:

Thus, the current in ammeter 4 is $11\text{ A}$ .