Chapter 6, Problem 10PEA

To determine

Whether switching on the toaster would trip the $15\text{A}$ circuit breaker.

Answer to Problem 10PEA

Switching on the toaster will trip the circuit breaker.

Explanation of Solution

Write the expression for Ohm’s Law.

    $V=IR$

Here, $V$ is the voltage, $I$ is the electric current and $R$ is the resistance.

Rewrite the above equation for current.

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

Write the expression for power.

    $P=VI$

Here, $P$ is the power, $V$ is the voltage and $I$ is the electric current.

Rewrite the above equation to calculate the current.

    $I=\frac{P}{V}$                                                                                                              (II)

Write the expression for the total current requirement by all the equipments.

    $I=I_1+I_2+I_3+I_4$                                                                                         (III)

Here, $I$ is the total current requirement, $I_1$ is the current in the toaster, $I_2$ is the current in the motor, $I_3$ is the current in the lightbulbs, $I_4$ is the current in the electric iron.

Conclusion:

Substitute $I_1$ for $I$, $120\text{V}$ for $V$ and $15\Omega$ for $R$ in equation (I) to find the current in the toaster.

    $\begin{array}{c}{I}_1=\frac{120\text{V}}{15\Omega }\\ =8\text{A}\end{array}$

Substitute $I_2$ for $I$, $0.20\text{hp}$ for $P$ and $120\text{V}$ for $V$ in equation (II) to find the current in the motor.

    $\begin{array}{c}{I}_2=\frac{0.20\text{hp}}{120\text{V}}\\ =\frac{0.20\text{hp}\left(\frac{746\text{W}}{1\text{hp}}\right)}{120\text{V}}\\ =1.24\text{A}\end{array}$

Calculate the total power of all the lightbulbs.

    $\begin{array}{c}P=3\times 100\text{W}\\ =300\text{W}\end{array}$

Substitute $I_3$ for $I$, $300\text{W}$ for $P$ and $120\text{V}$ for $V$ in equation (II) to find the current in the lightbulbs.

    $\begin{array}{c}{I}_3=\frac{300\text{W}}{120\text{V}}\\ =2.5\text{A}\end{array}$

Substitute $I_4$ for $I$, $600\text{W}$ for $P$ and $120\text{V}$ for $V$ in equation (II) to find the current in the electric iron.

    $\begin{array}{c}{I}_4=\frac{600\text{W}}{120\text{V}}\\ =5\text{A}\end{array}$

Substitute $8\text{A}$ for $I_1$, $1.24\text{A}$ for $I_2$, $2.5\text{A}$ for $I_3$ and $5\text{A}$ for $I_4$ in equation (III) to calculate the total current requirement by all the equipments.

    $\begin{array}{c}I=8\text{A}+1.24\text{A}+2.5\text{A}+5\text{A}\\ =16.74\text{A}\end{array}$

The total current requirement by all the equipments is greater than $15.0\text{A}$, therefore it will trip the circuit breaker.