Chapter 13, Problem 5SP

(a)

To determine

Current drawn by each of the appliances.

Answer to Problem 5SP

Current drawn by toaster is $5.22\text{A}$ , by Iron box is $10.44\text{A}$ and that by processor is $4.34\text{A}$

Explanation of Solution

Given info: Power of toaster is $600\text{W}$ , iron box is $1200\text{W}$ , processor is $500\text{W}$ , voltage supplied is $115\text{V}$

Write the equation to find the power of toaster.

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

Here,

$P_t$ is the power of toaster

$V$ is the voltage

$R$ is the resistance

Substitute $600\text{W}$ for $P_t$ and $115\text{V}$ for $V$ in equation to solve for $R$

$\begin{array}{c}R=\frac{{\left(115\text{V}\right)}^2}{600\text{W}}\\ =22.04\Omega \end{array}$

Write the equation to find the current.

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

Here,

$I$ is the current

$V$ is the voltage

$R$ is the resistance

Substitute $115\text{V}$ for $V$ and $22.04\Omega$ for $R$ to get $I$

$\begin{array}{c}I=\frac{115\text{V}}{22.04\Omega }\\ =5.22\text{A}\end{array}$

Write the equation to find the power of iron box.

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

Here,

$P_i$ is the power of iron box

$V$ is the voltage

$R$ is the resistance

Substitute $1200\text{W}$ for $P_i$ and $115\text{V}$ for $V$ in equation to solve for $R$

$\begin{array}{c}R=\frac{{\left(115\text{V}\right)}^2}{1200\text{W}}\\ =11.02\Omega \end{array}$

Write the equation to find the current.

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

Here,

$I$ is the current

$V$ is the voltage

$R$ is the resistance

Substitute $115\text{V}$ for $V$ and $11.02\Omega$ for $R$ to get $I$

$\begin{array}{c}I=\frac{115\text{V}}{11.02\Omega }\\ =10.43\text{A}\end{array}$

Write the equation to find the power of processor.

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

Here,

$P_p$ is the power of processor

$V$ is the voltage

$R$ is the resistance

Substitute $500\text{W}$ for $P_p$ and $115\text{V}$ for $V$ in equation to solve for $R$

$\begin{array}{c}R=\frac{{\left(115\text{V}\right)}^2}{500\text{W}}\\ =26.45\Omega \end{array}$

Write the equation to find the current.

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

Here,

$I$ is the current

$V$ is the voltage

$R$ is the resistance

Substitute $115\text{V}$ for $V$ and $26.45\Omega$ for $R$ to get $I$

$\begin{array}{c}I=\frac{115\text{V}}{26.45\Omega }\\ =4.34\text{A}\end{array}$

Conclusion:

Therefore, current drawn by toaster is $5.22\text{A}$ , by Iron box is $10.44\text{A}$ and that by processor is $4.34\text{A}$

(b)

To determine

Is there any problem if all these appliances are switched on?

Answer to Problem 5SP

Yes, there is problem if they are working together.

Explanation of Solution

If the three appliances are made working together then the total current delivered will be the sum of the current consumed by each of the appliance.

Write the equation to find the total current.

$I_{tot}=I_t+I_i+I_p$

Here,

$I_{tot}$ is the total current

$I_t$ is the current consumed by toaster

$I_i$ is the current consumed by iron box

$I_p$ is the current consumed by processor

Substitute $5.22\text{A}$ for $I_t$ , $10.43\text{A}$ for $I_i$ and $4.34\text{A}$ for $I_p$ to get $I_{tot}$

$\begin{array}{c}{I}_{tot}=5.44\text{A+10}\text{.02}\text{A+4}\text{.34}\text{A}\\ \text{=19}\text{.8}\text{A}\end{array}$

The maximum current allowed in the house is $15\text{A}$. If all these appliances are made working together then the circuit will get blown.

Conclusion:

Therefore, the circuit will get blown.

(c)

To determine

The resistance of the heating element.

Answer to Problem 5SP

The resistance of heating element is $11\Omega$

Explanation of Solution

Write the equation to find the power of iron box.

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

Here,

$P_i$ is the power of iron box

$V$ is the voltage

$R$ is the resistance

Substitute $1200\text{W}$ for $P_i$ and $115\text{V}$ for $V$ in equation to solve for $R$

$\begin{array}{c}R=\frac{{\left(115\text{V}\right)}^2}{1200\text{W}}\\ =11\Omega \end{array}$

Conclusion:

Therefore, the resistance of heating element is $11\Omega$