Chapter 13, Problem 5SP

A 850-W toaster, a 1200-W waffle iron, and a 1000-W coffeemaker are all connected to the same 115-V household circuit, fused at 20 A. (See figure 13.19 for a similar arrangement.)

a.    What is the current drawn by each of these appliances?

b.    If these appliances are all turned on at the same time, will there be a problem? Explain.

c.    What is the resistance of the heating element in the waffle iron?

To determine

The equivalent resistance of the two parallel segments.

Answer to Problem 5SP

The current drawn by the toaster is $7.39\text{A}$. The current drawn by waffle iron is $10.4\text{A}$. The current drawn by coffee maker is $8.70\text{A}$.

Explanation of Solution

Given info: The wattage of the toaster is $850\text{W}$. The wattage of the waffle iron is $1200\text{W}$. The wattage of the coffeemaker is $1000\text{W}$. The voltage is $115\text{V}$. The circuit is fused at $20\text{A}$.

Different instruments in parallel will draw current depending on the resistance or wattage of them.

Write the formula for the current drawn by the toaster.

$I_T=\frac{P_T}{V}$

Here,

$I_T$ is the current through toaster

$P_T$ is the power of the toaster

$V$ is voltage supply

Substitute $850\text{W}$ for $P_T$, $115\text{V}$ for $V$ to determine $I_T$.

$\begin{array}{c}{I}_T=\frac{850\text{W}}{115\text{V}}\\ =7.39\text{A}\end{array}$

Write the formula for the current drawn by the waffle iron.

$I_I=\frac{P_I}{V}$

Here,

$I_I$ is the current through waffle iron

$P_I$ is the power of the waffle iron

Substitute $1200\text{W}$ for $P_T$, $115\text{V}$ for $V$ to determine $I_T$.

$\begin{array}{c}{I}_I=\frac{1200\text{W}}{115\text{V}}\\ =10.4\text{A}\end{array}$

Write the formula for the current drawn by the coffee maker.

$I_C=\frac{P_C}{V}$

Here,

$I_C$ is the current through coffee maker

$P_C$ is the power of the coffee maker

Substitute $850\text{W}$ for $P_T$, $115\text{V}$ for $V$ to determine $I_T$.

$\begin{array}{c}{I}_C=\frac{1000\text{W}}{115\text{V}}\\ =8.70\text{A}\end{array}$

Conclusion:

The current drawn by the toaster is $7.39\text{A}$. The current drawn by waffle iron is $10.4\text{A}$. The current drawn by coffee maker is $8.70\text{A}$.

To determine

When all these appliances are on together whether there will be any problem.

Answer to Problem 5SP

When all appliances are on the circuit will get fused.

Explanation of Solution

Given info: The wattage of the toaster is $850\text{W}$. The wattage of the waffle iron is $1200\text{W}$. The wattage of the coffeemaker is $1000\text{W}$. The voltage is $115\text{V}$. The circuit is fused at $20\text{A}$.

The maximum total current that can flow in the circuit is the current at which the circuit will get fuse.

Write the formula for the total current when all the appliances are on.

$I=I_T+I_I+I_C$

Here,

$I_I$ is the current through waffle iron

$I_T$ is the current through toaster

$I_C$ is the current through coffee maker

$I$ is the total current

Substitute $7.39\text{A}$ for $I_T$, $10.4\text{A}$ for $I_I$, $8.70\text{A}$ for $I_C$ to determine $I$.

$\begin{array}{c}I=7.39\text{A}+10.4\text{A}+8.70\text{A}\\ =26.5\text{A}\end{array}$

The total current in the circuit when all appliances are on is $26.5\text{A}$ which is higher than $20\text{A}$. Thus when all appliances are on the circuit will get fused.

Conclusion:

When all appliances are on the circuit will get fused.

To determine

The resistance of the heating element of the waffle iron.

Answer to Problem 5SP

The resistance of the heating element of the waffle iron is $11\text{Ω}$.

Explanation of Solution

Given info: The wattage of the toaster is $850\text{W}$. The wattage of the waffle iron is $1200\text{W}$. The wattage of the coffeemaker is $1000\text{W}$. The voltage is $115\text{V}$. The circuit is fused at $20\text{A}$.

Write the formula for the resistance.

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

Here,

$R$ is the resistance

$I$ is the current

$P$ is the power

Substitute $1200\text{W}$ for $P$, $10.4\text{A}$ for $I$ to determine $R$.

$\begin{array}{c}R=\frac{1200\text{W}}{10.4{\text{A}}^2}\\ =11\text{Ω}\end{array}$

Conclusion:

The resistance of the heating element of the waffle iron is $11\text{Ω}$.