Chapter 18, Problem 30P

(a)

To determine

The current through the heater when it is operating.

Answer to Problem 30P

The answer is $0.956A$ .

Explanation of Solution

Given:

Voltage, $V=115V$

Power, $P=110W$

Formula used:

Expression for the power in terms of current and voltage:

  $P=IV$

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

Calculation:

Rearrange the expression, $P=IV$

Therefore, a new expression, $I=\frac{P}{V}$

Substitute the given values in the formula, $I=\frac{P}{V}$

  $I=\frac{110W}{115V}=0.956A$

Conclusion:

The current through the heater when it is operating is $0.956A$

(b)

To determine

The resistance through the heater when it is operating.

Answer to Problem 30P

The answer is $120\Omega$ .

Explanation of Solution

Given:

Voltage, $V=115V$

Power, $P=110W$

Formula used:

Expression for the power in terms of resistance and voltage:

  $P=\frac{V^2}{R}$

Where, $R$ is resistance, $P$ is power and $V$ is voltage.

Calculation:

Rearrange the expression, $P=\frac{V^2}{R}$

Therefore, a new expression, $R=\frac{V^2}{P}$

To calculate resistance, substitute the given values in the formula, $R=\frac{V^2}{P}$

  $R=\frac{{\left(115V\right)}^2}{110W}=\frac{13225V}{110W}=120.22\Omega \approx 120\Omega$

Conclusion:

The resistance through the heater when it is operating: $120\Omega$