Chapter 18, Problem 29P

(a)

To determine

The resistance and current through a light bulb.

Answer to Problem 29P

  $R=192\Omega$ and $I=0.625A$

Explanation of Solution

Given:

Voltage, $V=120V$

Power, $P=75W$

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.

Ohm’s Law: $I=\frac{V}{R}$

Where,

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

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

  $R=\frac{{\left(120V\right)}^2}{75W}=\frac{14,400V}{75W}=192\Omega$

To calculate current, substitute the given values in the formula, $I=\frac{V}{R}$

  $I=\frac{120V}{192\Omega }=0.625A$

Conclusion:

The resistance is $R=192\Omega$ and the current is $I=0.625A$ .

(b)

To determine

The resistance and current through light bulb .

Answer to Problem 29P

The answers are $R=32.7\Omega$ and $I=3.67A$

Explanation of Solution

Given:

Voltage, $V=120V$

Power, $P=440W$

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.

Ohm’s Law: $I=\frac{V}{R}$

Where,

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

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

  $R=\frac{{\left(120V\right)}^2}{440W}=\frac{14,400V}{440W}=32.7\Omega$

To calculate current, substitute the given values in the formula, $I=\frac{V}{R}$

  $I=\frac{120V}{32.7\Omega }=3.67A$

Conclusion:

The resistance is $R=32.7\Omega$ and the current is $I=3.67A$ .