Chapter 18, Problem 31P

(a)

To determine

Setting of a higher resistance.

Answer to Problem 31P

The answer is $850W$ .

Explanation of Solution

Given:

A $120V$ hair dryer has two settings: $850W,1250W$

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:

The resistance is inversely proportional to the power, a higher power setting to a lower resistance. According to this, resistance will be higher at $850W$ and lower at $1250W$ .

Conclusion:

Setting of a higher resistance: $850W$

(b)

To determine

Resistance at the lower setting.

Answer to Problem 31P

The answer is $16.94\Omega$ .

Explanation of Solution

Given:

A $120V$ hair dryer has two settings: $850W,1250W$

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:

The resistance is inversely proportional to the power, a higher power setting to a lower resistance. According to this, resistance will be higher at $850W$ and lower at $1250W$ .

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

  $R=\frac{{\left(120V\right)}^2}{850W}=\frac{14400V}{850W}=16.94\Omega$

Conclusion:

Resistance at the lower setting: $16.94\Omega$

(c)

To determine

Resistance at the higher setting.

Answer to Problem 31P

The answer is $11.52\Omega$ .

Explanation of Solution

Given:

A $120V$ hair dryer has two settings: $850W,1250W$ .

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:

The resistance is inversely proportional to the power, a higher power setting to a lower resistance. So, the lower power setting $850W$ corresponds to a higher resistance.

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

  $R=\frac{{\left(120V\right)}^2}{1250W}=\frac{14400V}{1250W}=11.52\Omega$

Conclusion:

Resistance at the higher setting: $11.52\Omega$