Chapter 18, Problem 44P

(a)

To determine

The resistance when every electrical thing is unplugged.

Answer to Problem 44P

  $R=\infty \Omega$

Explanation of Solution

Given info:

Root mean square Voltage, $V_{rms}=120V$

Formula used:

It is known that $V_{rms}=I_{rms}R$ , where, $V_{rms}$ is root mean square potential difference in volts, $I_{rms}$ is root mean square current in amperes, and $R$ is resistance.

Calculation:

According to the question, when everything is unplugged the zero current will pass through the home-circuit. Means that $I_{rms}=0$

So, by using the formula, $R=\frac{V_{rms}}{I_{rms}}=\frac{V_{rms}}{0}$

  $R=\infty \Omega$

Conclusion: The resistance is $R=\infty \Omega$ .

(b)

To determine

The resistance when there is a lone $75W$ light bulb is burning.

Answer to Problem 44P

The answer is $R=96\Omega$ .

Explanation of Solution

Given info:

Root mean square Voltage, $V_{rms}=120V$

Power of bulb, $P_{bulb}=75W$

Formula used:

It is known that $V_{rms}=I_{rms}R$ , where, $V_{rms}$ is root mean square potential difference in volts, $I_{rms}$ is root mean square current in amperes, and $R$ is resistance.

Calculation:

It is known that when two light bulbs are connected in series then the voltage will be the same for both. The resistance if the wires of the house is negligibledue to the resistance of light bulbs.

Now, the power of the light bulb:

  $P_{bulb}=\frac{V^2}{R}$

After rearranging this expression, $R=\frac{V^2}{P_{bulb}}$

According to the questions two lightbulbs are connected to the circuit. So, the new expression will be

  $R_{tot}=\frac{V^2}{2P_{bulb}}$

Now find the net resistance of the house-circuit by putting all the given values:

  $R_{tot}=\frac{{120}^2}{2\times 75}=96\Omega$

Conclusion: The resistance is $R=96\Omega$ .