Chapter 22, Problem 69A

a.

To determine

The resistance of the lamp when it is on.

Answer to Problem 69A

  $300\Omega$

Explanation of Solution

Given:

The current through the lamp, $I=0.40\text{A}$

The voltage across the lamp, $V=120\text{V}$

Formula used:

The electric resistance is given by,

  $R=\frac{V}{I}$

Where, I is the current and V is the potential difference.

Calculation:

Using the above formula and the given values, the resistance of the lamp will be,

  $\begin{array}{c}R=\frac{V}{I}\\ =\frac{120}{0.40}\\ =300\Omega \end{array}$

Conclusion:

Hence, theresistance of the lamp is $300\Omega$ .

b.

To determine

The lamp’s cold resistance.

Answer to Problem 69A

  $60\Omega$

Explanation of Solution

Given:

From part (a), theresistance of the lamp, $R=300\Omega$

The cold resistance, $R_c=\frac{1}{5}\text{R}$

Here, R is the resistance when the lamp is hot.

Calculation:

Using the above formula and the given values, the cold resistance of the lamp will be,

  $\begin{array}{c}{R}_c=\frac{1}{5}\left(R\right)\\ =\frac{1}{5}\left(300\Omega \right)\\ =60\Omega \end{array}$

Conclusion:

Hence, thecold resistance of the lamp is $60\Omega$ .

c.

To determine

The current through the lamp as it is turned on.

Answer to Problem 69A

  $2\text{A}$

Explanation of Solution

Given:

From part (b), the cold resistance of the lamp, $R_c=60\Omega$

The voltage across the lamp, $V=120\text{V}$

Formula used:

The electric resistance is given by,

  $R=\frac{V}{I}$

Where, I is the current and V is the potential difference.

Calculation:

When the lamp is turned on, the lamp will have cold resistance. Then, using the above formula and the given values, the current through the lamp as it is turned onwill be,

  $\begin{array}{c}I\text{'}=\frac{V}{R_c}\\ =\frac{120}{60}\\ =2\text{A}\end{array}$

Conclusion:

Hence, the current through the lamp as it is turned on is $2\text{A}$ .