Chapter 22, Problem 96A

a.

To determine

The current drawn when the bulb is lit.

Answer to Problem 96A

  $3\text{A}$

Explanation of Solution

Given:

The resistance of the bulb when it is not lit, $R_n=10.0\Omega$

The resistance of the bulb when it is lit, $R_l=40.0\Omega$

The voltage applied, $V=120\text{V}$

Formula used:

Ohm’s law,

  $V=IR$

Where, I is the current, V is the potential and R is the resistance.

Calculation:

Using the above formula and the value of the resistance of the bulb when it is lit, the current drawn when the bulb is lit will be,

  $\begin{array}{c}V=I_l{R}_l\\ 120=I_l\left(40\right)\\ I_l=3\text{A}\end{array}$

Conclusion:

Hence, the current drawn when the bulb is lit is $3\text{A}$ .

b.

To determine

The current drawn at the instant when the bulb is turned on.

Answer to Problem 96A

  $12\text{A}$

Explanation of Solution

Given:

The resistance of the bulb when it is not lit, $R_n=10.0\Omega$

The resistance of the bulb when it is lit, $R_l=40.0\Omega$

The voltage applied, $V=120\text{V}$

Formula used:

Ohm’s law,

  $V=IR$

Where, I is the current, V is the potential and R is the resistance.

Calculation:

When the bulb is turned on, the resistance of the bulb is the value of the resistance of the bulb when it is not lit.

Using the above formula and given values, the current drawn when the bulb turned on will be,

  $\begin{array}{c}V=I_n{R}_n\\ 120=I_n\left(10\right)\\ I_n=12\text{A}\end{array}$

Conclusion:

Hence, the current drawn when the bulb is turned on is $12\text{A}$ .

c.

To determine

The case when the bulb used the maximum power.

Answer to Problem 96A

When it is turned on

Explanation of Solution

Introduction:

The electric power $\left(P\right)$ across a circuit is given by,

  $P=IV$

Where, I is the current andV is the potential.

As can be seen from above formula, the power consumed is directly proportional to the current through the component. And,

From part (a), the current drawn when the bulb is lit, $I_l=3\text{A}$

From part (b), the current drawn when the bulb is turned on, $I_n=12\text{A}$

Conclusion:

Hence, the bulb will use more power when it is turned on.