Chapter 22, Problem 60A

a.

To determine

Power rating of the bulb.

Answer to Problem 60A

  $P=4.5\text{W}$

Explanation of Solution

Given:

Source voltage, $V=3\text{V}$

Current through the bulb, $I=1.5\text{A}$

Formula used:

The power is given as,

  $P=VI$

Where, $V$ is voltage across the load and $I$ is current through the load.

Calculation:

On substituting the given valuesin above relation,

  $\begin{array}{c}P=3\times 1.5\\ =4.5\text{W}\end{array}$

Conclusion:

Therefore, the power rating of the bulb is $P=4.5\text{W}$ .

b.

To determine

Energy converted by the bulb.

Answer to Problem 60A

  $E=2.97\text{kJ}$

Explanation of Solution

Given:

Source voltage, $V=3\text{V}$

Current through the bulb, $I=1.5\text{A}$

Time, $t=11\text{min}=660\text{s}$

Formula used:

The energy power relation is,

  $E=Pt$

Where, $E$ is energy, $P$ is power and $t$ is time.

Calculation:

As, $P=4.5\text{W}$ from previous subpart.

So, on substituting the values the energy converted by bulb is

  $\begin{array}{c}E=4.5\times 660\\ =2.97\text{kJ}\end{array}$

Conclusion:

Therefore, the energy converted by bulb is $E=2.97\text{kJ}$ .