Chapter 22.1, Problem 3PP

To determine

Rate of transformation of electrical energy to light.

Answer to Problem 3PP

  $P=62.5\text{W or 62}\text{.5}\text{J/s}$

Explanation of Solution

Given:

Applied voltage, $V=125\text{V}$

Current produced, $I=0.5\text{A}$

Formula Used:

The electrical power is given as,

  $P=VI$

Where, $P$ is electrical power, $V$ is applied voltage and $I$ is current produced.

Calculation:

Since electrical power is rate of transformation of electrical energy.

So, on substituting the given values in above relation,

  $\begin{array}{c}P=125\times 0.5\\ =62.5\text{W}\\ \text{= }62.5\text{J/s}\end{array}$

Conclusion:

Therefore, the rate of transformation of electrical energy to light by lightbulb is $P=62.5\text{W or 62}\text{.5}\text{J/s}$ .