Chapter 22.1, Problem 17PP

a.

To determine

Current through the lamp.

Answer to Problem 17PP

  $I=0.6\text{A}$

Explanation of Solution

Given:

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

Power of lamp, $P=75\text{W}$

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:

So, on substituting the given values in above relation,

  $\begin{array}{l}75=125\times I\\ \Rightarrow I=\frac{75}{125}\text{A}\\ =\text{0}\text{.6}\text{A}\end{array}$

Conclusion:

Therefore, the current through the lamp is $I=0.6\text{A}$ .

b.

To determine

Resistance of the lamp.

Answer to Problem 17PP

  $R=208.33\Omega$

Explanation of Solution

Given:

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

Power of lamp, $P=75\text{W}$

Formula Used:

Ohm’s law is given as,

  $V=IR$

Where, $V$ is applied voltage, $I$ is current and $R$ is resistance.

Calculation:

Since, from previous sub part $I=0.6\text{A}$

On substituting the values in above relation,

  $\begin{array}{l}125=0.6\times R\\ \Rightarrow R=\frac{125}{0.6}\text{A}\\ =208.33\Omega \end{array}$

Conclusion:

Therefore, the resistance of the lamp is $R=208.33\Omega$ .