Chapter 19, Problem 18P

To determine

The net resistance of the parallel combination of bulbs.

Answer to Problem 18P

The net resistance of parallel combination of the bulbs is $105.26\Omega$ .

Explanation of Solution

Given info:

Power of first bulb, $P_1=75\text{W}$

Power of second bulb, $P_2=40\text{W}$

Voltage of the supply, $V=110\text{V}$

Formula Used:

The expression to calculate the resistance of the first bulb is,

  $R_1=\frac{V^2}{P_1}$

The expression to calculate the resistance of the second bulb is,

  $R_2=\frac{V^2}{P_2}$

The expression to calculate the net resistance is,

  $R=\frac{R_1{R}_2}{R_1+R_2}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{R}_1=\frac{{\left(110\text{V}\right)}^2}{75\text{W}}\\ =161.33\Omega \end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{R}_2=\frac{{\left(110\text{V}\right)}^2}{40\text{W}}\\ =302.5\Omega \end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}R=\frac{\left(161.33\Omega \right)\left(302.5\Omega \right)}{161.33\Omega +302.5\Omega }\\ =105.26\Omega \end{array}$

Conclusion:

Therefore, the net resistance of parallel combination of the bulbs is $105.26\Omega$ .