Chapter 19, Problem 9Q

To determine

Change in overall resistance of the room’s electric circuit when an additional 100 W bulb is switched on.

Answer to Problem 9Q

Solution:The overall resistance of the room’s electric circuit will decrease when an additional bulb of $100\text{W}$ is switched on.

Explanation of Solution

Given:

One bulb of power, $P_1=60\text{W}$

Bulb of power, $P_2=60\text{W}$

Formula used:

The power of the electric circuit is given as

        $P=\frac{V^2}{R}$

Here,

        $P$ is the power

        $V$ is the voltage

        $R$ is the resistance

Calculation:

The resistance of the electric circuit when only the bulb of $60\text{W}$ is on

        $\begin{array}{c}60=\frac{V^2}{R}\\ R=0.016\frac{V^2}{60}\end{array}$…… (1)

When an additional bulb of $100\text{W}$ is switched on in house electric circuit, all the appliances are connected in parallel connection; therefore, voltage across all the appliance remains same as the new resistance of the electric circuit is calculated as

        $\begin{array}{c}100+60=\frac{V^2}{R_{\text{new}}}\\ R_{\text{new}}=0.00625\frac{V^2}{160}\end{array}$…… (2)

It’s clear from the Eq. (1)and Eq. (2)that the resistance of the electric circuit decreases when an additional bulb of $100\text{W}$ is switched on.