Question

An electric bulb P rated $220V,60W$ is working at full efficiency. Another identical bulb Q is connected across the mains as shown in the figure. The total power consumed is :

• A. $20W$
• B. $40W$
• C. $30W$
• D. $70W$

Answer 1

The correct option is C $30W$

Answer is C.

Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. In a series circuit, every device must function for the circuit to be complete.

In this case, a 60 W bulb is connected in series with an electric bulb P rated 220 V, 60 W.

The resistance across the bulb is calculated as follows.
P=VI, That is, $P=\frac{V^2}{R}.So,R=\frac{V^2}{P}$.

$R=\frac{{220}^2}{60}=806.67ohms$.

The bulb Q is also identical as bulb P. So it will also have the same resistance. The total resistance in the circuit is 806.67 + 806.67 = 1613.34 ohms.
Total power in the circuit is $P=\frac{V^2}{R}=\frac{{220}^2}{1613.34}=30W$.

Hence, the total power consumed is 30 W.