Question

Two electric bulbs rated at 25 W, 220 V and 100 W, 220 V are connected in series across a 220 V voltage source. The 25 W and 100 W bulbs now draw $P_1$ and $P_2$ powers respectively

• A. $P_1=16W$
• B. $P_1=4W$
• C. $P_2=16W$
• D. $P_2=4W$

Answer 1

Answer: (A), (D)

$P_1=16W$
$P_2=4W$

Here voltage across each bulb is $V=220V$.
Using $P=\frac{V^2}{R},$
The resistace of 25 W bulb is $R_1=\frac{{220}^2}{25}=1936\Omega$ and reisstance of 100 W bulb is $R_2=\frac{{220}^2}{100}=484\Omega$
As they are connected in series so equivalent resistance is $R_{eq}=R_1+R_2$
Current in the circuit is $I=\frac{V}{R_{eq}}=\frac{V}{R_1+R_2}=\frac{220}{1936+484}=0.09A$

Now the power in 25 W bulb is $P_1=I^2{R}_1=\left(0.09{)}^2\right(1936)=15.68\sim 16W$

Power in 100 W is $P_2=I^2{R}_2=\left(0.09{)}^2\right(484)=3.92\sim 4W$