Question

Two electric bulbs rated 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)

The correct options are
A $P_1=16W$
D $P_2=4W$

Given voltage of source $V_S=220V$
Using $P=V^2/R$ , the resistance of 25 W bulb is $R_1=\frac{{220}^2}{25}=1936\Omega$
and the resistance of 100 W bulb is $R_2=\frac{{220}^2}{100}=484\Omega$
As both bulbs are connected in series so current through each bulb will be equal. i.e
current, $I=\frac{V_S}{R_1+R_2}=\frac{220}{1936+484}=0.09A$
Now $P_1=I^2{R}_1=(0.09{)}^2\times 1936=15.68\sim 16W$ and
$P_2=I^2{R}_2=(0.09{)}^2\times 484=3.92\sim 4W$