Question

Two electric bulbs rated $25\, W , 220 \,V$ and $100 \,W , 200\, V$ are connected in series across $a$ $220 \,V$ source. The $25\, W$ and $100\, W$ bulbs now draw powers $P_{1}$ and $P_{2}$ respectively, then :
(1) $P_{1}=16\, W$
(3) $P_{2}=16 \,W$
(2) $P_{1}=4\, W$
(4) $P_{2}=4\, W$.

• A. 1 and 2 are correct
• B. 2 and 3 are correct
• C. 1 and 4 are correct
• D. 1, 2 and 3 are correct

Answer 1

Answer: (C)

Resistance of first bulb.
$R_{1}=\frac{V_{1}^{2}}{P_{1}}=\frac{(220)^{2}}{25}$
Resistance of second bulb
$R_{2}=\frac{V_{2}^{2}}{P_{2}}=\frac{(220)^{2}}{100}$
Net resistance in series combination will be
$R =R_{1}+R_{2}=$
$\frac{(220)^{2}}{25}+\frac{(220)^{2}}{100} =\frac{(220)^{2} \times 5}{100}=\frac{(220)^{2}}{20}$
Current through the circuit,
$I=\frac{V}{R}=\frac{220}{(220)^{2} / 20}=\frac{20}{220}=\frac{1}{11} A$
Power in $25 \,W$ bulb,
$P_{1}=I^{2} R_{1} $
$=\left(\frac{1}{11}\right)^{2} \times \frac{(220)^{2}}{25}=16\, W$
Power in $100\, W$ bulb,
$P_{2}=I^{2} R_{2}$
$=\left(\frac{1}{11}\right)^{2} \times \frac{(220)^{2}}{100}=4 \,W$