Question

Two electric bulbs, rated at (40 W, 220 V) and (160 W, 220 V) are connected in series at 110 V. If power drawn by bulb 1 (40 W) is $P_1$ & bulb 2 (160 W) is $P_2$, then

• A. $P_1=6.4W,P_2=3.2W$
• B. $P_1=1.8W,P_2=3.6W$
• C. $P_1=6.4W,P_2=1.6W$
• D. $P_1=3.2W,P_2=12.8W$

Answer 1

The correct option is C $P_1=6.4W,P_2=1.6W$

$P=I^{2}R$ & $P=\frac{V^2}{R}$ (rating)
$R_1=\frac{V^2}{P_1}=\frac{220\times 220}{40}{R}_2=\frac{V^2}{P_2}=\frac{220\times 220}{160}$
when in series
$I=\frac{V}{R_1+R_2}=\frac{110}{\frac{220\times 220}{40}+\frac{220\times 220}{160}}=\frac{1}{11+\frac{44}{16}}=\frac{1}{11+11/4}=\frac{4}{55}A$
$P_1=I^2{R}_1=\frac{4}{55}\times \frac{4}{55}\times \frac{220\times 220}{40}=\frac{16\times \text{/}{16}^4}{\text{/}405}=6.4W$
$P_2=I^2{R}_2=\frac{4}{55}\times \frac{4}{55}\times \frac{220\times 220}{160}=\frac{64}{4}=1.6W$