Question

Two bulbs rated $200\, W$, $220\, V$ and $100\, W$, $220\, V$ are connected in series. Combination is connected to $220 \,V$ supply. Power consumed by the circuit is?

• A. 80 W
• B. 67 W
• C. 76 W
• D. 65 W

Answer 1

Answer: (B)

Effective resistance in series is sum of individual resistances. Power consumed by a bulb, potential difference across which is V, is
$ P=\frac{{{V}^{2}}}{R} $
where, R is resultant resistance. In series combination,
$ R={{R}_{1}}+{{R}_{2}} $
$ \therefore $ $ P=\frac{{{V}^{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{{{V}^{2}}}{\frac{{{V}^{2}}}{{{P}_{1}}}+\frac{{{V}^{2}}}{{{P}_{2}}}} $
$ \Rightarrow $ $ \frac{1}{P}=\frac{1}{{{P}_{1}}}+\frac{1}{{{P}_{2}}} $
$ =\frac{1}{200}+\frac{1}{100} $
$ \Rightarrow $ $ P=67\,W $