A) 60 W
B) 180 W
C) 10 W
D) 20 W
Correct Answer: D
Solution :
Key Idea: In series order, the resistances of three bulbs must be added to give resultant resistance of the circuit Let\[{{R}_{1}},{{R}_{2}}\]and\[{{R}_{3}}\]are the resistances of three bulbs respectively. In series order \[R={{R}_{1}}+{{R}_{2}}+{{R}_{3}}\] but\[R=\frac{{{V}^{2}}}{P}\]and supply voltage in series order is the same as the rated voltage. \[\therefore \] \[\frac{{{V}^{2}}}{P}=\frac{{{V}^{2}}}{{{P}_{1}}}+\frac{{{V}^{2}}}{{{P}_{2}}}+\frac{{{V}^{2}}}{{{P}_{3}}}\] Or \[\frac{1}{P}=\frac{1}{60}+\frac{1}{60}+\frac{1}{60}\] Or \[P=\frac{60}{3}=20\,W\]
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