A) 60 W
B) 180 W
C) 10 W
D) 20 W
Correct Answer: D
Solution :
| [d] 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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