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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