A) 1 : 2
B) 2 : 1
C) 1 : 1
D) 1 : 4
Correct Answer: A
Solution :
In series, current is same in both the bulbs, hence \[P\propto R\,(P={{i}^{2}}R)\] \[\therefore \frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{2}\]
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