A) B draws more current than A
B) currents drawn are in the ratio of their resistances
C) both draw the same current
D) A draws more current than B
Correct Answer: A
Solution :
The power drawn by the bulb is \[P=\frac{{{V}^{2}}}{R}\] \[\Rightarrow \] \[R=\frac{{{V}^{2}}}{P}\] or \[R\propto \frac{1}{P}\] (as V is same in parallel) It means that greater power will have less resistance and therefore, draws more current. Hence, current flowing in bulb B will be more.
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