A) 1
B) 2
C) 0.5
D) 4
E) 0.25
Correct Answer: C
Solution :
Power of bulb, \[P=\frac{{{V}^{2}}}{R}\] \[\therefore \] \[\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{{{R}_{1}}}{{{R}_{2}}}\] Or \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{100}{200}=0.5\]You need to login to perform this action.
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