A) P/3
B) 3P
C) 9P
D) P/9
Correct Answer: C
Solution :
Let the resistance of each bulb is R. Then, total resistance in series\[=3R\] \[\therefore \] \[P=\frac{{{V}^{2}}}{3R}\] In parallel, total resistance \[=\frac{R}{3}\] \[\therefore \]Power dissipated \[P'=\frac{{{V}^{2}}}{R/3}\] \[=\frac{3{{V}^{2}}}{R}\] \[=9\left( \frac{{{V}^{2}}}{3R} \right)\] \[=9P\]You need to login to perform this action.
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