A) 40 W in series and 100 W in parallel
B) 100 W in series and 40 W in parallel
C) 40 W both in series and parallel will be uniform
D) 100 W both in series and parallel will be uniform
Correct Answer: A
Solution :
\[\because \] \[H={{I}^{2}}Rt\] \[\Rightarrow \] \[H\propto R\] (in series) Since, bulb having less power will have less resistance. Therefore, \[40W\]bulb will shine more brighter. Now, in parallel \[H=\frac{{{V}^{2}}}{R}t\] \[\Rightarrow \] \[H\propto \frac{1}{R}\] Since, bulb having more power will have less resistance, therefore \[100W\]bulb will shine more brighter.
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