A) Increase
B) Decrease
C) Remains the same
D) Not continuous
Correct Answer: A
Solution :
\[P=\frac{{{V}^{2}}}{R}\] ÞÞ \[P\propto \frac{1}{R}\] (V - constant) \ When one bulb will fuse out resistance of the series combination will be reduced. Hence from \[{{P}_{Consumed}}\propto \frac{1}{R}\] illumination will increase.You need to login to perform this action.
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