A) 1 : 2
B) 2 : 1
C) 4 : 9
D) 9 : 4
Correct Answer: D
Solution :
When all bulbs are glowing \[{{R}_{eq}}=\frac{R}{3}+\frac{R}{3}=\frac{2R}{3}\] \[Power\,{{P}_{1}}=\frac{3{{E}^{2}}}{2R}\]----------------- (i) In case two \[{{R}_{eq}}=\frac{3R}{2}\] \[Power\,{{P}_{2}}=\frac{2{{E}^{2}}}{3R}\]--------------- (ii) \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{3}{2}\times \frac{3}{2}=\frac{9}{4}\]You need to login to perform this action.
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