A) 2
B) 3
C) 4
D) 5
Correct Answer: D
Solution :
Current flowing through the lamp, \[I=\frac{V}{R}=\frac{60}{20}=3A\] The current in the lamp must remain 3A, when it is operated at 75 V supply. Let R be the resistance connected in series, then \[\frac{75}{20+R}=3\] \[\Rightarrow \] \[75=60+3R\] or \[3R=75-60\] or \[3R=15\] or \[R=5\Omega \]You need to login to perform this action.
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