A) 8.31 A
B) 6.82 A
C) 4.92 A
D) 2 A
Correct Answer: D
Solution :
6W and 6W are in series, so effective resistance is 12W which is in parallel with 3W, so \[\frac{1}{R}=\frac{1}{3}+\frac{1}{12}=\frac{15}{36}\] Þ \[R=\frac{36}{15}\] \ \[I=\frac{V}{R}=\frac{4.8\times 15}{36}=2\,A\]
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