A) 2 ampere
B) 4 ampere
C) 1 ampere
D) 6 ampere
Correct Answer: B
Solution :
\[6\Omega \] and \[2\Omega \]. are in parallel combination
Hence, \[{{R}_{AB}}=\frac{6\times 2}{6+2}=1.5\Omega \] \[V=IR\Rightarrow \frac{60}{15}=I\Rightarrow I=4A\] \[[\because \,\,{{R}_{eq}}=\frac{3\times 3}{3+3}=\frac{9}{6}=1.5\,\Omega ]\]
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