A) 1A
B) \[\frac{30}{7}A\]
C) \[\frac{5}{7}A\]
D) Information insufficient
Correct Answer: C
Solution :
Apply KCL at junction having potential V \[\frac{V-10}{2}+\frac{V-0}{6}+\frac{V-0}{2}=0\]\[\Rightarrow \]\[V=30/7volt\] Both\[3\Omega \]resistors are connected in series, so required current \[i=\frac{V-0}{6}=\frac{5}{7}amp\]You need to login to perform this action.
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