A) \[\frac{20}{7}V\]
B) \[\frac{40}{7}V\]
C) \[\frac{10}{7}V\]
D) zero
Correct Answer: D
Solution :
From the given figure, current through lower branch of resistances which are joined in series is \[{{i}_{1}}=\frac{10}{4+32}=\frac{10}{7}A\]You need to login to perform this action.
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