A) \[\frac{9V}{35}\]
B) \[\frac{5V}{18}\]
C) \[\frac{5V}{9}\]
D) \[\frac{18V}{5}\]
Correct Answer: B
Solution :
The given network is a balanced Wheatstone bridge. It's equivalent resistance will be \[R=\frac{18}{5}\Omega \] So current from the battery \[i=\frac{V}{R}=\frac{V}{18/5}=\frac{5V}{18}\]
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