| In the given circuit diagram, potential difference between A and C is \[\frac{V}{4},\] then current in branch BC is- |
 |
A) \[\frac{V}{2R}\]
B) \[\frac{V}{4r}\]
C) zero
D) \[\frac{3V}{4r}\]
Correct Answer: A
Solution :
| \[{{I}_{AC}}=\frac{V}{4R}=I\] |
| \[{{I}_{CD}}=\frac{3V}{4R}=3{{I}_{AC}}=3I\] |
 |
| Hence using KCL at junction C, |
| \[{{I}_{BC}}=3I-I=2I=\frac{V}{2R}.\] |
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