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, |
\[{{I}_{BC}}=3I-I=2I=\frac{V}{2R}.\] |
You need to login to perform this action.
You will be redirected in
3 sec