Question
Physics Question on Current electricity
In the following network potential at ′O′
A
4 V
B
3 V
C
6 V
D
4.8 V
Answer
4.8 V
Explanation
Solution
Let the potential at O is V0.
Application of Kirchhoff's first law at junction O gives
28−V0=4V0−4+2V0−2
=4V0−4+2V0−4
24(8−V0)=V0−4+2V0−4
16−2V0=3V0−8
16+8=5V0
V0=524=4.8V