NTA Abhyas JEE Main2020PhysicsCurrent ElectricityPractice
In the following network, the potential at O is
Options
- A4   V
- B3  V
- C6  V
- D4.8  V
Correct answer
D. 4.8  V
Step-by-step solution
Let the potential at O is V 0 . Application of Kirchhoff's first law at junction O gives 8 - V 0 2 = V 0 - 4 4 + V 0 - 2 2 = V 0 - 4 + 2 V 0 - 4 4 4 8 - V 0 2 = V 0 - 4 + 2 V 0 - 4 16 - 2 V 0 = 3 V 0 - 8 16 + 8 = 5 V 0 V 0 = 24 5 = 4 .8   V