UPSEE Solved Model Paper 3

Let the potential of the junction be V, then

6−V

2

+

4−V

4

+

8−V

4

= 0

12 - 2V + 4 -V + 8 - V = 0.24
= 4V
V = 6 volt
∴ Potential drop across capacitor = 6 - (- 10) = 16V
⇒ Charge on capacitor = 16 µC