IIT JEE , AIEEE Forum - Physics , Electricity

krishna.gopal

Let q charge enters in the circuit from P. It will be distributed in the circuit as shown in the figure . Applying KVL on the top loop.
(q/2-q')/C+(q-2q')/C+(q/2-q')/C-q'/C=0
2q-5q'=0
So q' =2q/5

If total potential difference from P to R is V then adding potential on one of the paths from P to R gives
V=q/2C+q'/C+q/2C
Putting q'=2q/C
V= 7q/5C
So q/V =5C/7
So net capacitance is 5C/7

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