Shown in the figure is network of capacitors & resistors. The potentials of some of the nodes are given. Find the(b) energy stored in the capacitor connected between R&S
(c) energy stored in the capacitor connected between P&Q

Explanation

At steady state, capacitor. offers infinite resistance to DC, therefore, no current passes through 1 1F capacitor and current passes through 1 Omega resistor across PQ. Since three identical resistors of resistance `(1)/(3)Omega`. each are connected in series. This series combination is connected in parallel with another resistor of 1Omega across PQ
.thereforeThe effective resistance between P and Q will be`(1)/(2)Omega`
Applying KCL at P, we obtain
`i_(1)  + i_(2) + i_(3) = 0` `implies(V_(P)-10)/(1)+(V_(P)-3)/(0.3)+(V_(p)-(-20))/(1)=0`
`impliesV_(p)[1+(1)/(0.3)+1]=10+10-20`
(a) `V_(p) = 0 `and `i__(1) = -10 amp`. `i_(2) = -10 amp`, `i_(3) = 20 amp`.The potential difference across PQ `V_(PQ)=R_(PQ)i_(3)`
`impliesV_(P)=(1//2)20=10V`
Since, `V_(p) = 0, V_(Q)= 0 - 10`
i.e.`V_(PQ) = +10 V`
The potential difference between R and `S = V_(RS)` Since `i_(PR)= I_(PQ)//2 = I_(3)//2``V_(R)=V_(P)-{l_(PR) (R_(PR) = - [10 xX 1//3)] = (10)/(3)V`
(b) The energy stored in the capacitor `(0:4muF)`The energy stored in the capacitor `(1 muF)`=`(1)/(2)(1XX10^(-6)(10)^(2)=50muJ`