In Fig. 25.35 the 2.0- mu F capacitor is charged to 150 V. while the 1.0- mu F c
ID: 2236461 • Letter: I
Question
In Fig. 25.35 the 2.0- mu F capacitor is charged to 150 V. while the 1.0- mu F capacitor is initially uncharged. Switch S is then closed. Find the total energy dissipated in the resistor as the circuit comes to equilibrium. (Hint: Think about charge conservation.) Solution When current stops flowing (at t = infinity), the potential difference across the capacitors is equal, but the total charge is just the initial charge. Thus, V1(infinity) = V2(infinity) V(infinity), and Q2(0) = Q1(infinity). Since Q = CV, C1V1(infinity) + C2V2(infinity) = C2V2(0) or V(infinity) ( )C = (C + ) = - V (0). The energy stored in the capacitors is initially U(0) =1/2C2V22(0) = 1/2 (2 mu F(150 V)2 = 22.5 mJ, and finally U(infinity) = 1/2(C1 + C2 )V2(infinity)(100 V)2 = 15.0 mJ. The difference, |Delta U| = 7.50 mJ, is dissipated in the resistor (except for a negligible amount of radiated energy).Explanation / Answer
The solution says is that the final voltage (which is same for each capacitor) = C2*V2(0)/(C1+C2)
<----- ( this step comes from the step Q = CV ,so, C1V1(infinity) +C2V2(infinity) = C2V2(0)
so, V1(infinity) = C2*V2(0)/(C2+C2) )
so,V1(infinity) = 2*150/(2+1) = 100 V <-------this is the doubt you had i guess
After that the next step is pretty evident from the answer ; ie he has calculated the difference between initial energy(U(0)) and final energy(U(infinity)) ..
-----------------------------HOPE THIS HELPS------------------------------