In the circuit shown in the figure, the capacitors are initially uncharged. The
ID: 1504503 • Letter: I
Question
In the circuit shown in the figure, the capacitors are initially uncharged. The switch is first thrown to position A and kept there for a long time. It is then thrown to position B. Let the charges on the capacitors be Q_1, Q_2, and Q_3 and the potential differences across them be V_1 V_2, and V_3 after the switch is thrown. a) Which of the following conditions must be true with the switch in position B? a)V1=V2=V3 b)V1 - V2 = V3 c) V1 + V2 = V3 d) Q1 + Q2 = Q3 e) V3 = V0 f) Q1 = Q2 = Q3 g) While the switch was in position A, the voltage drop across CI = the voltage drop across C2 h) While the switch was in position A, the voltage drop across B and the switch was VO.Explanation / Answer
When switch is kept at A for long time then C1 and C2 will fully charged.
and C3 will remain uncharged.
when switch is thrown to B then charge on C1 and C2 will flow to C3.
here C1 and C2 will act as source .
so charge will decrease on C1 and C2 and charge on C2 will increase so voltage.
Applying KVL for this loop:
V1 + V2 - V3= 0
V1 + V2 = V3
( or source C1 + C2 is connected in paralllel with C3 hence both will have same PD )