In the Figure ( Figure 1 ) , C 1 = 6.00 ? F , C 2 = 3.00 ? F , C 3 = 5.00 ? F .
ID: 2138027 • Letter: I
Question
In the Figure (Figure 1) , C1 = 6.00 ?F, C2 = 3.00 ?F, C3 = 5.00 ?F. The capacitor network is connected to an applied potential Vab. After the charges on the capacitors have reached their final values, the charge on C2 is 40.0 ?C.
A.What is the charge on capacitor
C1?
C3?
Vab?
In the Figure (Figure 1) , C1 = 6.00 ?F, C2 = 3.00 ?F, C3 = 5.00 ?F. The capacitor network is connected to an applied potential Vab. After the charges on the capacitors have reached their final values, the charge on C2 is 40.0 ?C. What is the charge on capacitor What is the charge on capacitor What is the applied voltageExplanation / Answer
C1 and C2 are in parallel and then it is in series with C3
Hence Ceq = (C1 + C2)||C3 = (9 uF) || (5 uF) = 45/14 uF = 3.214 uF
let potential be V.
Hence charge = (3.214 * V) uC
Charge on C3 = (3.214 * V)
Potential of C3 = (3.214 * V)/5 = 0.6428V
Potential across C1 and C2 = V - 0.6428V = 0.3572V
Charge on C2 = 0.3572V *3 = 40 (given)
Solving we get
V = 32.32 V
Hence potential difference applied is 32.32 V
a) Charge on capacitor C1 = Total Charge - Charge on C2 = 32.32 * 3.214 - 40 = 63.87 uC
b) Charge on C3 = 3.214 * 32.32 = 103.87 uC
c) Potential applied = 32.32 V