In the circuit shown in the figure (Figure 1) , the capacitors are all initially
ID: 1271693 • Letter: I
Question
In the circuit shown in the figure (Figure 1) , the capacitors are all initially uncharged and the battery has no appreciable internal resistance. Assume that E = 52.5V and R = 17.5 Omega .
A) After the switch S is closed, find the maximum charge on each capacitor.
B)After the switch S is closed, find the maximum potential difference across each capacitor.
C)After the switch S is closed, find the maximum reading of the ammeter A.
D)After the switch S is closed, find the time constant for the circuit.
Explanation / Answer
C20, C30 and C40 are in series, hence their capacitace
1/CP = 1/C20 + 1/C30 + 1/C40
= 1/20 + 1/30 + 1/40
= (6 + 4 + 3) / 120
CP = 120 / 13 pF
C10 and CP are in parallel, therefore net capacitance of the circuit
C = C10 + CP
= 10 + 120 / 13
= 250 / 13 pF
a. When long time has passed after closing the switch, there will be no current in the circuit and the capacitors will have full charge.
VC = V = 52.5 V
Charge on C10 q10 = C10 * V
= 10 pF * 52.5
= 525 pC
Charge on CP qP = CP * V
= (120 / 13) pF * 52.5
= 484.61 pC
Since charge in series remains same, hence
q20 = q30 = q40 = 484.61 pC
525 pC, 484.61 pC, 484.61 pC, 484.61 pC
b. Voltage across C10 = V = 52.5 V
Voltage across C20 = V20 = q20 / C20
= 484.61 pC / 20 pF
= 24.23 V
Voltage across C30 = V30 = q30 / C30
= 484.61 pC / 30 pF
= 16.15 V
Voltage across C40 = V40 = q40 / C40
= 484.61 pC / 40 pF
= 12.11 V
V10, V20, V30, V40 = 52.5 V, 24.23 V, 16.15 V, 12.11 V
c. Just after the switch is closed capacitance behaves as short circuit, hence maximum current
the circuit
I = V / R = 52.5 / 17.5
= 3 A
d. time constant ? = R * C
= 17.5 * (250 / 13 pF)
= 4375 * 10-12 / 13
= 3.365 * 10-10 s