In the circuit shown in the figure (Figure 1) , the capacitors are all initially
ID: 1267814 • 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.
Part A
After the switch S is closed, find the maximum charge on each capacitor.
Answer in the order indicated. Separate your answers using commas.
q10,q20,q30,q40 =
Part B
After the switch S is closed, find the maximum potential difference across each capacitor.
Answer in the order indicated. Separate your answers using commas.
V10,V20,V30,V40 =
PART C
After the switch S is closed, find the maximum reading of the ammeter A.
IA =
Part 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 = 50 V
Charge on C10 q10 = C10 * V
= 10 pF * 50
= 500 pC
Charge on CP qP = CP * V
= (120 / 13) pF * 50
= 461.54 pC
Since charge in series remains same, hence
q20 = q30 = q40 = 461.54 pC
500 pC, 461.54 pC, 461.54 pC, 461.54 pC
b. Voltage across C10 = V = 50.0 V
Voltage across C20 = V20 = q20 / C20
= 461.54 pC / 20 pF
= 23.08 V
Voltage across C30 = V30 = q30 / C30
= 461.54 pC / 30 pF
= 15.38 V
Voltage across C40 = V40 = q40 / C40
= 461.54 pC / 40 pF
= 11.53 V
V10, V20, V30, V40 = 50.0 V, 23.08 V, 15.38 V, 11.53 V
c. Just after the switch is closed capacitance behaves as short circuit, hence maximum current
the circuit
I = V / R = 50 / 20
= 2.5 A
d. time constant ? = R * C
= 20 * (250 / 13 pF)
= 5000 * 10-12 / 13
= 3.846 * 10-10 s