E&M. 2. A 6.0 F capacitor, C 1 , is initially charged using a 30 V battery. C 1
ID: 2076756 • Letter: E
Question
E&M. 2. A 6.0 F capacitor, C1 , is initially charged using a 30 V battery. C1 is then inserted in the circuit represented above with a resistor of resistance R and the 12.0 F capacitor C2, which is initially uncharged. The switch S in the circuit is initially open.
Please note that A&B were answered in another question but C,D,&E were not.
6.0 HF E&M.; 2. A 6.0 LLF capacitor, C1, is initially charged using a 30 v battery. ch is then inserted in the circuit represented above with a resistor of resistance R and the 12 HF capacitor C2, which is initially uncharged The switch S in the circuit is initially open. (a) Calculate the charge e on ci before the switch is closed. The switch is now closed. (b) q2 be the charge on capacitor C2 at any time t after the switch is closed. Write, but do NoT solve, Let a differential equation for the charge q2 as a function of the time t Write your equation in terms of the charge from part (a), C1, C2,R, and fundamental constants, as appropriate.Explanation / Answer
A) Q=C1V1 = 6*30 = 180 uC
B) q1/C1 - q2/C2 = iR = R dq2/dt
Q/C1 - q2/C1 - q2/C2 = R dq2/dt
C) Q1/C1 = Q2/C2
Q1 =180*6/18 = 60uC
Q2= 180-60 = 120 uC
D) final voltage = 60uC/6uF = 10V
energy dissipated = 0.5*(6*30^2 - 6*10^2 - 12*10^2)
= 1800 uJ
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