Q.8 T 30 s wo capacitors are charged by a 12.0 V battery. There is a 5.0 Q resis
ID: 2036521 • Letter: Q
Question
Q.8 T 30 s wo capacitors are charged by a 12.0 V battery. There is a 5.0 Q resistor in eries between the capacitors as shown in the Figure. After switch S1 has been closed and S2 open, S, S2 12.0 V 3.00 ?F 6.00 ? 5.00 ? 2.000 6.00 ?F What is the time constant of the circuit? After the switch S1 has been closed for one time constant, (i) What is the voltage across the 3.00 ?F capacitor? (ii) What is the voltage across the 6.00 ?F capacitor? (ii) What is the power across the two resistors? How long will it take to charge the capacitors to 90% of their maximum charge? (a) (b) (c) (d) What is the maximum charge on both the capacitors after the switch has been (e) If S1 is now open and S2 is closed, how long will it take for the charge to become ( What is the direction and magnitude of the maximum current? closed for a long time? one-fifth of its maximum value?Explanation / Answer
8. given two capacitors
C1 = 3 uF
C2 = 6 uF
V = 12 V
r = 5 ohm
after S1 is closed, S2 is open
a. time constant = Reff*Ceff
Reff = r + 2 = 7 ohm
Ceff = C1C2/(C1 + C2) = 18/9 = 2 uF
hence
t' = 7*2*10^-6 = 14*10^-6 s
b. after S1 is cloased for one time constant
t = t' = 14*10^-6 s
i. voltage across 3 uF capacitor = V1
now,
q1 = qo*(1 - e^(-t/t'))
now, q1 = C1V1
V1 = q1/C1
V1 = (qo/C1)(1 - e^(-t/t'))
here qo = Ceff*V = 2*10^-6*12 = 24*10^-6 C
hence
V1 = 8*(1 - e^(-1)) = 5.056964470628 V
ii. voltage across C2 = V2
V2 = (qo/C2)(1 - 1/e) = 4(1 - 1/e) = 2.5284822353142 V
iii. power across two resistors = P
P = (12 - V1 - V2)*I
I = (12 - V1 - V2)/(R1 + R2)
hence
P = (12 - V1 - V2)^2/(R1 + R2) = 2.784040112296 W
c. time taken for 90 % charge = t
0.9 = 1 - e^(-t/t')
t = 32.2361*10^-6 s
d. maximum charge on both capacitors = qo = Ceff*12 = 24*10^-6 C
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