circuit shown in Figure 28.5 is constructed using a 6V battery, a 3k ohm resisto
ID: 1427936 • Letter: C
Question
circuit shown in Figure 28.5 is constructed using a 6V battery, a 3k ohm resistor, and a 6k ohm resistor. Initially both switches are open and the capacitor is uncharged. Switch A is closed and the capacitor begins to charge. After 1/2 second the potential difference across the capacitor is 3.78 V. What is the capacitance of the capacitor? After several minutes switch A is opened. What is the approximate potential difference across the capacitor? Switch B is now closed and the capacitor begins to discharge. What is the characteristic time for discharging this capacitor? will the current through the 6 k ohm resistor be after 2 seconds? What will the charge on the capacitor be after 3 seconds?Explanation / Answer
a) at time t
v = vmax*(1 - e^(-t/T))
3.78 = 6*(1 - e^(-t/T))
3.78/6 = 1 - e^(-t/T)
e^(-t/T) = 1 - 3.78/6
-t/T = ln(0.37)
T = -t/ln(0.37)
R*C = -0.5/ln(0.37)
C = 0.503/R
= 0.503/(3*10^3)
= 1.68*10^-4 F
b) VC = Vmax = 6 volts
c) Time constant of the ckt, T = R*C
= 6*10^3*1.68*10^-4
= 1.008 s
d) I = Imax*e^(-t/T)
= (Vmax/R)*e^(-2/1)
= (6/(6*10^3))*e^(-2)
= 1.35*10^-4 A
e) Q = Qmax*e^(-t/T)
= (C*Vmax)*e^(-3/1)
= (1.68*10^-4*6)*e^(-3)
= 5.02*10^-5 C
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