4. The Capacitor shown is initially uncharged, when the switch is closed as show

Question

4. The Capacitor shown is initially uncharged, when the switch is closed as shown. 400 60 V 300 F-4300 what is the time constant, , for charging? What is the charge on the capacitor once it is fully charged? Sometime later the switch is moved to position b. How long after the switch is moved does it take for the capacitor to discharge to less than 0.6v? What is the maximum current in the 4300 resistor? How much energy is dissipated through the resistor when the capacitor is discharging? a. b. c. d. e.

Explanation / Answer

a) time constant is T = R*C = 400*300*10^-6 = 0.12 sec

b) Q = C*V = 300*10^-6*60 = 0.018 C

c) v = V*e^(-t/T)

60-0.6 = 60*e^(-t/(4300*300*10^-6))

t = 0.0129 sec

d) maximum current is i = e/R = 60/4300 = 0.0139 A

e) P = i^2*R = 0.0139^2*4300 = 0.83 J/sec

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