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RC Circuit - Capacitor Discharge Time, Time Constant, etc.

(Please see the attachment for detailed problem)

1) In a circuit shown below it took capacitor 0.138 seconds to discharge from its maximum charge Qmax to half of that maximum charge once the battery was removed.

a) How much time would it take that capacitor to discharge from its maximum charge to 1/16th of that maximum charge?
b) What is the time constant of this circuit?
c) What is the capacitance of the capacitor?
d) The battery is reconnected and the capacitor charges up again. When the capacitor is fully charged what is the voltage drop across the capacitor?
e) What is the charge Qmax on the capacitor?
f) How far apart are capacitors plates if each is a circle with 2 cm diameter?
g) With the battery positioned as shown in the figure above is the left plate of the capacitor charged positively or negatively?
h) What are the magnitude and the direction of the electric field inside this capacitor
i) A proton is placed in the space between the capacitor plates. What are the direction and magnitude of the force acting on that proton inside the capacitor?
j) If initially proton had velocity straight down the page will it be deflected from its straight path by the electric field of the capacitor? If so which way?
k) What is the magnitude and direction of that proton's acceleration (proton's mass mp =1.6726×10^-27 kg)


Solution Preview

Please refer to the attachment.

a) While discharging, charge Q on the capacitor at any instant t is given by :

Q = Qmax e-t/RC where Q0 = The initial charge and RC = τ = Time constant of the circuit

For Q = Qmax/2, t = 0.138 sec. Hence, Qmax/2 = Qmax e-(0.138)/τ

Or e(0.138)/τ = 2 or loge2 = 0.138/τ or 0.693 = 0.138/τ

Or τ = 0.2 sec

For Q = Qmax/16 we have : Qmax/16 = Qmax e-t/0.2

Or et/0.2 = 16 ...

Solution Summary

The expert examines capacitor discharge time, and time constants for RC circuits.