The capacitor has a capacitance of 10 mu F. The resistor has a resistance of 20

Question

The capacitor has a capacitance of 10 mu F. The resistor has a resistance of 20 OHM, Originally, the capacitor has a charge Q =1.0 mu C. When the circuit is closed, the capacitor discharges, its charge flowing from the top plate to the bottom plate, passing through the resistor. Because the resistor and the capacitor share common electrodes in quasi-equilibrium, any instant of time, the voltage V = Q/C across the capacitor due to the remaining charge must be the same as the voltage drop V = IR across the resistor. Moreover, the current I and the charge Q are related in the sense that I = - dq/dt. For the example above, what is the RC time-constant (in seconds)? How long must we wait before the charge on the capacitor is only 0.5 mu C? What is the current in the resistor at a time t? The rate of dissipation of energy in the resistor is the product IV, and both I and V decrease as the the capacitor discharges. When the capacitor is fully discharged, how much energy will have been dissipated in the resistor? Haw does this energy loss compare with Q^2/2C where Q is the initial charge on the capacitor?

Explanation / Answer

7)
A)
Time constant = R*C
                               =20*10*10^-6
                                =2*10^-4 s
B)
use:
Q' = Q*e^(-t/tau)
0.5 = 1*e^(-t/(2*10^-4))
t/(2*10^-4) = 0.693
t = 1.386*10^-4 s
C)
Q' = Q*e^(-t/tau)
dQ/dt = -Q/tau *e^(-t/tau)
i = -dQ/dt
    = Q/tau *e^(-t/tau)
    = 1*10^-6 / 2*10^-4 * e^ (-t/0.0002)
     = 0.005 * e^ (-t/0.0002)
D)
energy dissipated = initial energy stored in capacitor
    =0.5*Q^2/C
    = 0.5*(1*10^-6)^2 / (10*10^-6)
    = 5*10^-8 J
E)
They must be equal

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