Finding the time constant, and the maximum charge on the capacitor, in a series RC circuit

Question

Consider a series RC circuit for which C= 6 µF, R=2*10⁶ Ω, E=20 V

How do I find:

a) The time constant of the circuit

b) The maximum charge on the capacitor after a switch in the circuit closes

c) If the battery is immediately taken out of the circuit once the capacitor reaches its maximum charge, how long would it take for the capacitor to drain to one-tenth its maximum charge?

Answer 1

Let

Q(t) be the charge on the capacitor at time t,

V(t) be the voltage across the capacitor at time t,

I(t) be the current in the circuit at time t.

(a)

By definition, the time constant is

RC = 6μF × 2×10⁶Ω = 12 s

(b)

Once the switch closes, charge starts flowing from the battery to the capacitor.

By definition of capacitance, at every point in time

Q(t) = CV(t)

Asymptotically the circuit approaches the state when V(t) = E.

While that would happen only after an infinite amount of time,

that maximum voltage implies maximum charge of

CE = 6μF × 20V = 120 μC

(c)

Once the battery is removed, the initial charge of CE on the capacitor

starts moving from the positive terminal of the capacitor to its negative terminal

though the resistor.

That means the current flowing through the capacitor is from its negative

terminal to its positive terminal, i.e., the current is negative.

Thus we have the following equations

dQ/dt = -I(t) (by definition of current)

RI(t) = V(t) (by Ohm's and Kirchoff's laws)

Q(t) = CV(t) (by definition of capacitance)

Substituting the last two equations into the first

dQ/dt = -Q/RC

This equations can be solved by separation of variables

∫dQ/Q = ∫-dt/RC

The solution is

ln(Q) = -t/RC + D

where the constant of integration, D, is determined from the initial condition, Q(0) = CE.

Hence,

D = ln(CE)

ln(Q) = -t/RC + ln(CE)

Q(t) = CEe^-t/RC

The capacitor reaches one-tenth of its initial charge at time t when

Q(t) = 0.1EC

That is

CEe^-t/RC = 0.1EC

Simplifying

e^-t/RC = 0.1

-t/RC = ln(0.1)

t = -RCln(0.1) = -12×(-2.3) = 27.63 s