A discharged capacitor C = 2700F is connected in a series with the resistor R = 1 Omega . (a) Find the general formula for the voltage drop V_(cap) = f(t) across the capacitor if the circuit is connected at time t = 0 to a battery supplying voltage V_(bat) = 2.7V. (b) How long does it take for the voltage accross the capacitor to reach 1.45V? The differential equation governing the voltage drop on the capacitor in this circuit is:

A) (Vc) = 2.7 x (1 - e^(-t/2700)).

B) t = 2079 seconds

Explanation:

We are given;

Vb = 2.7 V

R = 1 ohm

C = 2700 F

A) First of all, let's write the separable equation as;

d(Vc)/dt = (1/(RC)) x ((Vb) - (Vc))

Let's separate variables to get;

d(Vc)/((Vb) - (Vc)) = (1/(RC)) dt

If we integrate both sides, we'll get;

-ln((Vb) - (Vc)) = (1/(RC))t + K

Where K is the constant

So,

(Vb) - (Vc) = k(e^(-t/(RC)))

Note, that, I replaced e^(-K) with k

The constant is arbitrary at this point, thus;

(Vc) = (Vb) - k*e^(-t/(RC))

Now at time t = 0, we have (Vc) = 0, hence;

0 = (Vb) - k*1

So, k = (Vb)

We now have;

(Vc) = (Vb) x (1 - e^(-t/(RC))).

Plugging in the relevant values;

(Vc) = 2.7 x (1 - e^(-t/2700)).

(b) We need to find the time it takesfor the voltage accross the capacitor to reach 1.45V. Thus;

1.45 = 2.7 x (1 - e^(-t/2700))

So,

e^(-t/2700) = 1 - (1.45 / 2.7)

e^(-t/2700) = 0.463

-t/2700 = ln(0.463)

t = -2700 x In(0.463)

Due to the negative sign, well rewrite as;

t = 2700•In(1/0.463)

t = 2700 x 0.77 = 2079 seconds.