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.

science is a systematic and logical approach to discovering how things in the universe work. true to this definition, science aims for measurable results through testing and analysis. science is based on fact, not opinion or preferences. the process of science is designed to challenge ideas through research.

explanation: