Question

Find the charge on the capacitor in an LRC-series circuit when L=1/4 h, R=20 Ω , C=1/300 f, E(t)=0 V, q(0)=4 C, and I(0)=0 A. Is the charge on the capacitor ever equal to zero?

Short answer

The charge of capacitor cannot be zero ever.

Step-by-step solution

Definition of the capacitor in an LRC-series circuit:

A capacitor stores energy in the electric field (E) between its plates, depending on the voltage across it, and an inductor stores energy in its magnetic field (B), depending on the current through it.

In LCR series electric circuits:

Consider the second order linear differential equation of LRC series circuit as,

L(d2q/dt2)+R(dq/dt)+q/C=E(t)

(1/4)(d^2q/dt^2)+20(dq/dt)+q/(1/300)=0

(d^2q/dt^2)+80(dq/dt)+1200q=E(t)

The initial conditions are,

q(0)=4, q'(0)=0

The auxiliary equation is,

m^2+80m+1200=0

(m+20)(m+60)=0

m=-20, m=-60

The roots are real and complex. So,

q(t)=c1exp(-20t)+c2exp(-60t)

q'(t)=-20c1exp(-20t)-60c2exp(-60t)

Now,

q(0)=4

Implies,

c1+c2=4 ..... (1)

And

q'(0)=0

Gives,

-20c1-60c2=0

c1+3c2=0

From equation (1), c1=4-c2

So,

4-c2+3c2=0

c2=-2

From equation (1),

c1=4-c2

=4-(-2)

=6

Thus, q(t)=6exp(-20t)-2exp(-60t) is the charge on the capacitor in the given LRC series circuit.

Use the initial condition:

Now,

q(t)=0

6exp(-20t)-2exp(-60t)=0

2exp(-60t) [3 exp(40t)-1]=0

Since,

exp(-60t)≠ 0;

Therefore,

3 exp(40t)-1=0

t=-1/40 ln^3

<0

But time can not be negative, so

q(t)≠ 0

Hence, the charge as the capacitor is never zero.