Question

Find the charge on the capacitor and the current in the given LRC-series circuit. Find the maximum charge on the capacitor. L = 5/ 3 h, R = 10 V, C = 1/30 f, E(t) = 300 V, q(0) = 0 C, i (0) = 0 A

Short answer

The maximum charge on the capacitor is 10.432 at t=π /3.

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)

(5/3)(d²q/dt²) +10(dq/dt)+30q=300

(5)(d²q/dt²) +30(dq/dt)+90q=900

(d²q/dt²)+6(dq/dt)+18q=180 ..... (2)

Thus, the linear second-order differential equation is (d^2q/dt^2)+6(dq/dt)+18q=180.

m²+6m+18=0

m=[-6±(36-4 (1) (18))^1/2] /2

=[-6±(-36)^1/2] /2

[-6± i6] /2

=3± i3

Thus, the complementary function is,

q_c(t)=e^-3t(c₁cos3t+c₂sin3t)

Use the method of undetermined coefficients to find a particular solution.

Assume particular solution as,

q_p=A

Thus,

d²q_p/dt²=6(dq_p/dt)+18q_p=180

Substitute, dq_p/dt=0, and d²q_p/dt² =0 in the above equation.

0+6(0)+18q_p=180

18q_p=180

q_p=180 /80

q_p=10

Thus, the solution is,

q(t)=q_c+q_p

Therefore, the general solution of above differential equation is,

q(t)=e^-3t [c₁cos3t+c₂sin3t]+10 ..... (3)

Differentiate q(t) with respect to t.

q'(t)=e^-3t [-3c₁cos3t-3c₂sin3t-3c₁sin3t+3c₂cos3t]

q'(t)=e^-3t [-3c₁+3c₂)cos3t-(3c₂+3c₁) sin3t] ..... (4)

Substitute the current i(t)=q'(t) in the equation (4).

i(t)=e^-3t [-3c₁+3c₂)cos3t-(3c₂+3c₁) sin3t]

The charge on the capacitor:

Now use the initial condition, q(0)=0 C, i(0)=0 A for finding the current i(t) and q(t).

Substitute q(0)=0 C in the solution q(t).

q(0)=e⁰ [c₁ cos0+c₂ sin0]+10

0= [c₁+c₂x 0]+10

c₁=-10

Substitute i(0)=0 in the equation of i(t).

i(0)=e [(-3c₁+3c₂) cos(0)-(-3c₂+3c₁) sin(0]

0= -3c₁+3c₂

c₂=c₁= -10

Substitute the values of c₁ and c₂ into the equation (3) as below.

q(t)=e^-3t [-10cos3t-10sin3t]+10

Therefore, the charge on the capacitor is,

q(t)=10-10e^-3t [cos3t-sin3t]

Substitute the values of c₁ and c₂ in the equation of i(t).

i(t)= e^-3t [-3 (-10)+3 (-10))cos3t-(3 (-10)+3 (-10)) sin3t]

= e^-3t [(30-30)cos3t-(-30-30) sin3t]

=60 e^-3t sin3t

Therefore, the current in the LRC series circuit is i(t)=60 e^-3t sin3t.

Since the maximum charge occurs when,

i(t)=0

Which implies,

60 e^-3t sin3t =0

sin3t=0

sin3t=sin nπ

3t=nπ

t= nπ /3

Here, n=1,2,3,...

So,

t=π /3, 2π/3, 3π/3, ...

Calculate the charges on the capacitor at t=π /3, 2π/3, 3π/3, ....

q (π /3 )=10-10e^-π [cosπ -sinπ ]

=10.432

q (2π /3 )=10-10e^-2π [cos2π -sin2π ]

=9.971

q (3π /3 )=10-10e^-3π [cos3π -sin3π ]

=10.0008

.

.

.

Hence, the maximum charge on the capacitor is 10.432 at t=π/3.