Question

Kirchhoff’s voltage law states that the sum of the voltage drops across a resistor, R, an inductor, L, and a capacitor, C, in an electrical circuit must be the same as the voltage source, E(t), applied to that RLC circuit. Applying the additional fact that the current, I, is related to the charge, q, by the relationship I=dq/dt, the resulting ODE model for the charge in a circuit is: L (d^2 q)/(dt^2 )+R dq/dt+1/C q=E(t). If a 100 sin⁡(60t)source is connected to an RLC circuit with a 1/20 henry inductor, L, a 1 ohm resistor, R , and a 1/130 farad capacitor, C, find the charge q(t) given that q(0)=0 and I(0)=q'(0)=0.

Answer 1

Given that,

L = 0.05 H , R = 1 ohm , C = 1/130 F

E(t) = 100sin(60t)

And the differential equation is,

L(d^2 q/dt^2)+R*dq/dt+q/C = E(t)

0.05*D²q + Dq + 130*q = 100*sin(60t)

The complementary solution is,

0.05*D²q + Dq + 130*q = 0

0.05*x² + x + 130 = 0

x = (-1+5i)/2 , x = (-1-5i)/2

q_c = e^-t(A*cos5t+B*sin5t) --(1)

The particular integral solution is,

q_p = P*cost+Q*sint

but , 0.05*D²q + Dq + 130*q = 100*sin(60t)

Dq_p = -P*sint+Q*cost --(2)

D²q_p = -P*cost-Q*sint

substitute the values in the equation (2);

0.05(-P*cost-Q*sint)+(-P*sint+Q*cost)+130(P*cost+Q*sint) = 100*sin(60t)

sint(-0.05*Q-P+130*Q)+cost(-0.05*P+Q+130*P) = 100*sin(60t)

= 60; 129.95*Q-P = 100 and 129.95*P+Q = 0

P = 0.005922 , Q = -0.7695

Therefore , q_p = 0.005922*cos(60t)-0.7695*sin(60t)

The solution is, q(t) = e^-t(A*cos5t+B*sin5t) + 0.005922*cos(60t)-0.7695*sin(60t)

that implies, q'(t) = -e^-t(A*cos5t+B*sin5t)+e^-t(-A*sin5t+B*cos5t)-0.35532sin(60t)-46.17*cos(60t)

and given that q(0)=0 and I(0)=q'(0)=0,substituting in above equations

q(0) = A + 0.005922 = 0

A = -0.005922

q'(0) = -A+B-46.17 = 0

B = 46.17+A = 46.1759

Therefore the final solution is; q(t) = e^-t(-0.005922*cos5t+46.1759*sin5t) + 0.005922*cos(60t)-0.7695*sin(60t)