Dr. Kevin G. TeBeest
Assoc. Prof. of Applied Mathematics
Kettering University

RLC Example 2:    Solution with Transient and Steady State Terms

>	restart; de := 2*diff(q(t),t$2) + 24*diff(q(t),t) + 272*q(t) = 1200*sin(8*t);

(1)

Solve the above ODE, subject to initial conditions   q(0) = 4  and  q'(0) = 36 . Store the result in Q:

>	Q := rhs( dsolve( {de, q(0) = 4, D(q)(0) = 36 }, q(t) ) );

(2)

Simplify the solution, and store the result in Q: 

(3)

(4)

Plot the solution: 

Current is time rate of change of charge dQ/dt:

>	i := diff(Q,t);

(5)

Steady State Current: 

>	iss := 24*cos(8*t) + 32*sin(8*t);

(6)

>	plot( [iss,i], t=0..2, title=`Ex 2: RLC Circuits`, ytickmarks=5,thickness=5,titlefont=[HELVETICA,BOLD,16],color=[red,blue],legend=["Steady State Current iss(t)     ", "Current i(t)"],legendstyle=[font=[HELVETICA,BOLD,14]]);

 

The complementary solution qc(t) represents damped oscillation while the particular solution qp(t) is simple harmonic. In this case we say the complementary solution is transient and the particular solution is steady. So after a sufficient amount of time has passed (about 1/2 second in this example), the charge q has reached steady state, and we may approximate the charge q(t) by the particular solution: