Question

Find the steady-state current in an LRC series-circuit when L=1/2 h, R=20Ω, C=0.001 f, and E(t) =100 sin60t + 200 cos40t V.

Short answer

The steady state current in LRC circuit is,

$i_p\left(t\right)=-\frac{45}{13}\sin 60t-\frac{30}{13}\cos 60t+\frac{40}{17}\sin 40t+\frac{160}{17}\cos 40t$

Step-by-step solution

Oscillation of spring-mass system:

As in the case of forced oscillations of a spring-mass system with damping, you can call ${\mathit{Q}}_{\mathbf{p}}$the steady state charge on the capacitor of the LRC circuit. Since $\mathit{I}\mathbf{=}\mathit{Q}\mathbf{\text{'}}\mathbf{=}\mathit{Q}{\mathbf{\text{'}}}_{\mathbf{c}}\mathbf{+}\mathit{Q}{\mathbf{\text{'}}}_{\mathbf{p}}$and $\mathit{Q}{\mathbf{\text{'}}}_{\mathbf{c}}$also tends to zero exponentially as $\mathit{t}\mathbf{\to }\mathbf{\infty }$, we say that ${\mathit{I}}_{\mathbf{c}}\mathbf{=}\mathit{Q}{\mathbf{\text{'}}}_{\mathbf{c}}$is the transient current and ${\mathit{I}}_{\mathbf{p}}\mathbf{=}\mathit{Q}{\mathbf{\text{'}}}_{\mathbf{p}}$is the steady state current.

Finding the steady-state current in an LRC series-circuit:

In the LRC series electric circuit, you have

$L\left(\frac{d^{2}q}{d{t}^2}\right)+R\left(\frac{dq}{dt}\right)+\frac{q}{c}=E\left(t\right)$

So that,

$\frac{1}{2}\left(\frac{d^{2}q}{d{t}^2}\right)+20\left(\frac{dq}{dt}\right)+0.001q=100\sin 60t+200\cos 40t$

Simply,

$\frac{1}{2}\left(\frac{d^{2}q}{d{t}^2}\right)+40\left(\frac{dq}{dt}\right)+2000q=200\sin 60t+400\cos 40t$

Assume that the particular solution is,

$q_p\left(t\right)=A\cos 60t+B\sin 60t+C\cos 40t+D\sin 40t$

And substitute into the differential equation to get the values of $A,B,C,$and $D$. Therefore,

$q_p\left(t\right)=-\frac{3}{52}\cos 60t-\frac{1}{26}\sin 60t+\frac{1}{17}\cos 40t+\frac{4}{17}\sin 40t$

The steady state current is,

localid="1668585064685" $i_p\left(t\right)=-\frac{45}{13}\sin 60t-\frac{30}{13}\cos 60t+\frac{40}{17}\sin 40t+\frac{160}{17}\cos 40t$

This is the final result.