Question

Use problem 54 to show that 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 V, is given by ip(t)= 4.160 sin(60t – 0.588).

Short answer

The steady state current in LRC Circuit is $i_p\left(t\right)=4.160\sin \left(60t-0.588\right)$.

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 Q′_c also tends to zero exponentially as $\mathit{t}\mathbf{\to }\mathbf{\infty }$, you 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.

The steady-state current in an LRC-series circuit:

According to results of Example 10, and recall Problem 54,

$i\left(t\right)=\raisebox{1ex}{$E_0$}\!\left/ \!\raisebox{-1ex}{$Z$}\right.\sin \left(\gamma t-\phi \right)$

Where $\phi ={\tan }^{-1}\left(\raisebox{1ex}{$X$}\!\left/ \!\raisebox{-1ex}{$R$}\right.\right)$. From the given $\gamma =60,E_0=100,$and $R=20$. Also,

$\begin{array}{rcl}X& =& L_{\gamma }-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_{\gamma }$}\right.\\ & =& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(60\right)-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$0.01$}\right.\left(60\right)\\ & =& \frac{40}{3}\end{array}$

And

role="math" localid="1668583677144" $\begin{array}{rcl}Z& =& (X^2+R^2)\\ & =& \left[{\left(\raisebox{1ex}{$40$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}^2+{\left(20\right)}^2\right]\\ & =& 20{\left(\raisebox{1ex}{$13$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$

Thus,

$\begin{array}{rcl}\raisebox{1ex}{$E_0$}\!\left/ \!\raisebox{-1ex}{$Z$}\right.& =& \raisebox{1ex}{$15$}\!\left/ \!\raisebox{-1ex}{$\sqrt{13}$}\right.\\ & =& 4.160\end{array}$

And

$\begin{array}{rcl}\phi & =& {\tan }^{-1}\left[\frac{\left(\raisebox{1ex}{$40$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}{20}\right]\\ & =& {\tan }^{-1}\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)\\ & =& 0.588\end{array}$

Hence,

$i_p\left(t\right)=4.160\sin \left(60t-0.588\right)$

This is the the steady state current in LRC Circuit.