Question

Show that the amplitude of the steady-state current in the LRC –series circuit in Example 10 is given by E₀/Z, where Z is the impedance of the circuit.

Short answer

The amplitude of the steady-state current is $\raisebox{1ex}{$E_0$}\!\left/ \!\raisebox{-1ex}{$Z$}\right.$.

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 $\mathit{Q}{\mathbf{\text{'}}}_{\mathbf{c}}$also tends to zero exponentially as localid="1668585106723" role="math" $\mathit{t}\mathbf{\to }\mathbf{\infty }$, you say that localid="1668582564199" ${\mathit{I}}_{\mathbf{c}}\mathbf{=}\mathit{Q}{\mathbf{\text{'}}}_{\mathbf{c}}$is the transient current and localid="1668585153519" ${\mathit{I}}_{\mathbf{p}}\mathbf{=}\mathit{Q}{\mathbf{\text{'}}}_{\mathbf{p}}$is the steady state current.

Showing that the amplitude of the steady-state current in the LRC –series circuit in is given by EoZ:

As you clearly see that,

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

So that the amplitude is given by,

$\begin{array}{rcl}\raisebox{1ex}{$E_0$}\!\left/ \!\raisebox{-1ex}{$Z$}\right.{\left(\raisebox{1ex}{$R^2$}\!\left/ \!\raisebox{-1ex}{$Z^2$}\right.+\raisebox{1ex}{$X^2$}\!\left/ \!\raisebox{-1ex}{$Z^2$}\right.\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}& =& \raisebox{1ex}{$E_0$}\!\left/ \!\raisebox{-1ex}{$Z$}\right.{\left(\raisebox{1ex}{$R^2+X^2$}\!\left/ \!\raisebox{-1ex}{$Z^2$}\right.\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ & =& \raisebox{1ex}{$E_0$}\!\left/ \!\raisebox{-1ex}{$Z^2$}\right.{\left(R^2+X^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ & =& \raisebox{1ex}{$E_0$}\!\left/ \!\raisebox{-1ex}{$Z^2$}\right.{\left(Z^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ & =& \raisebox{1ex}{$E_0$}\!\left/ \!\raisebox{-1ex}{$Z$}\right.\end{array}$

Hence, the amplitude of the steady-state current is $\raisebox{1ex}{$E_0$}\!\left/ \!\raisebox{-1ex}{$Z$}\right.$.