Physics - EMI/AC Question with Solution | TestHub

In the initial LR circuit, after a long time, the current reaches its steady-state value $I_0=E/R$.

$I_0=10\text{V}/10\Omega =1\text{A}$. This is the initial current in the inductor when the RLC circuit is formed. The initial charge on the capacitor is $q(0)=0$. For a damped RLC circuit, the charge on the capacitor is given by $q(t)=Q_0{e}^{-\alpha t}\cos ({\omega }^{\prime }t+\varphi )$, where $\alpha =R/(2L)$ and ${\omega }^{\prime }=\sqrt{\frac{1}{LC}-{\alpha }^2}$. First, calculate $\alpha$: $\alpha =10/(2\times 0.5)=10{\text{s}}^{-1}$. Next, calculate ${\omega }^{\prime }$: $\frac{1}{LC}=\frac{1}{0.5\times 20\times 10^{-6}}=\frac{1}{10\times 10^{-6}}=10^5{\text{s}}^{-2}$.${\omega }^{\prime }=\sqrt{10^5-(10{)}^2}=\sqrt{100000-100}=\sqrt{99900}\text{rad/s}$. Using initial conditions: $q(0)=0⟹Q_0\cos (\varphi )=0⟹\varphi =\pi /2$. The current is $i(t)=-dq/dt=-Q_0{e}^{-\alpha t}[-\alpha \cos ({\omega }^{\prime }t+\varphi )-{\omega }^{\prime }\sin ({\omega }^{\prime }t+\varphi )]$. At $t=0$, $i(0)=I_0=-Q_0[-\alpha \cos (\pi /2)-{\omega }^{\prime }\sin (\pi /2)]=-Q_0[0-{\omega }^{\prime }]=Q_0{\omega }^{\prime }$. Therefore, the maximum charge $Q_0=I_0/{\omega }^{\prime }=1\text{A}/\sqrt{99900}\text{rad/s}$.$Q_0=1/(10\sqrt{999})\approx 1/(10\times 31.60696)\approx 0.0031638\text{C}$. In microcoulombs, $Q_0\approx 3163.80\mu \text{C}$.