Let $p$ the initial phase of $v$

$\frac{1}{C} Q + R \frac{d Q}{d t} + L \frac{d^2 Q}{d t^2}= v e^{i ( \omega t + p )}$

$\frac{1}{C} + R \lambda + L \lambda^2 = 0$

$Q = X e^{i (\omega t + p)}$

$\frac{1}{C} X + R i \omega X + L ( i \omega )^2 X = v$

$X = \frac{v}{\frac{1}{C} - L \omega^2 + i R \omega} \equiv |X| e^{i \theta}$

Note that $X$ is always in III and IV quadrant and $Q$ is before $V$ by $\theta$. When $R=0$, $Q$ is before $V$ by 0 or 180 degree.

$\begin{aligned}|X| = \frac{v}{\sqrt{\left( \frac{1}{C} - L \omega^2 \right)^2 + (R \omega)^2}}\\V_C = \frac{\frac{1}{C \omega} \cdot v}{\sqrt{\left( \frac{1}{C \omega} - L \omega \right)^2 + R^2}}\\I = \frac{d Q}{d t} = i \omega X e^{i (\omega t + p)} = \frac{v}{R + i \left( L \omega - \frac{1}{C \omega} \right)} e^{i (\omega t + p)}\\|I| = \frac{v}{\sqrt{ \left( \frac{1}{C \omega} - L \omega \right)^2 + R^2}}\\V_R = \frac{R v}{\left(\frac{1}{C \omega} - L \omega \right)^2 + R^2}\\\frac{d^2 Q}{d t^2} = - \omega^2 X e^{i (\omega t+p)}\\V_L = \frac{L \omega v}{\sqrt{\left(\frac{1}{C \omega} - L \omega \right)^2 + R^2}}\end{aligned}$

Note that

$(V_C - V_L)^2 + V_R^2 = v^2$

The general solution is for some constants $c_1$ and $c_2$ and $\lambda_1$ the eigen value of smaller absolute value

$Q = X e^{i ( \omega t + p )} + c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}$

$I = i \omega X e^{i ( \omega t + p )} + c_1 \lambda_1 e^{\lambda_1 t} + c_2 \lambda_2 e^{\lambda_2 t}$

When capacitor does not exist, aka, $C$ is infinite,

$L \omega = R \sqrt{\left( \frac{v}{V_R} \right)^2 - 1}$

Note that $I$ is before $Q$ (therefore $V_C$ ) 90 degree.

Typical initial condition is that $Q$ and $I$ are zero because no charge in the capacitor. If there is no capacitor, initial condition about $Q$ can be omit, so instead, the initial condition is $I =0$ only

$\begin{aligned}0 = X e^{i p} + c_1 + c_2\\0 = i \omega X e^{i p} + c_1 \lambda_1 + c_2 \lambda_2\\c_1 = - \frac{i \omega X e^{i p} - \lambda_2 X e^{i p}}{\lambda_1 - \lambda_2}\\c_2 = - \frac{i \omega X e^{i p} - \lambda_1 X e^{i p}}{\lambda_2 - \lambda_1}\\c_1 \lambda_1 e^{\lambda_1 t} + c_2 \lambda_2 e^{\lambda_2 t} = \frac{i \omega X \lambda_1 e^{\lambda_1 t} e^{i p} - \lambda_2 X \lambda_1 e^{\lambda_1 t} e^{i p}}{\lambda_2 - \lambda_1} + \frac{-i \omega X \lambda_2 e^{\lambda_2 t} e^{i p} + \lambda_1 X \lambda_2 e^{\lambda_2 t} e^{i p}}{\lambda 2 - \lambda 1} \lambda_1 e^{\lambda_1 t} \\= \frac{i \omega X ( \lambda_1 e^{\lambda_1 t} - \lambda_2 e^{\lambda_2 t} ) e^{i p} + \lambda_1 X \lambda_2 ( e^{\lambda_2 t} - e^{\lambda_1 t} ) e^{i p}}{\lambda_2 - \lambda_1} \\= \frac{i \omega X ( \lambda_1 e^{\lambda_1 t} - \lambda_2 e^{\lambda_2 t} ) e^{i p} + \frac{X}{L C} ( e^{\lambda_2 t} - e^{\lambda_1 t} ) e^{i p}}{\lambda_2 - \lambda_1}\end{aligned}$

And:

$\begin{aligned}I = \left( i \omega X e^{i \omega t} + \frac{i \omega X \left( \lambda_1 e^{\lambda_1 t} - \lambda_2 e^{\lambda_2 t} \right) + \frac{X}{L C} \left( e^{\lambda_2 t} - e^{\lambda_1 t} \right)}{\lambda_2 - \lambda_1} \right) e^{i p}\\ \mid \frac{I}{\omega X} \mid = \mid i e^{i \omega t} + \frac{i \left( \lambda_1 e^{\lambda_1 t} - \lambda_2 e^{\lambda_2 t} \right) + \frac{1}{L C \omega} \left( e^{\lambda_2 t} - e^{\lambda_1 t} \right)}{\lambda_2 - \lambda_1} \mid \\< \mid i e^{i \omega t} \mid + \mid \frac{i \left( \lambda_1 e^{\lambda_1 t} - \lambda_2 e^{\lambda_2 t} \right)}{\lambda_2 - \lambda_1} \mid + \mid \frac{\frac{1}{L C \omega} \left( e^{\lambda_2 t} - e^{\lambda_1 t} \right)}{\lambda_2 - \lambda_1} \mid\end{aligned}$

Note that

$\begin{aligned}\mid e^{\lambda_2 t} - e^{\lambda_1 t} \mid < 2\\ \mid \lambda_1 e^{\lambda_1 t} - \lambda_2 e^{\lambda_2 t} \mid = \mid \lambda_1 e^{\lambda_1 t} - \lambda_1 e^{\lambda_2 t} + \lambda_1 e^{\lambda_2 t} - \lambda_2 e^{\lambda_2 t} \\< \mid \lambda_1 (e^{\lambda_1 t} - e^{\lambda_2 t}) \mid + \mid (\lambda_1 - \lambda_2) e^{\lambda_2 t} \mid \\< 2 |\lambda_1| + |\lambda_1 - \lambda_2|\end{aligned}$

Therefore,

$\mid \frac{I}{\omega X} \mid < 2 + 2 \mid \frac{\lambda_1}{\lambda_2 - \lambda_1} \mid + \frac{2}{L C \omega \mid \lambda_2 - \lambda_1 \mid}$

Transformer example. $R = 1 . 5 \Omega , C = \infty$ , rating 3000 watt 220 V, then $L \Omega = \frac{220^2}{3000} = 16 .13 \Omega , v = 220 \cdot \sqrt{2} = 311$

$\begin{array}{l}\lambda_1 = 0\\\lambda_2 = - \frac{R}{L}\\\omega |X| = \frac{v}{\sqrt{(L \omega)^2 + R^2}}\\I = \omega |X| \left( i e^{i ( \omega t + \theta + p )} - i e^{-\frac{R}{L} t} e^{i ( \theta + p )} \right)\\\frac{d I}{d t} = \omega |X| \left(- \omega e^{i (\omega t + \theta + p)} + i \frac{R}{L} e^{- \frac{R t}{L}} e^{i ( \theta + p )} \right)\end{array}$

Take the real parts,

$\begin{array}{l}\Re(I) = \omega |X| \left( - \sin(\omega t + \theta + p) + e^{\frac{-R}{L} t} \sin(\theta + p) \right)\\\Re \left( \frac{d I}{d t} \right) = \omega |X| \left( - \omega \cos(\omega t + \theta + p) - \frac{R}{L} e^{\frac{-R}{L} t} \sin(\theta + p) \right)\end{array}$

Steady energy power:

$\begin{array}{l}E_R = \omega |X| \cdot \sin(\omega t + \theta + p) \cdot \omega |X| \cdot \sin(\omega t + \theta + p) \cdot R = \frac{v^2 R}{(L \omega)^2 + R^2} \sin^2(\omega t + \theta + p)\\E_L = \omega^3 |X|^2 L \sin(\omega t + \theta + p) \cdot \cos(\omega t + \theta + p) \cdot L = \frac{v^2 L \omega}{(L \omega)^2 + R^2} \cdot \frac{\sin(2 ( \omega t + \theta + p )}{2}\\E_C = 0\end{array}$

If $\theta + p$ is near zero, i.e. the voltage starts applied at peak, it is safe to start the transformer. If $\theta + p$ is near $\frac{\pi}{2}$ or $\frac{3 \pi}{2}$ , i.e. the voltage starts applied at zero, then it will trigger huge current and fries the electronic parts. See for example of 240 VAC 50 Hz 0.02 R 0.03 L.

The solutions are:

1. change v gradually so that $\frac{2 v}{L \omega}$ is small aka "soft start". For example, to minimize the current  less than 10A, change the voltage by steps of $5 L \omega$ and wait for current stabilizing then repeat.
2. Change $R$ gradually by some negative coefficient variable resistance. See The Simplest Soft Starter