The circuit equations for the $L C$ circuit are: $$ L \frac{\mathrm{d} i}{\mathrm{d} t}+\frac{1}{C} \int i \mathrm{d} t=0 $$

$$\therefore \quad L \frac{\mathrm{d}^{2} i}{\mathrm{d} t^{2}}+\frac{1}{C} i(t)=0$$

Taking laplace transform of the above equation, $\left(S^{2} L+\frac{1}{C}\right) I(s)=0$

$$\begin{array}{ll}{\therefore \qquad i(t)} & {=E_{\mathrm{dc}} \sqrt{\frac{c}{L}} \sin \omega_{0} t} \\ {\text { where }} & {\omega_{0}=\sqrt{\frac{1}{L C}}}\end{array}$$

Therefore, the peak commutation current is

$$I_{C_{\text { (peak) }}}=E_{\text { dc }} \cdot \sqrt{C / L}$$

For this Class B commutation method, the peak discharge current of the capacitor is assumed to be twice the load-current $I_{L}$ and the time for which the SCR is reverse biased is approximately equal to one-quarter period of the resonant circuit.

Therefore, $$I_{C_{\text { (peak) }}}=2 I_{L}=E_{\text { dc }} \sqrt{C / L}$$

And $$\qquad t_{\text { off }}=\frac{\pi}{2} \sqrt{L C}$$