Snubber circuit is a series connection of $R_s$ and $C_s$ and is connected parallel to SCR.L is connected in series with load resistor $R_1$ for $di/dt$ protection.

The capacitor $C_s$ behaves like a short circuit and SCR is in the forward blocking state offers a very high resistance as shown in fig. b.

The voltage equation for fig.b is

$$V_{s}=i\left(R_{s}+R_{L}\right)+L \frac{d i}{d t}-----(1)$$

Apply laplace and inverse laplace transform and Simplifying, the solution of equation (1) is

$$i=I\left(1-e^{-t / \tau}\right)$$

Where $I = \frac {V_S}{R_s} +R_L$ and $\zeta = \frac{L}{R_s}+R_L-----(2)$

Differentiate equation (2) wrt 't' we get

$$\frac{d i}{d t}=I\left[0-e^{-t / \tau} \cdot-\frac{1}{\tau}\right]$$

$$\frac{d_i}{d_t}=I e^{-t / \tau} \cdot \frac{1}{\tau}-----(3)$$

Substituting I and $\tau$ values in equation(3), we get

$$\frac{d i}{d t}=\frac{V_{s}}{R_{s}+R_{L}} e^{-t / \tau} \cdot \frac{1/L}{R_{s}+R_{L}}$$

$$\frac{d i}{d t}=\frac{V_{s}}{R_{S}+R_{L}} e^{-t / \tau} \cdot \frac{R_{s}+R_{L}}{L}$$

$$\frac{d i}{d t}=\frac{V_{s}}{L} e^{-t / z}$$

The value of $d_i/d_t$ is maximum when $t=0$

$$\left(\frac{d i}{d t}\right)_{\operatorname{max}}=\frac{V_{s}}{L} e^{-0 / \tau}$$

$$\left(\frac{d i}{d t}\right)_{\max }=\frac{V_{s}}{L}$$

$$\therefore L=\frac{V_{\mathrm{s}}}{\left(\frac{d i}{d t}\right)_{\mathrm{max}}}$$

The voltage across SCR is given by,

$$V_{A K}=i R_{s}-----(4)$$

Differentiate equtaion (4) w.r.t 't'

$$\frac{d V_{A K}}{d t}=R s \frac{d i}{d t}$$

$$\left(\frac{d V_{A K}}{d t}\right)_{\max }=R_{S} \cdot\left(\frac{d i}{d t}\right)_{\max }$$

Substituting $\left(\frac{d i}{d t}\right)_{\operatorname{max}}$ value in above equation

$$\left(\frac{d V_{A K}}{d t}\right)_{\max }=R s \cdot \frac{V_{s}}{L}$$

$$\therefore R_{s}=\frac{L}{V_{s}}\left(\frac{d V_{A x}}{d t}\right)_{\max }$$

Capacitor $C_s$ can be calculated by using the formula

$$C_{S}=\frac{1}{2 L}\left[\frac{0.564 \mathrm{V_m}}{\frac{d v}{d t}}\right]^{2}$$

Where $V_m$ is the peak input voltage.