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I have few queries in analyzing the curves. Below is the flyback converter with RCD snubber circuit and waveforms.

When the FET is turned ON, there is no primary current flowing through snubber as the diode is reverse biased. In ics waveform below, that is current in snubber cap is -Vcs/Rs. Since the diode is reverse biased how does this current appear?

When the FET is turned OFF, the parasitic cap of FET charges rapidly. When Vsw reaches Vin+Vcs the diode is forward biased. The primary current flows through snubber circuits. Here till the snubber cap is charged Vsw will be constant to Vin+Vcs. Why is it constant? and then it suddenly changes to Vin+nVo and settles.

Please help me understand the waveforms.

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In ics waveform below, that is current in snubber cap is -Vcs/Rs. Since the diode is reverse biased how does this current appear?

You should read the article more carefully: -

The snubber capacitor is slowly discharged by the snubber resistor, 
as shown by the negative current, −VCs / Rs.

That's the current that flows that you are referring to I believe.

Vsw will be constant to Vin+Vcs. Why is it constant? and then it suddenly changes to Vin+nVo and settles.

When the transistor deactivates, the transformer primary leakage inductance (\$L_{\ell}\$) charges capacitor Cs via the flyback/snubber diode. The leakage inductance is initially carrying the full drain current and, that unusable energy needs to be "handled" by the snubber.

So, for a short period of time, Cs charges whilst \$L_{\ell}\$ energy is spent. It then remains at a constant value. In fact it is slightly discharged by Rs if you want the fine detail. What happens next is best handled by the words in the linked page: -

Once the leakage energy is fully transferred to the clamp capacitor
at time t = T3, the primary current goes to zero, and the snubber diode 
gets reverse biased as the transistor voltage drops from Vin + VCs 
to Vin + (N1 / N2) × Vo.

There is a little bit of fine detail omitted by the article but, that's the gist of it.

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