i dont know hoe to apply kvl to the circuit?
KVL has nothing to do with linearity -- all it says is that the algebraic sum of all voltages around a closed loop equals zero -- that is it. Simlarly, KCL says that the algebraic sum of all currents flowing into a node equals zero. There is no mention of linearity here at all. Given any point in time, both of these laws must hold no matter what the circuit elements are.
As I posted before, linearity only comes in when you are trying to solve for voltages and currents that is contained in these equations. See the attached circuit.
You can easily write the KCL equation to solve for the unknown node voltage \[v_x\]:
\[ I_1 = i_{D1} + \frac{v_x}{R}\]
Now we note that \[i_{D1}=I_s (e^{v_x/n V_T} - 1)\]. If you substitute with the values given above, you get
\[ 500mA = I_s (e^{v_x/n V_T} - 1) + v_x/1\]
This is a nonlinear equation, but it is nonlinear because of the diode. Just because you have a nonlinear device does not mean you cannot apply KCL.
Here \[I_s\] is the saturation current, \[V_T\] is the thermal voltage (about 26mV at room temperature), \[n\] is the emission coefficient (usually 1 to 2). Now, you can see that you have to solve this numerically -- that is, try to get a value of the voltage \[v_x\] so the right hand and left hand sides are equal. If you use \[n=1, V_T=0.026V, and I_s=6.9\mu A\]. So the equation to solve is
\[ 500mA = 6.9\mu A (e^{v_x/0.026V} - 1) + v_x\]
Solving this I get, \[v_x=271mV\]. Again, KCL has to do with writing down the first equation. Its actual solution (be it linear or nonlinear) is not a limitation of KCL itself at all.
I hope this is helpful.
By the way, this is how spice would solve the circuit shown. Just put the circuit in
and do an operating point (.op) analysis. Just put the diode of your choice in there in case you don't have the same model available.
Best regards,
v_c