Current Equivalence at a Circuit Node

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Dustinsfl
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Given the circuit below:

4BotojH.png


Why does KCL equate to
\[
\frac{e - v_c}{R_1} = C\frac{dv_c}{dt} + f(v_c) + i_L
\]
 
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dwsmith said:
Given the circuit below:

4BotojH.png


Why does KCL equate to
\[
\frac{e - v_c}{R_1} = C\frac{dv_c}{dt} + f(v_c) + i_L
\]

KCL says that in any node current in equals current out.

In the node at the top, the current through $R_1$ is coming in, which must therefore be equal to the current going out and into the capacitor plus the current through the non-linear resistor plus the current through the coil.
 
I like Serena said:
KCL says that in any node current in equals current out.

In the node at the top, the current through $R_1$ is coming in, which must therefore be equal to the current going out and into the capacitor plus the current through the non-linear resistor plus the current through the coil.

I still don't quite understand. Can you be more specific by node at the top and the currents you mention?