The following simple circuit has \$V1\$ as a \$DC \$ voltage supply and \$C1\$ is a linear capacitor and \$C2\$ is non-lineaer capacitor.

It is known that:

\$q_1 = C_1\cdot V_{c1} \$ , hence a linear capacitor

\$q_2 = \alpha\cdot \sqrt{V_{c2}} \$ , non-linear capacitor

Since both capacitors are in series, thus the same current flows through them, is it correct to clime they has the same amount of charge (i.e. \$q_1 = q_2\$) after some \$t \to \infty\$?


simulate this circuit – Schematic created using CircuitLab


Of course, both capacitors will have the same current flowing through them and therefore they will end up having the same charge. Everything else would violate KCL.

The difference will be the voltage of the capacitors and the fact that the voltage ratio is not constant but dependent on the nonlinearity.

Your circuit does not contain a resistor, therefore it is problematic as the charging of the capacitors will happen instantly.

  • \$\begingroup\$ could you please explain how would it violate KCL? \$\endgroup\$ – ThunderWiring Sep 15 '16 at 9:31
  • \$\begingroup\$ "The algebraic sum of all currents entering and leaving a node must equal zero". When you integrate the current you get the charge. Therefore the same current results in the same charge. \$\endgroup\$ – Mario Sep 15 '16 at 9:36
  • \$\begingroup\$ I agree, however, since we have the non-linearity, i don't understand why it still holds? \$\endgroup\$ – ThunderWiring Sep 15 '16 at 9:41
  • 1
    \$\begingroup\$ KCL does not depend on linearity. \$\endgroup\$ – Mario Sep 15 '16 at 9:48

Lets assume both capacitors start off fully discharged then they each have a charge of zero coulombs.

Now with the capacitors in series we apply some voltage and the capacitors charge.

Charge is the integral of current with respect to time.

$$Q = \int I dt$$

The same current has flown in both capacitors at all times so they must have equal charge. Because one capacitor is non-linear they will not have the same voltage across them however.


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