I'm trying to come up with the state-space equations for this circuit, but I think I'm making a mistake when writing KVL through a loop with a current source.


simulate this circuit – Schematic created using CircuitLab

Is it valid to write for the top loop:

V_L + V_C = 0

where V_L is the voltage across the inductor and V_C is the voltage across the capacitor.

Only writing KVL for the bottom loop doesn't give me enough equations to solve for the state space variables (VL, VC, and V2C)

  • \$\begingroup\$ No. A current source is an open circuit, not a closed one. To find out the voltage across L1, the current and the value of L1 is enough to know. To find out about the voltage across C, you have to calculate a current divider between C and 2C+R1. \$\endgroup\$ – Janka Jun 19 '17 at 0:19
  • \$\begingroup\$ You cannot assume the voltage across the current source is zero, so the voltage must be included when doing KVL. V_L + V_C + V_I1 = 0. V_I1 can we written in terms of I1 and the other impedances. \$\endgroup\$ – mkeith Jun 19 '17 at 1:08
  • \$\begingroup\$ @Janka, an ideal current source isn't an open circuit (if it were, there would be no current through). However, if you zero an ideal current source, you have an open circuit. Amir, the voltage across the current source is independent of the current through, i.e., it is entirely determined by the connected circuit; you cannot set it zero but must leave it as a variable. Do keep in mind that the mesh current in the top loop is completely determined by the current source so that might be another approach to try. \$\endgroup\$ – Alfred Centauri Jun 19 '17 at 2:14
  • \$\begingroup\$ Is \$I_1\$ a function of time, or not? And is 2C really \$C_2\$? Or do you mean it is twice the value of the other capacitor? \$\endgroup\$ – jonk Jun 19 '17 at 3:02
  • \$\begingroup\$ Amir, your current source determines the current in \$L_1\$, whether time-varying or not. The energy in \$L_1\$ isn't an independent state variable. So assuming \$I_1\$ is an input (it must be, I believe), you only have two independent state variables to worry over, not the three you listed at the end of your question. \$\endgroup\$ – jonk Jun 19 '17 at 15:18

The equation you wrote is not valid as you are neglecting the voltage of current source. Instead, you can use KCL to solve the circuit.

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