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When using kirchhoff with a loop with resistors and current source, how should I use kirchhoff voltage law?

for example in this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

I can not use kirchhoff voltage law

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  • \$\begingroup\$ " loop with resistors and current source" -> L1 is not a resistor. \$\endgroup\$ – Rev1.0 Jan 27 '16 at 19:01
  • \$\begingroup\$ sorry, I have edited \$\endgroup\$ – gbox Jan 27 '16 at 19:04
  • \$\begingroup\$ The very same. The sum of voltages in the loop is 0. \$\endgroup\$ – Eugene Sh. Jan 27 '16 at 19:05
  • \$\begingroup\$ @EugeneSh. but \$V=I_1R_1\$ and I and R are non zero \$\endgroup\$ – gbox Jan 27 '16 at 19:10
  • \$\begingroup\$ Right. It's on the resistor. It means that the very same voltage will be on the current source (with an opposite sign). \$\endgroup\$ – Eugene Sh. Jan 27 '16 at 19:12
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You can't use KVL to analyze this circuit.

You probably know you can't use the basic nodal analysis (which applies KCL at every node) in a circuit with voltage sources, and you have to use the modified nodal analysis instead.

Similarly you can't use basic mesh analysis (KVL on every minimal loop) on a circuit with current sources. You'd have to use a modified mesh analysis instead.

That would mean creating a "supermesh" combining two meshes that share the current source. But your circuit doesn't have a second mesh, so you can't do that.

Instead, you should use KCL (or just simple inspection) to analyze your circuit.

Edit

This doesn't mean that KVL doesn't apply to this circuit. It does. KVL tells you that the voltage across the current source is equal to the voltage across the resistor. But you don't know either of these variables, so KVL doesn't help you get the other one.

Once you apply KCL at one of the nodes, you'll know the current through the resistor. And that will tell you the voltage across the resistor. Then KVL will let you get the voltage across the source. But you had to use KCL first to make KVL useful.

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