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I am a beginner at circuit analysis and cannot find a good answer to this question. I know that for a current source voltage is dependent on the rest of the circuit, but the same can be said of a voltage source and current. When we do KCL we consider the current on a voltage source, but when we do KVL we do not consider voltage on a current source.

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    \$\begingroup\$ Current is not "across", it's through. \$\endgroup\$
    – Hearth
    Commented Apr 18, 2023 at 5:41

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I'm not sure why you think KCL concerns itself with voltages anywhere, or KVL somehow needs to know anything about current anywhere. When applying KCL, we consider only current. When applying KVL, we consider only voltage.

Often we use a combination of Ohm's law and KVL in a single step, which might give the impression that we are considering current during an application of KVL:

schematic

simulate this circuit – Schematic created using CircuitLab

To save steps in the algebra, you might write KVL applied to this loop as:

$$ V_1 - (I_3\times R_2) -V_3 = 0 $$

The term \$I_3\times R_2\$ is an application of Ohm's law mixed in with KVL. The truth is, though, that KVL in and of itself doesn't say anything more than:

$$ V_1 - V_2 -V_3 = 0 $$

There is absolutely no consideration or mention of any current anywhere in that expression.

A similar mix of laws might be used to save steps in an application of KCL. Here we know the voltage across R1, and can use that information to our advantage:

schematic

simulate this circuit

KCL applied to node A, if we sneak in a cheeky application of Ohm's law, would say something like:

$$ \frac{4V}{R_2} = I_2 + I_3 $$

Again, this gives the impression that KCL somehow concerns itself with voltage, since there is a voltage term in that equation, but in reality it doesn't. KCL's only claim is that:

$$ I_1 = I_2 + I_3 $$

This is regardless of R1, or the voltage across it, or R2, or V3 or anything else.

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  • \$\begingroup\$ RE "When applying KCL, we consider only current.", Typically the next step is to write the currents as a function of the node voltages, and then the node voltages become the unknowns in the system of variables we need to solve. \$\endgroup\$
    – The Photon
    Commented Apr 18, 2023 at 17:48

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