# How does Kirchhoff's current law take resistance into consideration? [closed]

I just drew a small diagram on what I mean.

Based on Kirchhoff's current law, in a junction, the current entering a junction must equal to the current leaving the junction.

In the picture the originating wire has 240V voltage, 1ohm resistance, so 240A current.

The wire splits into three wires. Until each wire reaches the resistors, they have 80V, 1ohm, 80A each. However, after resisting the current with 10ohm, they reduce from 80A to 8A.

When merging the branches in the junction, the current leaving the junction will have only 24A, which isn't equal to the 240A that was entering it.

What am I missing here?

• The example you created is not one that can actually exist. Apr 18 at 15:09
• Can you draw it as a circuit? Apr 18 at 15:11
• A voltage source does not force a specific current through a circuit - the current will be determined by the resistance of the circuit. Your 240 V source may be capable of delivering 240 A, but the 13.3 Ohm resistance in the rest of the circuit will limit the current actually drawn, and the current passed by each part will determine the voltage across that part. Apr 18 at 16:57

Draw it in a more practical circuit, where the circuit is complete and it will make more sense. The block diagram does not describe a practical circuit. It does not make sense that the block would send out a current without return path.

Series and parallel resistors will affect the total resistance in the circuit. The resistors will also affect what the voltage is in each node.

But yes, Kirchoffs law will apply also in circuits with resistors.

In the picture the originating wire has 240V voltage, 1ohm resistance, so 240A current.

Wires have resistance, they do not have voltage or current, they carry current and drop voltage dependent on the current and resistance.

So when you say a wire has 240 V, where is that voltage coming from? There must be a voltage or current source somewhere, and for current to flow there must be a closed loop which you do not have.

Then you say there are 10 $$\\Omega\$$ resistors with 80 V across them and 8 A through each one, where do you get this? You can't just arbitrarily assign voltages and currents in a circuit. Even if you had a closed loop the currents and voltages are going to depend on the sources and resistances in the circuit.

As others have suggested you need to draw a complete closed circuit with voltage and/or current sources along with the resistances. Only then can you start to apply Kirchhoff's laws.