After Mr. Ohm yesterday, let’s see what Mr. Kirchhoff has to say.

Kirchhoff’s First Law is about current. In a nutshell – what goes in, must come out:

I’ve drawn resistors, but that’s really irrelevant here – each component could be anything.

The current I1 must be the same as the current I2, It does not matter what I3 is in this case (assuming the two pins next to it are not feeding or drawing any outside current).

Also, if I3 is zero, i.e. if the pins are not connected to anything at all, then the current through the leftmost resistors is identical. In this case, they are essentially in series, with the resistor on the right not doing anything at all (no current = no voltage drop = Ohm’s Law).

This makes it possible to reason about that point in the middle, where the three resistors meet. The currents at that point *must* cancel out: that’s what Kirchhoff’s Current Law says.

Suppose all three resistors are 1 kΩ, and the current I1 is 1 mA:

If the two pins on the right are left open, no current will flow there. So the same current I1 (which is also the same as I2) will flow through both resistors on the left. Total voltage drop from top to bottom will be 1 mA x 1 kΩ = 1V on the top resistor and another 1V on the bottom one, for a total voltage of 2V across the left two pins.

Or to put it more practically: if you place a 2V supply across those left two pins, then 1 mA will flow. The voltage in the center point will be halfway, i.e. at 1V.

What will happen when we short the pins on the right?

Again, there’s 1 mA flowing in from the top, so there will be 1 mA coming out the bottom. The bottom-left and right resistor will *together* see a current of 1 mA going through them. Since they are both the same 1 kΩ, it should not come as a surprise that each resistor will get half the current, i.e. 0.5 mA each. Total voltage from top to bottom will be 1.5V.

This isn’t such a great example in terms of practical use, since normally the reasoning goes the other way around: what current will flow when I apply voltage X to the entire circuit?

*That’s the other version of this law, described tomorrow.*

I had problems in the beginning to understand this law, especially with more complex scenarios including 3 to 4 different power sources.

But an easy hint: Resistors in series are Voltage dividers the current stays the same. Resistors in parallel are Current dividers the voltage stays the same.

This simple rules often helped me to survive school tests.

28 April 2013at11amNice tip. For me, personally, the above three laws feel easier, but that’s just me – use whatever works!

28 April 2013at10pmHi Jcw

Imho it depends, with three resistors everything is easy, but when you have to solve a circuit with eight to ten unknowns you have to very careful setting up your system equations. On the other hand reducing the circuit to a resistor when possible then calculating current and afterwards handle yourself backward seemed much more natural to me, you even could approximate the outcome using mental arithmetic.

29 April 2013at9am