Yesterday’s post was about Kirchoff’s Current Law.

Now let’s look at the second version of that law, Kirchhoff’s Voltage law:

This law says something about the voltages in a circuit: voltages cancel out, *regardless* of which way you follow a path in the circuit, i.e.:

```
V1 - V2 - V3 = 0 (hence V1 = V2 + v3)
```

But also:

```
V3 - V4 - V5 = 0 (hence V3 = V4 + V5)
```

And:

```
V1 - V2 - V4 - V5 = 0 (hence V1 = V2 + V4 + V5)
```

The signs are based on how the arrows are drawn: voltages are always *relative*, so it all depends on the direction of each of those arrows. Since V2 and V3 are opposite to the direction of “following” a single path through the circuit, they end up negative in the above equation. Same for V4 and V5.

Now *this version* of Kichhoff’s law is really useful. If each of the resistors has the same value (it doesn’t matter what the value is), then we can reason about the voltages in different parts of the circuit, even when only V1 is known at the start.

Let’s look at this circuit again in terms of voltages, with V1 set to 3V:

Since V1 = V2 + V3, and all resistors have the same value, it’s a safe guess that the voltage over both will be the same. A more exact reasoning is to include the currents, and since they are the same for the two resistors on the left, we know that the voltages will be the same (Ohm’s law, regardless of the value of these resistors – as long as they are the same).

Perhaps a bit more surprising is that V4 is 0V. This is again a consequence of Ohm’s law: since the pins on the right are not connected, no current will flow. With no current, the voltage over a resistor is always 0V (Ohm’s law, again).

Since V4 is 0V, and V2 = V4 + V5 as we saw above, V5 must be 1.5V, the same as V2.

Note how a voltage over a resistor can be zero, and over two open pins can be non-zero.

If we use a multi-meter with a really high input resistance (any modern multi-meter will do, usually they have 10 MΩ or more), to make sure no current starts flowing, then we can indeed measure the voltage to be half of the 3V between top and bottom, i.e. 1.5V.

The main point to take away from this, is that the three laws are intimately inter-connected. Given known resistors, which is often the case, Ohm’s law *dictates* the relationship between voltage and current, and with the two variants of Kirchhoff’s law, it is usually possible to “reason your way into” the different points in a circuit, given an externally applied power supply.

One last example: what is the equivalent resistance of two resistors of 1000 Ω in *parallel?* Well, say we apply voltage V1, then the current through one resistor will be V1/1000. The same current will go through the other resistor, since it too has V1 on both pins. So the total current of the two in parallel will be 2 x V1 / 1000 = V1 / 500. What is the resistance when we apply V1 and see a current V1 / 500 flowing? Again, Ohm’s law: R = U / I, i.o.w. R = V1 / (V1 / 500) = 500. So the general answer is: the equivalent resistance of two resistors in parallel is *half* their resistance. A direct consequence of Ohm’s and Kirchhoff’s laws.

*Get used to the three laws, and with a bit of practice, the rest should fall into place!*

I don’t remember Kirchoff – but they taught us Thévenin’s

29 April 2013at12pm