Example on Kirchhoff's Laws
Extension only
This example shows how Kirchhoff 1 and II can be used to solve circuits that can't be solved by simple series and parallel circuit techniques.
What are the currents I1, I2 and I3 in this circuit?
Kirchhoff I : I1 + -I2 + I3 = 0 (This is because I1 and I3 are coming into the top junction and I2 is leaving. Currents coming in are positive and currents going out are negative)
Kirchhoff II: Sum of pds round a closed loop is 0.
We will do a clockwise journey round loop A, travelling from positive to negative.
Kirchhoff II states Spd = 0 :
I1 x 0 ohms + - I3 x 6 ohms + -3.0 V = 0
- I3 x 6 ohms = 3.0 V
I3 = - 0.5 A
The minus sign tells us that the current is flowing against the direction indicated by the arrow. This does not matter, as long as we are consistent.
Now let's do loop B:
I2 x 0 ohms + I3 x 6 ohms + [I1 + I3] x 2 ohms + -5 V = 0
-0.5A x 6 ohms + [I1+ -0.5 A] x 2 ohms = 5 V
[I1 +-0.5 A] x 2 ohms = 5 V + 3 V = 8 V
[I1+ -0.5 A] = 4.0 A
I1 = 4.5 A
So now we can work out I2 with Kirchhoff I:
I1 + -I2 + I3 = 0
4.5 A + -I2 + -0.5 A = 0
I2 = 4.0 A
You may well have found this quite hard and these problems make me scratch my head a bit. Remember that you will NOT encounter a problem like this at AS, and it's only likely that you will come across this in Advanced Extension Awards. However the example shows how Kirchhoff's Laws can be used to solve difficult circuit problems.