# 4.13: Kirchhoff’s Rules

- Page ID
- 5979

There are two h’s in his name, and there is no *tch* sound in the middle. The pronunciation is approximately keerr–hhofe.

The rules themselves are simple and are self-evident. What has to be learned, however, is the art of using them.

- K1: The net current going into any
*point*in a circuit is zero; expressed otherwise, the sum of all the currents entering any point in a circuit is equal to the sum of all the currents leaving the point. - K2: The sum of all the EMFs and \(IR\) products in a
*closed circuit*is zero. Expressed otherwise, as you move around a closed circuit, the potential will sometimes rise and sometimes fall as you encounter a battery or a resistance; but, when you come round again to the point where you started, there is no change in potential.

In the above circuit, the 24 V battery is assumed to have negligible internal resistance. Calculate the current in each of the resistors.

The art of applying Kirchhoff’s rules is as follows.

- Draw a large circuit diagram
*in pencil*. - Count the number of independent resistors. (Two is series with nothing in between don’t count as independent.) This tells you how many independent equations you can obtain, and how many unknowns you can solve for. In this case, there are five independent resistors; you can get five independent equations and you can solve for five unknowns.
- Mark in the unknown currents. If you don’t know the directions of some of them, don’t spend time trying to think it out. Just make a wild guess. If you are wrong, you will merely get a negative answer for it. Those who have some physical insight might already guess (correctly) that I have marked
*I*_{5}in the wrong direction, but that doesn’t matter. - Choose any closed circuit and apply K2. Go over that closed surface
*in ink*. Repeat for several closed circuits until the entire diagram is inked over. When this happens, you cannot get any further independent equations using K2. If you try to do so, you will merely end up with another equation that is a linear combination of the ones you already have, - Make up the required number of equations with K1.

Let us apply these to the present problem. There are five resistors; we need five equations. Apply K2 to OACBO. Start at the negative pole of the battery and move counterclockwise around the circuit. When we move up to the positive pole, the potential has gone up by 24 V. When we move down a resistor in the direction of the current, the potential goes down. For the circuit OACBO, K2 results in

\[24-3I_1-2I_3=0\nonumber\]

Now do the same this with circuit OADBO:

\[24-I_2-8I_4=0,\nonumber \]

and with circuit ACDA:

\[3I_1+4I_5-I_2=0\nonumber \]

If you have conscientiously inked over each circuit as you have done this, you will now find that the entire diagram is inked over. You cannot gain any further independent equations from K2. We need two more equations. Apply K1 to point C:

\[I_1=I_3+I_5,\nonumber \]

and to point D:

\[I_4=I_2+I_5\nonumber \]

You now have five independent linear equations in five unknowns and you can solve them. (Methods for solving simultaneous linear equations are given in Chapter 1, Section 1.7 of Celestial Mechanics.) The solutions are:

\[I_1=+4.029A,\,I_2=+4.380A,\,I_3=+5.956A,\,I_4=+2,453A,\,I_5=-1.927A.\nonumber\]