Kirchhoff’s First Law

The physicist Gustav Robert Kirchhoff (1824-1887) was a researcher and experimentalist in electricity back in the time before radio, before electric lighting, and before much was understood about how currents flow.

Kirchhoff reasoned that current must work something like water in a network of pipes, and that the current going into any point has to be the same as the current going out. This is true for any point in a circuit, no matter how many branches lead into or out of the point. Two examples are shown in Fig. 5-5.

In a network of water pipes that does not leak, and into which no water is added along the way, the total number of cubic feet going in has to be the same as the total volume going out. Water can’t form from nothing, nor can it disappear, inside a closed system of pipes. Charge carriers, thought Kirchhoff, must act the same way in an electric circuit.
This is Kirchhoff’s First Law. An alternative name might be the law of conservation of current.

Example 1:
Refer to Fig. 5-5A. Suppose all three resistors have values of 100 Ω, and that I1 =
2.0 A while I2 = 1.0 A. What is the battery voltage?

First, find the current I drawn from the battery. It must be 3.0 A;
I = I1 + I2
= 2.0 +1.0
= 3.0 A.
Next, find the resistance of the whole combination. The two 100-Ω resistors in series give a value of 200 Ω, and this is in parallel with 100 Ω. You can do the calculations and find that the total resistance, R, across the battery, E, is 66.67 Ω. Then
E = IR
= 66.67 x 3.0
= 200 volts. (Some battery.)

Example 2
In Fig. 5-5B, suppose each of the two resistors below point Z has a value of 100 Ω, and all three resistors above Z are 10.0 Ω. The current through each 100-Ω resistor is 500 mA. What is the current through any of the 10.0-Ω resistors, assuming it is equally distributed? What is the voltage, then, across any of the 10. 0-Ω resistors?

The total current into Z is 500 mA + 500 mA = 1.00 A. This must be divided three ways equally among the 10-Ω resistors. Therefore, the current through any one of them is 1.00/3 A = 0.333 A = 333 mA.

The voltage across any one of the 10.0-Ω resistors is found by Ohm’s Law:
E = IR
= 0.333 x 10.0 = 3.3 Volt