# Kirchhoff’s First Law

The physicist Gustav Robert Kirchhoff (1824–1887) was a researcher and experimentalist in a time when little was understood about how electric currents flow. Nevertheless, he used certain commonsense notions to deduce two important properties of dc circuits.

Kirchhoff reasoned that dc ought to behave something like water in a network of pipes, and that the current going into any point ought to be the same as the current going out of that point. This, Kirchhoff thought, must be true for any point in a circuit, no matter how many branches lead into or out of the point.

**Kirchhoff ’s First Law. At A, the current into point X or point Y is the same as the current out of that point. That is, I = I1 + I2. At B, the current into point Z equals the current flowing out of point Z. That is, I1 + I2 = I3 + I4 + I5. Illustration for Quiz Questions 13 and 14.
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Two examples of this principles are shown in above figure. Examine illustration A. At point X, I, the current going in, equals I1 + I2, the current going out. At point Y, I2 + I1, the current going in, equals I, the current going out. Now look at illustration B. In this case, at point Z, the current I1 + I2 going in is equal to the current I3 + I4 + I5 going out. These are examples of Kirchhoff ’s First Law.We can also call it Kirchhoff ’s Current Law or the principle of conservation of current.

#### Problem 1

**Refer to above figure A. Suppose all three resistors have values of 100 Ω, and that I1 = 2.0 A and I2 = 1.0 A. What is the battery voltage?**

First, find the current I drawn from the battery: I = I1 + I2 = 2.0 + 1.0 = 3.0 A. Next, find the resistance of the entire network. The two 100-Ω resistances 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, connected across the battery is 66.67 Ω. Then E = IR = 66.67 × 3.0 = 200 V.

#### Problem 2

**In above figure B, suppose each of the two resistors below point Z has a value of 100 Ω, and all three resistors above point Z have values of 10.0 Ω. Suppose the current through each 100-Ω resistor is 500 mA. What is the current through any one of the 10.0-Ω resistors, assuming that the current through all three 10.0-Ω resistors is the same? What is the voltage across any one of the three 10.0-Ω resistors?**

The total current into point Z is 500 mA + 500 mA = 1.00 A. This is divided equally among the three 10-Ω resistors. Therefore, the current through any one of them is 1.00/3 A = 0.333 A. The voltage across any one of the 10.0-Ω resistors can thus found by Ohm’s Law: E = IR = 0.333 × 10.0 = 3.33 V.