**Introduction**

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 electric currents flow.

**Kirchhoff’s Current Law**

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 (Fig. 12-13).

**Fig. 12-13** . Kirchhoff’s current law. The current entering point Z is equal to the current leaving point Z. In this case, *I* _{1} + *I* _{2} = *I* _{3} + *I* _{4} + *I* _{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 meters going in has to be the same as the total volume going out. Water cannot 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.

**Kirchhoff’s Current Law Practice Problem**

**Kirchhoff’s Current Law Practice Problem**

**Problem**

In Fig. 12-13 , suppose that each of the two resistors below point *Z* has avalué of 100 ohms and that all three resistors above *Z* have values of 10.0 ohms. The current through each 100-ohm resistor is 500 mA (0.500 A). What is the current through any of the 10.0-ohm resistors, assuming that the current is equally distributed? What is the voltage, then, across any of the 10.0-ohm resistors?

**Solution**

The total current into *Z* is 500 mA + 500 mA = 1.00 A. This must be divided three ways equally among the 10-ohm 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-ohm resistors is found by Ohm’s law: *E* = *IR* = 0.333 × 10.0 = 3.33 V.** **

**Kirchhoff’s Voltage Law**

The sum of all the voltages, as you go around a circuit from some fixed point and return there from the opposite direction, and taking polarity into account, is always zero. At first thought, some people find this strange. Certainly there is voltage in your electric hair dryer, radio, or computer! Yes, there is—between different points in the circuit. However, no single point can have an electrical potential with respect to itself. This is so simple that it’s trivial. A point in a circuit is always shorted out to itself.

What Kirchhoff was saying when he wrote his voltage law is that voltage cannot appear out of nowhere, nor can it vanish. All the potential differences must balance out in any circuit, no matter how complicated and no matter how many branches there are.

Consider the rule you’ve already learned about series circuits: The voltages across all the resistors add up to the supply voltage. However, the polarities of the emfs across the resistors are opposite to that of the battery. This is shown in Fig. 12-14. It is a subtle thing, but it becomes clear when a series circuit is drawn with all the components, including the battery or other emf source, in line with each other, as in Fig. 12-14.

**Kirchhoff’s Voltage Law Practice Problem**

**Kirchhoff’s Voltage Law Practice Problem**

**Problem**

Refer to the diagram of Fig. 12-14 . Suppose that the four resistors have values of 50, 60, 70, and 80 ohms and that the current through them is 500 mA (0.500 A). What is the supply voltage *E?*

**Solution**

Find the voltages *E* _{1} , *E* _{2} , *E* _{3} , and *E* _{4} across each of the resistors. This is done using Ohm’s law. In the case of *E* _{1} , say, with the 50-ohm resistor, calculate *E* _{1} = 0.500 × 50 = 25 V. In the same way, you can calculate *E* _{2} = 30 V, *E* _{3} = 35 V, and *E* _{4} = 40 V. The supply voltage is the sum *E* _{1} + *E* _{2} + *E* _{3} + *E* _{4} = 25 + 30 + 35 + 40 V = 130 V.

Practice problems of these concepts can be found at: Direct Current Practice Test

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