Two different examples are given below. Notice that the value of the current through every circuit element is given. In addition, the directions of the currents through resistors D, E and F are given. The goal of the two examples is to determine the directions of the currents through batteries B and H as well as through resistors A, C and G.
Kirchoff's Node Law states that the sum of the currents into a node equals the sum of the currents out of the node. A node is a location where two or more wires are connected together. The circuit at right is identical to the one above. The flashing dot is a node. The circuit elements that are connected to this node have been highlighted. The circuit element that is colored green has a current with a known value and direction. Battery H and resistor G are colored red to remind you that you are trying to determine the directions of the currents through them.
To apply Kirchoff’s Node Law, you must separate the currents that are directed into the node from those that are directed away from it. For the circuit above, the current through resistor D is directed into the node. The directions of the currents through battery H and resistor G are unknown. We will make a random guess that the currents through both are away from the node. This guess is purely random. We could just as easily have guessed the opposite direction for one, the other or both. The analysis below will tell us whether the guesses are correct or incorrect. Kirchoff’s Node Law gives:
Total Current Toward Node = Total Current Away from Node
ID = IG+ IH
where ID, IG and IH represent the currents through D, G and H respectively.
The values of all three currents are given in the circuit diagram. When substituting these values into the Kirchoff’s Node Law equation, remember that we guessed that the currents through the red circuit elements were both directed away from the node. This guess could easily be incorrect. Hence, put a plus/minus sign in front of these current values. Substituting the values from the circuit diagram gives:
4A = ±6A + ±2A.
There is only one choice for the plus/minus signs that makes this equation true:
4A = +6A + -2A.
This means that IG is positive (+6A) and IH is negative (-2A). A positive IG means that we correctly guessed that the current through resistor G is directed away from the node. In other words the current through resistor G is downward. A negative IH means that our guess about the direction of the current through battery H was incorrect. Hence, the actual current through battery H must be directed toward the node, or upward.
The circuit at right is identical to the circuit in the previous example, except that the circuit elements attached to a different node are highlighted. As before, the circuit elements that are colored green have currents with known values and directions. The red resistors also have known current values but are colored differently to remind you that you are trying to determine the directions of the currents though them.
Total Current Directed Toward Node = Total Current Directed Away From Node
IE + IF = IA + IB + IC
1A + 3A = ±1A + ±8A + ±3A.
1A + 3A = -1A + +8A + -3A.
Hence, IA = -1A, IB = +8A and IC = -3A.