Resistors in Parallel The formula for the total parallel resistance is as follows: Two Resistors in Parallel Typically, it is more convenient to consider only two resistors at a time because this setup occurs in common practice. Any number of resistors in a circuit can be broken down into pairs. Therefore, the most common method is to use the formula for two resistors in parallel. Current Source A current source is an energy source that provides a constant value of current to a load even when the load changes in resistive value. The general rule to remember is that the total current produced by current sources in parallel is equal to the algebraic sum of the individual sources. Kirchhoff’s Current Law Kirchhoff’s current law can be stated as: The sum of the currents into a junction or node is equal to the sum of the currents flowing out of that same junction or node. A junction can be defined as a point in the circuit where two or more circuit paths come together. In the case of the parallel circuit, it is the point in the circuit where the individual branches join. General formula IT = I1 + I2 + I3 Refer to Figure 10-91 for an illustration. Point A and point B represent two junctions or nodes in the circuit with three resistive branches in between. The voltage source provides a total current IT into node A. At this point, the current must divide, flowing out of node A into each of the branches according to the resistive value of each branch. Kirchhoff’s current law states that the current going in must equal that going out. Following the current through the three branches and back into node B, the total current IT entering node B and leaving node B is the same as that which entered node A. The current then continues back to the voltage source. Figure 10-92 shows that the individual branch currents are: I1 = 5 mA I2 = 12 mA The total current flow into the node A equals the sum of the branch currents, which is: IT = I1 + I2 Substitute I T = 5mA + 12mA IT = 17mA The total current entering node B is also the same. Figure 10-93 illustrates how to determine an unknown current in one branch. Note that the total current into a junction of the three branches is known. Two of the branch currents are known. By rearranging the general formula, the current in branch two can be determined. General formula IT = I1 + I2 + I3 Substitute 75mA = 30mA + I2 + 20mA Solve I2 I2 = 75mA – 30mA – 20mA I2 = 25mA