WINDSHIELD GROUP 8S2487 - Caterpillar

Series-Parallel Circuits
Illustration 6 g01070324
A series-parallel circuit is composed of a series section and a parallel section. All of the rules previously discussed regarding series circuits and parallel circuits are applicable in solving for unknown circuit values.Although some series-parallel circuits appear to be very complex, the series parallel circuits are solved quite easily by using a logical approach. The following tips will make solving series-parallel circuits less complicated:
Examine the circuit carefully. Then determine the path or paths that current may flow through the circuit before returning to the source.
Redraw a complex circuit to simplify the appearance.
When you simplify a series parallel circuit, begin at the farthest point from the voltage source. Replace the parallel resistor combinations one step at a time.
A correctly redrawn series parallel (equivalent) circuit will contain only ONE series resistor in the end.
Apply the simple series rules for determining the unknown values.
Return to the original circuit and plug in the known values. Use Ohm's Law to solve the remaining values.Solving a Series-Parallel Problem
Illustration 7 g01070325
The series parallel circuit, as shown in Illustration 7, shows a 2Ohms resistor in series with a parallel branch that contains a 6Ohms resistor and a 3Ohms resistor. To solve this problem it is necessary to determine the equivalent resistance for the parallel branch. Using the following equation, solve for the parallel equivalent (Re) resistance:1/Re = 1/R2 + 1/R31/Re = 1/6 + 1/3 or1/Re = .1666 +.3333 = .50 or1/Re = 1/.50 or Re = 2 ohmsIllustration 7 has been redrawn (See Illustration 8) with the equivalent resistance for the parallel branch. Solve circuit totals by using simple Ohm's Law rules for series circuits.
Illustration 8 g01070328
Using the rules for series circuits, the total circuit resistance can now be calculated by using the equation Rt = R1 + Re or Rt = 2 + 2 or 4 ohms.The remaining value that is unknown is current. Again, using Ohm's Law Circle, current can be calculated by the equation:I = E/R orI = 12/4 orI = 3 ampIllustration 9 shows all the known values.
Illustration 9 g01070330
Circuit calculations indicate that the total current flow in the circuit is 3 amps. Since all current flow that leaves the source must return, you know that the 3 amps must flow through R1. It is now possible to calculate the voltage drop across R1 by using the equation E = I × R, or E = 3A × 2Ohms, or E1 = 6 volts.If 6 volts is consumed by resistor R1, the remaining source voltage (6V) is applied to both parallel branches. Using Ohm's Law for the parallel branch reveals that 1 amp flows through R2 and 2 amps flow through R3 before combining into the total circuit current of 3 amps returning to the negative side of the power source.Other Methods and Tips for Solving Complex Series Parallel Circuits
As stated earlier, complex circuits can be easily solved by carefully examining the path for current flow and then draw the circuit again. No matter how complex a circuit appears, drawing an equivalent circuit