# Mixed Series and Parallel Circuits

Part 1: Fundamental Concepts: Mixed Series and Parallel Circuits

Up to this point, the circuits we have looked at have either been composed of all series or all parallel connected resistors.  Most circuits are composed of a number of different components connected in both series and parallel.  One possible circuit that has a mixture of series and parallel resistors is shown below in Figure 1.

Figure 1. Circuit with a mixture of series and parallel resistors.

What if you wanted to find the voltage drop across each resistor and the current through it?  Although it may not seem obvious at first, you can apply the rules you already know about series and parallel circuits and how to combine series and parallel resistors to find the solution.  Let’s start by trying to figure out how much current is flowing from the 6V source.  To do that, we need to know the total value of the resistance connected to the voltage source.  To figure this out, the parallel connected groups of resistors is a good place to start.  The following figure has their combined values labeled.  The use of “||” in the figure means that the values on either side of it are in parallel.  Recall that to find the parallel combinations value, use the reciprocal, of the sum of the reciprocals, of each value that is in parallel.

Figure 2. Finding the equivalent value of the parallel groups.

With the values of the parallel groups found, they can be treated as a single resistance value.  This allows the value of the series combination of R2, and the two parallel groups, to be found as shown in the following figure.  In this case, the resistance values are simply added together.

Figure 3. Finding the equivalent value of the center branch.

The entire group of resistors composed of R2 through R7, appears to the circuit to be a single value of 567.69Ω.  Knowing this allows them to be placed in parallel with R8 and a value found for the entire group of resistors composed of R2 through R8.  This is shown below in Figure 4.

Figure 4. Finding the equivalent value of all but R1.

The combined value of the resistance R2 through R8, as shown in Figure 4, appear as a single value of 332.06Ω in series with R1.  This only leaves one more addition, shown below in Figure 5, to find the total resistance connected to the 6V voltage source.

Figure 5. Finding the equivalent value of all resistors.

With the total load resistance known, the current flowing from the 6V source can be found.  The current flowing from the 6V source is given by Ohm’s Law (V = IR) so 6 = I * 432.06 or I = 0.01389A.

Because all of the current leaving the 6V source flows through R1, the voltage drop across R1 can be found.  V = IR, therefore V = 0.01389 * 100 or V = 1.389V.

The voltage on one side or R1 is 6V because that is the value of the voltage source.  The drop across R1 is 1.389V, therefore, the voltage on the other side of R1 is 4.611V (6-1.389 = 4.611).  This means that the voltage across the two parallel branches, the one composed of R2 and the resistors that follow it, and the one with just R8, have 4.611V across them.  This is shown in the following figure.

Figure 6. Voltage present across both branches.

With the voltage across R8 known, the current through it can be found.  Knowing V= IR yields 4.611 = I * 800 or I = 0.00576A.  The total current delivered from the source must divide between the two branches.  Therefore, the current flowing through the branch that contains R2 can be found by I = 0.01389-0.00576 or I = 0.00813A.  This is shown in Figure 7.

Figure 7. The branch currents.

With the current through the branch that starts with R2 known, the voltages that appear across R2, the parallel combination of R3 and R4, as well the parallel combination of  R5, R6, and R7, can be found.  This is a simple matter because the resistance of these parallel combinations is known (see Figure 2).

For R2:

V = IR

V = 0.00813 * 200

V = 1.626V

For R3 || R4:

V = IR

V = 0.00813 * (300 || 400)

V = 0.0813 * 171.43

V = 1.394V

For R5 || R6 || R7:

V = IR

V = 0.00813 * (500 || 600 || 700)

V = 0.00813 * 196.26

V = 1.596

Figure 8, shown below, annotates the schematic with the voltage drops calculated above.

Figure 8. The first Branch voltage drops.

With the voltage drops available across the parallel groups of resistors, the current through each resistor can be found by repeated application of Ohm's law as follows.

For R3:

I = 1.394/300 or I = 0.00465

For R4:

I = 1.394/400 or I = 0.00349

For R5:

I = 1.596/500 or I = 0.00319

For R6:

I = 1.596/600 or I = 0.00266

For R7:

I = 1.596/700 or I = 0.00228

There are other approaches that could have been taken to find the voltages and currents present in this circuit, particularly after the total current from the voltage source was found.  The exact technique used is not important.  The goal was to show that, with some patience, any arbitrarily complex circuit composed of series and parallel connected resistors can be solved by repeated application of a few basic principles.