Tristan’s Guide to:
Solving Series-Parallel Circuits.
Version: 1.0
Written in 2006
Written By: Tristan Miller
Tristan@CatherineNorth.com
Series-Parallel.
Once you understand Series Circuits, and Parallel Circuits, you can easily understand
how to solve a Series-Parallel Circuit. A Series-Parallel Circuit is not any harder. In
fact, it is the exact same, only combined.
The trick to solving these circuits is how you look at them.
Wow, look at that mess. I know what you’re thinking… Stop swearing!
All you have to do is break the circuit down, into several smaller and simpler Series, and
Parallel Circuits.
Let’s start this one, by identifying all of the Series Circuits.
I’ve circled all of the resistors that are in Series, with a red circle.
I circled 3 series circuits. A, B, and C.
We know that the Total Resistance of a Series Circuit is equal to all of the resistor values
added together, so we can simplify this drawing a little bit.
RT A = R1 + R2
RT B = R3 + R4
RT C = R9 + R10 + R11
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As you can see, I re-drew the diagram replacing the resistors that I had circled before,
with the Total Resistance of the Series Circuits.
Now let’s identify some of the Parallel Branches.
There are 2 Parallel Branched that I circled, I labeled them D and E.
We can once again simplify them, and re-draw the circuit.
RT D = 1 ÷ ( (1÷RT A) + (1÷RT B) )
RT E = 1 ÷ ( (1÷R6) + (1÷R7) )
The new circuit looks like this.
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From here, we can identify another simple Series circuit, labeled F.
RT F = RT D + R5 + RT E + R8
Once redrawn, you can see that there is only a Parallel Circuit left.
The total resistance of the Original Circuit is equal to G.
RT = G = 1 ÷ ( (1÷RT F) + (1÷RT C) )
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Example:
Let’s find RT first.
R2 and R3 are in series,
we can just add them together.
R2 + R3 = 300 Ohms + 100 Ohms
= 400 Ω.
R1 is in parallel with R2 and R3.
1 ÷ ( (1÷R1) + (1÷(R2+R3)) )
= 1÷ ( ( 1÷40Ω) + (1÷400Ω) )
= 1÷ ( 0.025 + 0.0025 )
= 1÷ 0.0275
= 36.36363636Ω
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R5 and R6 are also in parallel.
1 ÷ ( (1÷R5) + (1÷R6) )
= 1 ÷ ( (1÷400Ω) + (1÷800Ω) )
= 1 ÷ ( 0.0025 + 0.00125 )
= 1 ÷ 0.00375
= 266.6667Ω
Now we have 3 resistances in series,
add them together to get the Total
Resistance.
RT = (R1&R2&R3) + R4 + (R5&R6)
= 36.363636Ω + 120Ω + 266.666667Ω
= 423.030303Ω
Let’s now find the Current that travels through R1.
We’ll start by finding the Voltage Across R1. We’ll use the Voltage Divider Formula for
this.
R1
VR1 = ----- x E
RT
Our R value is going to be the Resistance Value of R1 in parallel with R2 and R3. We
calculated it to be 36.363636 Ohms.
Enter the numbers we know into the formula, and solve for VR1
36.363636Ω
VR1 = ----------------- x 12V
423.030303Ω
VR1 = 0.085960 x 12V
VR1 = 1.03152V
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Now we know the Resistance of R1 and the voltage across R1. Using this formula gives
us the Current.
Volts
Current = ------------Resistance
VR1
IR1 = ------R1
1.03152V
IR1 = ----------40 Ohms
IR1 = 0.025788A or 25.788mA
How about the Current in R4?
It’s the same process. Find the voltage across R4, and multiply it by R4’s Resistance.
R4
VR4 = ----- x E
RT
VR4
IR4 = ------R4
120Ω
VR4 = ---------------- x 12 VR4 = 3.404011V
423.030303Ω
3.404011V
IR4 = -------------120 Ohms
IR1 = 0.028367A or 28.367mA
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You can use the same process to calculate the Voltage, or Current or Resistance in any
Series-Parallel Circuit.
Do not look at it as one complex gigantic mess of lines and scribbles, identify the smaller,
simpler ‘Sub Circuits’ and solve them one at a time.
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