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# Simplifying resistor networks

A strategy for combining series and parallel resistors to reduce the complexity of a circuit. Written by Willy McAllister.
Complicated resistor networks can be simplified by identifying series and parallel resistors within the larger context of the circuit. This article describes a systematic way to simplify a circuit, using this example,
We have a voltage source connected to a resistor network. The two small circles at the left end represent the ports of the resistor network.
Let's say we want to figure out the current demand placed on the voltage source by the resistor network. The answer is not immediately obvious. But, we have some tools at our disposal: we know how to compute the equivalent resistance of series and parallel resistors. With these tools we can simplify the resistor network until the problem is easy to solve.
Strategy for simplifying a resistor network
• Begin as far away as possible from the circuit location in question.
• Replace series or parallel resistors with their equivalent resistor.
• Continue, moving left until a single equivalent resistor represents the entire resistor network.
The location in question is the input voltage source, so we start the simplification process way over on the far right, and work our way toward the source.
Simplifying a circuit is a process of many small steps. Consider a chunk of circuit, simplify, then move to the next chunk. Tip: Redraw the schematic after every step so you don't miss an opportunity to simplify.
Step 1. The shaded resistors, 2, \Omega and 8, \Omega, are in series.
Looking into the shaded area from the perspective of the arrows, the two series resistors are equivalent to a single resistor with resistance of
\Omega.

The two resistors can be replaced by their equivalent resistance:
Key insight: From outside the shaded box, the two series resistors and the equivalent resistor are indistinguishable from each other. The exact same current and voltage exist in both versions.
Step 2. We now find two 10, \Omega resistors in parallel at the new far right of the circuit.
These two resistors can be replaced by their parallel combination.
The resulting equivalent resistor is:
\Omega.

Again, looking into the shaded box from the left, the current and voltage with the equivalent resistor is still indistinguishable from the entire original circuit.
Step 3. A pattern is emerging. We are working through the schematic from right to left, simplifying and redrawing as we go. Next up we find two series resistors, 1, \Omega and 5, \Omega.
These series resistors can be replaced by an equivalent resistance of:
\Omega.

Step 4. This step is a bit more challenging. We have three resistors in parallel.
These three resistors can be replaced by their parallel combination.
The resulting equivalent resistor is:
\Omega.

Step 5. We are down to the last two series resistors,
We are left with a single 3, \Omega resistor. It represents the entire network as far as the voltage source is concerned. The current demanded from the voltage source is,
i, equals, start fraction, start text, V, end text, divided by, 3, \Omega, end fraction
We started with 7 resistors and simplified down to 1, a significant reduction in complexity. Not too shabby.
Key idea: The strategy for simplification is to start at a point in the circuit farthest away from the component of interest.
In this example, we were asked about the current load on the voltage source on the far left, so we started at the far right end of the circuit and worked to the left. Working in this 'backwards' direction may initially feel awkward, given our ingrained habit of reading from left to right.
It is common in electronics to start at the output end of a circuit (usually drawn on the right) and work back to the input. A left-to-right reading bias can get in the way if you always look at the 'normal' left end of the schematic first. For now, just remember you may have a left-to-right habit you may want to break.
Not all simplifications get down to a single resistor at the end. (The circuit may not be made entirely of resistors.) But always take the opportunity to simplify if the chance is presented.
Just for fun ... Here is an animation of the circuit simplification,

## Exceptions

Certain resistor configurations cannot be simplified using the strategy described above, and are therefore treated separately. Examples are described in the next article on Delta Wye Transformation.

## Want to join the conversation?

• For which applications the DC and AC current is used. Describe the devices to convert AC into DC voltage and vice versa with working function for each.
• This sounds like a homework problem. What ideas do you have so far?
• I've just watched a video about the direction of the current and now I am confused. The first image of this article shows the current "i" going out of the positive side of the battery? shoundn't it be the other way around? Can someone explain it to me?
(1 vote)
• Whoever posed the problem (me) chose to ask a question about current and pointed the current arrow as given.
• How do you simplify a complex circuit with diagonal paths that cut off multiple resistors?
(1 vote)
• R1 is a 3 ohm resistance connected in parallel to R2, a 6 ohm resistance. Both resistances are connected with a 4 ohm resistance, R3. Current of 0.8A flows through R1. THEN find the potential diffrence across R3.
(1 vote)
• Hello, I have some questions.

In the original diagram, there are two resistors, one of 12 ohms and the other of 4 ohms. They appear to be in parallel, are they initially in parallel? And if so, Can I start simplifying the entire circuit by replacing those two resistors with one resistor of 3 ohms and then proceed? I tried but it gives me a different result.

I understand that starting from the far right is part of the strategy, I just want to know if replacing first resistors of 12 and 4 would be possible.
(1 vote)
• where can i find questions related to finding resistors and currents from circuits for class 10 from
(1 vote)
• Hello Hemant,

I have found these books to be good. If you dig you can find pdf copies for download:

If you are interested in electronics I recommend you get a copy of "The Art of Electronics" by Horowitz and Hill. It's one of the most used books in my collection!

Regards,

APD
(1 vote)
• What would happen if you added a five ohm resister in between the 12 ohm resister and the 4 ohm resister?
(1 vote)
• Hello James,

The overall resistance of the circuit would be lower with a resulting increase in current flow.

Recommend you work the problem with the additional resistor and see the results. It may be useful to first work the steps from in the animation...

Regards,

APD
(1 vote)
• please give me more questions of circuit
(1 vote)
• The circuit is sq. Shaped-abcd in which a and b are the opp.sides. Ad=0.5ohm,db=1.5ohm,bc=2.5ohm,ac=0.5ohm. Find effective resistance
(1 vote)
• Find the effective resistance between which two corners of the square? You might try drawing your circuit in this schematic tool https://spinningnumbers.org/circuit-sandbox/index.html

Draw your circuit. Click on the "link" icon and copy/paste the resulting big URL back here as a comment.
(1 vote)
• In this circuit, is R7 neglected, if so, why?
http://i.imgur.com/ec5jOqt.png