Simplifying resistor networks

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.

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, $2\mathrm{\Omega }$‍ and $8\mathrm{\Omega }$‍, are in series.

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 $10\mathrm{\Omega }$‍ resistors in parallel at the new far right of the circuit.

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, $1\mathrm{\Omega }$‍ and $5\mathrm{\Omega }$‍.

Step 4. This step is a bit more challenging. We have three resistors in parallel.

Step 5. We are down to the last two series resistors,

You can do this one in your head:

We are left with a single $3\mathrm{\Omega }$‍ resistor. It represents the entire network as far as the voltage source is concerned. The current demanded from the voltage source is,

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,

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.