If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Course: Electrical engineering > Unit 2

Lesson 2: Resistor circuits
Science
Electrical engineering
Circuit analysis
Resistor circuits
© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

Simplifying resistor networks

Google Classroom
Learn how to combine series and parallel resistors to reduce the complexity of a circuit.
Complicated resistor networks can be simplified by combining series and parallel resistors. Consider this example circuit:
The diagram shows a voltage source connected to a resistor network. The two small circles at the left end represent the ports of the resistor network.
Imagine we want to calculate how much current flows from the voltage source. The answer is not immediately obvious, since there are many resistors and branches. However, by following a systematic process, we can combine the resistors until the resistor network is reduced to one equivalent resistor.
Strategy for simplifying a resistor network
  • Begin as far away as possible from the circuit location in question (which in this case is the voltage source).
  • Replace series or parallel resistors with their equivalent resistor.
  • Continue one step at a time until a single equivalent resistor represents the entire resistor network.

Example

The location in question is the input voltage source, so we start the simplification process 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, 2Ω and 8Ω, are in series.
Looking into the shaded area from the perspective of the arrows, the two resistors can be replaced by their series combination.
The equivalent resistance of the shaded region is
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
Ω.

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 10Ω resistors in parallel at the new far right of the circuit.
These two resistors can be replaced by their parallel combination.
The equivalent resistance of the shaded region is
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
Ω.

Again, looking into the shaded box from the left, the current and voltage with the equivalent resistor is still indistinguishable from the original circuit.
Step 3. A pattern is emerging. We're working through the schematic from right to left, simplifying and redrawing as we go. Next we find two series resistors, 1Ω and 5Ω.
The equivalent resistance of the shaded region is
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
Ω.

Step 4. Now we have three resistors in parallel.
The equivalent resistance of the shaded region is
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
Ω.

Step 5. We're down to the last two resistors, which are in series.
You can do this one in your head:
We're left with a single 3Ω resistor. It represents the entire network as far as the voltage source is concerned. The current demanded from the voltage source is:
I=V3Ω
We started with 7 resistors and simplified down to 1, a significant reduction in complexity.
Just for fun, here's an animation of the circuit simplification:
Khan Academy video wrapper
See video transcript

Want to join the conversation?

Log in