Reducing a Mixed Circuit, Step by Step

You’ll be able to

Reduce a combination circuit to a single equivalent resistance
Work back outward to find the current and voltage for each element
Verify a solution using Kirchhoff's rules

Collapse inward, then expand outward

A circuit that is neither purely series nor purely parallel is solved by a fixed two-phase procedure. Collapse: find the innermost group that is clearly series or clearly parallel, replace it with one equivalent resistor, and repeat until a single resistance remains. Then expand: use the source voltage and total resistance to find total current, and work back out one stage at a time, applying the two rules at each step.

The two rules you expand with

SERIES: current is the same, voltages add · PARALLEL: voltage is the same, currents add

Every step of the expansion phase uses one of these. Naming which one you are using prevents applying the wrong one.

Recognizing what is in parallel

Two elements are in parallel when both ends are connected to the same pair of nodes — not merely when they look side by side in the drawing. A wire with no component in it is a single node however long it is drawn, so elements can be in parallel while appearing far apart. Redrawing the circuit with nodes labeled is often the fastest route through a confusing diagram.

Two sanity checks

A parallel combination is always smaller than the smallest resistor in it — adding a path can only make it easier for current to flow. And the voltages around any loop must sum to zero. Both checks take seconds and catch most arithmetic errors. If your parallel equivalent came out larger than one of its members, you have almost certainly added resistances instead of reciprocals.

Worked example

A 6.0 Ω resistor is in series with a parallel combination of 12 Ω and 4.0 Ω, all across a 24 V battery. Find the total current and the voltage across the parallel section.

1.Collapse the parallel pair: 1/R = 1/12 + 1/4.0 = 1/12 + 3/12 = 4/12, so R = 3.0 Ω.
2.Check: 3.0 Ω is smaller than 4.0 Ω, the smaller member. Good.
3.Total resistance = 6.0 + 3.0 = 9.0 Ω.
4.Total current: I = 24/9.0 = 2.67 A. This flows through the 6.0 Ω resistor and into the parallel section.
5.Voltage across the parallel section: V = IR = (2.67)(3.0) = 8.0 V. Across the 6.0 Ω: (2.67)(6.0) = 16 V, and 16 + 8 = 24. Loop checks.

Answer: Total current 2.67 A, with 8.0 V across the parallel section and 16 V across the series resistor. The branch currents are 8.0/12 = 0.67 A and 8.0/4.0 = 2.0 A, which sum to 2.67 A as required.

Watch out

For a parallel combination, do not forget the final reciprocal. Computing 1/12 + 1/4 = 1/3 and reporting 1/3 Ω instead of 3 Ω is one of the most common errors in the unit, and the "smaller than the smallest" check catches it instantly.

Checkpoint

Three 9.0 Ω resistors are connected in parallel. The equivalent resistance is:

Checkpoint

An additional resistor is connected in parallel with an existing network. The total resistance of the circuit:

Checkpoint

Two resistors are correctly described as being in parallel when:

On the exam

Write the equivalent resistance at every stage of the collapse and keep those values — you need them on the way back out. Discarding intermediate results means recomputing them, which is where time is lost on circuit free responses.

Answer the 3 checkpoints as you read.