Series and parallel

YOU WILL BE ABLE TO

Identify series and parallel connections by current and voltage sharing.
Compute equivalent resistance for series and parallel combinations.
Reduce a resistive network step by step to a single equivalent.
Use the two-resistor parallel shortcut fluently.

Real circuits contain many resistors, but a source connected to any purely resistive network can't tell the difference between the network and one well-chosen single resistor — its equivalent resistance. Finding that equivalent is the first systematic skill of circuit analysis, and it rests on two patterns.

Elements are in series when they carry the same current — connected end to end, with nothing branching off between them. Voltages along a series chain add, so resistances add too.

R_{eq} = R_1 + R_2 + \cdots + R_N

series

One current, three resistors: R_eq = R_1 + R_2 + R_3.

Elements are in parallel when they share the same pair of nodes, and therefore the same voltage. Currents through parallel branches add, which means conductances add — and resistance combines by reciprocals.

\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_N}

parallel

One voltage, two paths: currents add, conductances add.

For exactly two resistors the reciprocal formula collapses to the product-over-sum shortcut — worth committing to memory, along with two of its corollaries: N equal resistors R in parallel give R/N, and a parallel combination is always smaller than its smallest member.

R_{eq} = \frac{R_1 R_2}{R_1 + R_2}

two in parallel

WORKED EXAMPLE

Step-by-step reduction

A 6 kΩ and a 3 kΩ resistor are in parallel; that combination is in series with 4 kΩ.
Parallel first: 6k·3k/(6k+3k) = 18/9 = 2 kΩ.
Then series: 2 kΩ + 4 kΩ = 6 kΩ.

R_eq = 6 kΩ

WORKED EXAMPLE

Equal resistors

Four 1 kΩ resistors are connected in parallel.
N equal resistors: R/N = 1000/4 = 250 Ω.
Sanity check: 250 Ω is smaller than the smallest member (1 kΩ). ✓

250 Ω

PRACTICE

Compute the equivalent of 100 Ω, 220 Ω, and 330 Ω in series.

Two 1 kΩ resistors are in parallel. What's the equivalent, and why might a designer do this instead of using one 500 Ω part?

Find R_eq for 60 Ω, 30 Ω, and 20 Ω in parallel.

A 2 kΩ resistor is in series with the parallel combination of 6 kΩ and 12 kΩ. The whole network sits across 12 V. Find R_eq and the current drawn from the source.