Series & Parallel Resistor Networks

Everything in this topic descends from two statements that are not about resistors at all. First, in the steady state charge does not accumulate anywhere, so whatever current enters a junction leaves it. If two elements meet at a point that nothing else touches, every coulomb through the first must go through the second: they are in series and carry one common current I. Second, electric potential is single-valued — a node has one potential, whichever wire you arrive on. So two elements whose ends are tied to the same pair of nodes must have the same potential difference V across them: they are in parallel. Now apply V = IR. In series, Vₜₒₜₐₗ = IR₁ + IR₂ = I(R₁ + R₂), so R_eq = R₁ + R₂ = 400 Ω for a 300 Ω and a 100 Ω. In parallel, Iₜₒₜₐₗ = V/R₁ + V/R₂ = V(1/R₁ + 1/R₂), so 1/R_eq = 1/300 + 1/100 = 1/75 and R_eq = 75 Ω. The current constraint builds resistance; the voltage constraint builds conductance.

Students lose more marks to misreading a diagram than to the algebra. Two resistors drawn end to end are not automatically in series, and two drawn side by side are not automatically in parallel: what matters is what else is connected to the point between them. Work with nodes instead of shapes. Give every stretch of ideal wire a letter, remembering that a wire with nothing in it is one node however long or bent it is drawn. Then two elements are in parallel exactly when they carry the same pair of node letters at their ends, and in series exactly when they share a node that no third element touches. A resistor drawn in line with another but with the supply lead joining between them fails that test — a third element touches the shared node, so the two currents differ and R₁ + R₂ is simply wrong. Redrawing the network with the nodes spread out costs thirty seconds and removes the guesswork.

Reduction is a two-pass procedure. Pass one collapses the network to R_eq, always starting at the end furthest from the source, because only there can you be sure nothing else joins the junction you are about to absorb. Take a 12 V ideal source driving a 10 Ω resistor that feeds two parallel branches: a lone 60 Ω, and a 20 Ω in series with a 40 Ω. The far branch reduces first: 20 + 40 = 60 Ω. Now 60 Ω sits across 60 Ω, giving 60 × 60/120 = 30 Ω. That block is in series with the 10 Ω, so R_eq = 40 Ω. Pass two unwinds it, restoring one layer at a time. The source current is I = 12/40 = 0.30 A. The 10 Ω drops 0.30 × 10 = 3.0 V, leaving 9.0 V across the block, so each 60 Ω branch carries 9.0/60 = 0.15 A. Those add back to 0.30 A, and the drops 3.0 + 9.0 V add back to 12 V — the audit that tells you the unwinding was done correctly.

Two shortcuts save the round trip through the current. For a series pair carrying one current, the potential splits in proportion to resistance: V₂ = Vᵢₙ R₂/(R₁ + R₂). Across 12.0 V, an 8.2 kΩ above a 3.3 kΩ puts 12.0 × 3.3/11.5 = 3.44 V at the tap. For two parallel branches sharing one voltage, the current splits in inverse proportion, which makes the formula look inverted: I₂ = Iₜₒₜₐₗ R₁/(R₁ + R₂), with the other branch's resistance on top. Sending 1.2 A into a 20 Ω beside a 30 Ω puts 1.2 × 30/50 = 0.72 A through the 20 Ω and 1.2 × 20/50 = 0.48 A through the 30 Ω — more current down the easier path, and the two still sum to 1.2 A. The voltage divider carries one condition: it is exact only while nothing is drawn from the tap, which is exactly what a real meter cannot promise.

Before computing, bracket. Adding a branch in parallel can only lower R_eq, and adding an element in series can only raise it — so n branches obey Rₘᵢₙ/n ≤ R_eq < Rₘᵢₙ, and a chain obeys R_eq > Rₘₐₓ. Those bounds catch most arithmetic slips on sight. The extremes are more useful still. A branch of near-zero resistance dominates a parallel group absolutely: 0.10 Ω across 1.0 kΩ gives 0.10 × 1000/1000.1 = 0.09999 Ω, which is what a short circuit means quantitatively. An open branch has infinite resistance, contributes zero conductance, and simply drops out of the sum. Apply this to a 10 Ω in series with the parallel pair R₂ and 40 Ω: as R₂ shrinks towards a short the block vanishes and R_eq approaches its floor, 10 Ω; as R₂ opens the block climbs towards 40 Ω and R_eq approaches its ceiling, 50 Ω, without ever arriving. Knowing the floor and the ceiling turns a formula into a prediction.

An ideal voltmeter has infinite resistance and an ideal ammeter zero; real ones do not, and the network you measure is the one the meter created. Put a voltmeter of resistance R_V across the lower arm of a divider and that arm becomes R₂ in parallel with R_V, so the reading falls to V_read/Vₜᵣᵤₑ = 1/(1 + Rₚ/R_V), where Rₚ = R₁R₂/(R₁ + R₂) is what you see looking back from the tap. With R₁ = R₂ = 1.0 MΩ across 12 V, the true tap is 6.00 V and Rₚ = 0.50 MΩ; a 10 MΩ meter reads 6.00/1.05 = 5.71 V, 4.8% low, and that error has nothing to do with the meter's calibration. An ammeter fails the mirror way: its resistance R_A adds in series, so a 1.0 Ω meter inserted in a 10 Ω loop reads 10/11 of the true current, 9.1% low. The working rule is R_V of order a hundred times Rₚ, and R_A of order a hundredth of the loop resistance.