2 resistors counted
Series resistors share the same current and add directly. The equivalent resistance is the simple sum, so adding more resistors in series always increases total resistance — and the largest individual resistor dominates the chain.
R_eq = R₁ + R₂ + R₃ + … + Rₙ
Parallel resistors share the same voltage and provide multiple paths for current. The reciprocals of the resistances sum to the reciprocal of the equivalent — the equivalent is always smaller than the smallest branch, and adding more branches lowers it further.
1/R_eq = 1/R₁ + 1/R₂ + … + 1/Rₙ
Two resistors can be wired two fundamental ways. In series they sit end-to-end on a single current path: the same electrons flow through every resistor, voltage drops add up across each one, and the resistances simply add. In parallel they sit across the same two nodes: every resistor sees the same voltage, current splits between them inversely proportional to resistance, and the reciprocals add. This calculator takes any number of resistor values, applies the series sum or parallel reciprocal-sum, and reports the single equivalent resistance you could swap in to replace the whole network without changing the circuit's external behavior. The math falls out of Kirchhoff's voltage and current laws and is the bedrock of every basic DC circuit analysis.
You have two resistors, R₁ = 100 Ω and R₂ = 200 Ω. Find the equivalent resistance when they are wired in series and when they are wired in parallel.
The two answers — 300 Ω in series, 66.67 Ω in parallel — differ by a factor of 4.5 from exactly the same two resistors. The wiring topology controls the equivalent resistance every bit as much as the component values.
Three rules of thumb make resistor networks much faster to reason about. First, parallel resistance is always less than the smallest individual resistor: if you have a 100 Ω and a 1 MΩ resistor in parallel, the equivalent is essentially 100 Ω because the high-value branch barely conducts. Second, series resistance is always greater than the largest individual resistor: stacking resistors only adds. Third, identical resistors in parallel give R/N, where N is how many you have — four 1 kΩ resistors in parallel make a 250 Ω equivalent. Real circuits frequently mix series and parallel sub-networks (ladder networks, voltage dividers, R-2R DAC ladders); analyze them by collapsing one cluster at a time, replacing each collapsed cluster with its single equivalent, until the whole network reduces to one resistance.
Add the individual resistances directly: R_eq = R₁ + R₂ + R₃ + … + Rₙ. Each resistor carries the same current, voltage drops sum across the chain, and the total resistance is always larger than the largest individual resistor. Two 100 Ω resistors in series make 200 Ω.
Sum the reciprocals of each resistance and then invert the result: 1/R_eq = 1/R₁ + 1/R₂ + … + 1/Rₙ, so R_eq = 1 / (1/R₁ + 1/R₂ + … + 1/Rₙ). Each resistor sees the same voltage, current splits between the branches, and the equivalent is always smaller than the smallest individual resistor.
1/R_eq = 1/R₁ + 1/R₂ + … + 1/Rₙ. For the two-resistor case there is a shortcut: R_eq = (R₁ · R₂) / (R₁ + R₂), often called the product-over-sum formula. The general reciprocal-sum form works for any number of resistors.
Adding a second path for current to flow can never reduce the total current at a given voltage — it can only add to it. Since R = V/I, a higher total current at the same voltage means a lower equivalent resistance. Each extra parallel branch is another route for charge to take, so the more branches you add, the easier it is for current to flow, and the lower R_eq drops.
Yes — most real circuits do. Identify the smallest series or parallel sub-cluster, replace it with its single equivalent, redraw, and repeat until the whole network collapses to one resistance. This calculator handles a single homogeneous group; for mixed networks (e.g., (R₁ + R₂) ∥ R₃) collapse the inner cluster first, then feed its equivalent into the outer one.
Each additional parallel branch adds another path, so total current from the source increases (the equivalent resistance drops). If the supply voltage stays the same, total current = V / R_eq, and R_eq shrinks as you add branches. Individual branch currents stay the same — the new branch simply draws extra current of its own without affecting the others.
It is half of one of them. Two 100 Ω resistors in parallel give R_eq = 100/2 = 50 Ω; in general, N identical resistors of value R in parallel give R/N. This is one of the most useful shortcuts in circuit design: a 250 Ω equivalent from a stock of 1 kΩ parts is four 1 kΩ in parallel.
Yes. A 0 Ω branch is a perfect short across the two nodes — all the current flows through that wire and none flows through the other branches, so the equivalent resistance is 0 Ω regardless of what else is connected in parallel. This calculator flags a 0 Ω row in parallel mode rather than silently computing an answer.
Reference: Kirchhoff's voltage and current laws (Robert Kirchhoff, 1845). Series sum and parallel reciprocal-sum formulas appear in every introductory circuit-analysis text, including Sedra and Smith's Microelectronic Circuits and Horowitz and Hill's The Art of Electronics.
Two formulas cover every linear resistor combination — every other network reduces to repeated applications of these:
Where:
For exactly two resistors in parallel there is a useful shortcut: Req = (R₁ · R₂) / (R₁ + R₂), the “product over sum” form. Beyond two resistors, fall back to the general reciprocal-sum.
Canonical Pair
Combine R₁ = 100 Ω and R₂ = 200 Ω in both configurations and compare.
Series = 300 Ω, Parallel ≈ 66.67 Ω.
Parallel resistance (66.67 Ω) is below the smaller of the two individual resistors (100 Ω), as it must be for any pair of positive resistors in parallel.
LED Current Limiting
A red LED (Vf ≈ 2 V, If = 20 mA) on 5 V drops 3 V across the series resistor; pair two common E12 values, 220 Ω + 470 Ω, to land at ~690 Ω.
Series equivalent = 690 Ω, LED current ≈ 4.3 mA (safe).
Real LED Vf shifts with current; this is the textbook first-order sizing. For a brighter LED, halve the series resistor or move to a constant-current driver.
Parallel Divider
Five identical 1 kΩ chips in parallel — the equivalent is R/N = 1000/5.
Equivalent resistance = 200 Ω, power capability ≈ 1.25 W.
Splitting load across N parallel resistors is the standard way to extend power-handling without paying for a high-wattage part.
