Equivalent Resistance Calculator: Series and Parallel Circuits

Unplug one bulb from an old-fashioned string of Christmas lights and the whole string goes dark. Unplug one lamp from your living room wall socket and every other lamp, TV and charger in the house keeps running without a flicker. Both are real circuits obeying the same laws of physics — the only difference is how the components are wired together, and that single choice changes everything about how current, voltage and resistance behave.

This page is an equivalent resistance calculator and an interactive series/parallel circuit simulator. Build a circuit with two to four resistors, switch between series and parallel wiring, and watch the current and voltage in every single resistor update live — while the bar chart makes the single most important fact in circuit analysis impossible to miss: series resistances always add up, parallel resistance is always less than the smallest resistor in the circuit.

What Is Equivalent Resistance in Series and Parallel Circuits?

Equivalent resistance ($R_{eq}$) is the single resistor value that would draw the same current from the supply as the whole network of resistors combined. It lets you replace a complicated tangle of resistors with one imaginary resistor for the purposes of calculation.

The formula depends entirely on how the resistors are wired:

$R_{series} = R_1 + R_2 + \cdots + R_n$

$\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}$

Series means the resistors are chained end-to-end in a single loop — there is only one path for current to take, so it must pass through every resistor in turn. Parallel means each resistor is wired directly across the same two points, giving current multiple independent paths to choose between.

A resistor's resistance itself comes from electrons colliding with the atomic lattice as they're driven through the material by an electric field — see our electric field post for that underlying picture, and our Ohm's law calculator if you need a refresher on $V = IR$ before diving into how multiple resistors combine.

Why Does Resistance Add in Series but Shrink in Parallel?

Think of resistance as a narrow section of pipe restricting the flow of water. Connect two narrow pipe sections end-to-end (series) and the water has to squeeze through both restrictions one after another — the total restriction is obviously greater than either one alone. This is why series resistances simply add.

Now instead connect two pipes side-by-side, both leading to the same destination (parallel). Water can take either path, so more total water flows through for the same push (pressure) than through a single pipe — the combined restriction to flow is less than either pipe alone. This is why parallel resistance is always smaller than the smallest individual resistor — a fact so counter-intuitive to students meeting it for the first time that it is worth stating twice: adding a second resistor in parallel makes the combined resistance go down, not up.

An even sharper version of this rule: two equal resistors $R$ in parallel always give $R/2$. Three equal resistors in parallel give $R/3$. Adding more parallel paths always eases the flow.

Interactive Equivalent Resistance Calculator

Choose series or parallel wiring, pick how many resistors (2 to 4), and drag each resistor's value and the supply voltage. The top panel draws the live circuit schematic with the current through every resistor labelled directly on the diagram. The bottom panel is the whole lesson in one glance: watch the green "Equivalent" bar tower above every individual resistor in series mode, then switch to parallel and watch it drop below every one of them.

The Math: Kirchhoff's Laws and the Resistance Formulas

Both resistance formulas are direct consequences of two conservation laws first stated by Gustav Kirchhoff in 1845.

Kirchhoff's Current Law (KCL): the total current flowing into any junction equals the total current flowing out — charge cannot pile up or vanish at a wire junction.

Kirchhoff's Voltage Law (KVL): the sum of voltage drops around any closed loop equals the total voltage supplied — energy is conserved as charge travels around the circuit.

Deriving the Series Formula

In a series circuit there are no junctions between the resistors, so by KCL the same current $I$ flows through every one of them. By Ohm's law, each resistor's voltage drop is $V_i = IR_i$. By KVL, these drops must sum to the supply voltage:

$V = V_1 + V_2 + \cdots + V_n = IR_1 + IR_2 + \cdots + IR_n = I(R_1 + R_2 + \cdots + R_n)$

Since $V = IR_{eq}$ by definition of the equivalent resistor, dividing through by $I$ gives $R_{eq} = R_1 + R_2 + \cdots + R_n$ — the series formula.

Deriving the Parallel Formula

In a parallel circuit every resistor connects the same two nodes, so by KVL each one sees the same voltage $V$. By Ohm's law, each resistor draws its own current $I_i = V/R_i$. By KCL, the total current from the supply is the sum of the branch currents:

$I = I_1 + I_2 + \cdots + I_n = \frac{V}{R_1} + \frac{V}{R_2} + \cdots + \frac{V}{R_n} = V\left(\frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}\right)$

Since $I = V/R_{eq}$, dividing through by $V$ gives $\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots + \dfrac{1}{R_n}$ — the parallel formula. For exactly two resistors this rearranges into the often-quicker "product over sum" form:

$R_{parallel} = \frac{R_1 R_2}{R_1 + R_2}$

How Do You Find the Current and Voltage in Each Resistor?

1. Find $R_{eq}$ using the series or parallel formula above.
2. Find the total current from the supply: $I_{total} = V_{supply} / R_{eq}$.
3. Series: every resistor carries $I_{total}$. Find each one's voltage drop with $V_i = I_{total} \cdot R_i$.
4. Parallel: every resistor sees $V_{supply}$. Find each one's current with $I_i = V_{supply} / R_i$.

Why Does One Broken Bulb Kill an Entire String of Christmas Lights?

The Christmas-lights problem from the opening is a wiring choice, not a manufacturing flaw — old-style string lights are built in series on purpose, because it used to be the cheaper way to wire them. Click any bulb below to break it and let the wiring decide what happens next.

In series mode, breaking one bulb opens the only path current has — the loop stops conducting and every bulb goes dark at once, not just the one you clicked. In parallel mode, each bulb is its own branch across the same two supply wires, so breaking one only removes that branch; the other three keep drawing their normal current, completely unaffected by the failure next to them.

Why Is Household Wiring Parallel and Not Series?

Household electrical wiring is built entirely from parallel circuits, for two decisive reasons — the fault finder above already demonstrated the first one directly.

• Independence. Every outlet and light fixture is wired across the same two supply rails (live and neutral), so each one sees the full 120 V or 230 V supply voltage regardless of what else is switched on or what fails elsewhere on the circuit. Turning off the kitchen light does not dim the bedroom lamp, and a blown bulb in one fixture does not black out the rest of the house.
• Consistent voltage rating. Every appliance is designed to run at the full mains voltage. Wiring appliances in series would divide that voltage between them, exactly as the series formula predicts, starving each one of the voltage it needs to work properly.

Old-style Christmas lights (and a few still sold today, built in series deliberately for cost) are the rare exception, which is why a single failed bulb famously darkens the whole string.

What Is a Voltage Divider and How Does It Work?

Not every resistor network is purely series or purely parallel. The most common "mixed" circuit in electronics — and one of the most frequently tested applications in exams — is the voltage divider: two resistors in series across a supply, with the output tapped from the junction between them.

$V_{out} = V_{in} \cdot \frac{R_2}{R_1 + R_2}$

A potentiometer is a voltage divider you can adjust continuously: a single resistive track with a sliding contact that splits it into two resistances, $R_1$ above the wiper and $R_2$ below. Drag the wiper below and watch $V_{out}$ follow it — move it up and $R_2$ grows toward the full track resistance, pulling $V_{out}$ toward $V_{in}$; move it down and $R_2$ shrinks toward zero, pulling $V_{out}$ toward zero.

One caveat worth knowing for exams: the formula above only holds when nothing draws current from the output. Connect a load resistor across $V_{out}$ and it sits in parallel with $R_2$, lowering the effective resistance on that side of the divider and pulling $V_{out}$ down from what the unloaded formula predicts — a phenomenon called loading the divider. Real sensor circuits are designed around this: either the load resistance is kept much larger than $R_2$ so the sag stays negligible, or the output is buffered with an op-amp that draws essentially no current at all.

Worked Examples for Physics Exams

Example 1: Three resistors in series

A 100 Ω, 220 Ω and 330 Ω resistor are connected in series across a 12 V battery. Find the equivalent resistance, the current, and the voltage across each resistor.

$R_{eq} = 100 + 220 + 330 = 650$ Ω. Current: $I = 12 / 650 \approx 0.0185$ A ≈ 18.5 mA (the same through all three). Voltage drops: $V_1 = 0.0185 \times 100 \approx 1.85$ V, $V_2 = 0.0185 \times 220 \approx 4.06$ V, $V_3 = 0.0185 \times 330 \approx 6.09$ V. Check: $1.85 + 4.06 + 6.09 = 12.00$ V ✓ — the drops sum to the supply voltage, confirming KVL. Try it in the simulator: Series mode, R₁=100, R₂=220, R₃=330, V=12.

Example 2: The same three resistors in parallel

Now wire the same 100 Ω, 220 Ω and 330 Ω resistors in parallel across the same 12 V battery.

$\dfrac{1}{R_{eq}} = \dfrac{1}{100} + \dfrac{1}{220} + \dfrac{1}{330} \approx 0.01 + 0.00455 + 0.00303 = 0.01758$, so $R_{eq} \approx 56.9$ Ω — far smaller than even the smallest 100 Ω resistor. Branch currents: $I_1 = 12/100 = 0.12$ A, $I_2 = 12/220 \approx 0.0545$ A, $I_3 = 12/330 \approx 0.0364$ A. Total current: $0.12 + 0.0545 + 0.0364 \approx 0.211$ A — matching $I_{total} = V/R_{eq} = 12/56.9 \approx 0.211$ A, confirming KCL.

Example 3: Two equal resistors in parallel

Two 100 Ω resistors are connected in parallel. Find the equivalent resistance.

Using the product-over-sum shortcut: $R_{eq} = \dfrac{100 \times 100}{100 + 100} = \dfrac{10{,}000}{200} = 50$ Ω — half of a single resistor, matching the general rule that $n$ equal resistors $R$ in parallel give $R/n$.

Example 4: Voltage divider output

A voltage divider has a 1000 Ω track and a 9 V supply, with the wiper set 60% of the way up. Find $R_1$, $R_2$ and $V_{out}$.

$R_2 = 0.6 \times 1000 = 600$ Ω, $R_1 = 1000 - 600 = 400$ Ω. Current: $I = 9/1000 = 0.009$ A. $V_{out} = 0.009 \times 600 = 5.4$ V. Try it in the simulator: drag the wiper to 60% of the way up the track (defaults: 1000 Ω track, 9 V supply) and check the readout matches 5.4 V.