Series and parallel resistance calculator

List the resistors and a source voltage to get the total resistance, the total current and the voltage across each resistor. Work a string in series, a bank in parallel, or one series group feeding one parallel group, and see every step.

Resistors in series

List the resistances in ohms, separated by commas or spaces, and the source voltage.

Total resistance
650.00 Ω
Total current
18.46 mA
  1. Series resistances add: R = 100.00 + 220.00 + 330.00 = 650.000 Ω.
  2. Ohm's law for the whole string: I = V ÷ R = 12.000 ÷ 650.000 = 0.018462 A.
  3. The same current flows through every resistor, so each drop is V = I × R. The drops add back up to 12.000 V.
ResistorResistanceVoltage across itCurrent through it
R1100.00 Ω1.846 V18.46 mA
R2220.00 Ω4.062 V18.46 mA
R3330.00 Ω6.092 V18.46 mA

Resistors in parallel

Total resistance
68.75 Ω
Total current
174.55 mA
  1. Add the reciprocals: 1 ÷ R = 1 ÷ 100.00 + 1 ÷ 220.00 = 0.014545 S.
  2. Invert: R = 1 ÷ 0.014545 = 68.750 Ω. The total is always smaller than the smallest branch.
  3. Check with product over sum: (100.00 × 220.00) ÷ (100.00 + 220.00) = 68.750 Ω.
  4. Every branch sees the full source voltage, so each branch current is I = V ÷ R. The branch currents add up to 174.55 mA, the total current.
BranchResistanceVoltage across itCurrent through it
R1100.00 Ω12.000 V120.00 mA
R2220.00 Ω12.000 V54.55 mA

Series and parallel together

One series group in line with one parallel group, on a single source.

Series group
147.00 Ω
Parallel group
132.00 Ω
Total resistance
279.00 Ω
Total current
43.01 mA
Voltage across the parallel group
5.677 V
  1. Series group: 47.00 + 100.00 = 147.000 Ω.
  2. Parallel group: 1 ÷ (1 ÷ 220.00 + 1 ÷ 330.00) = 132.000 Ω.
  3. The two groups are in line with each other, so they add: 147.000 + 132.000 = 279.000 Ω.
  4. Total current: I = 12.000 ÷ 279.000 = 0.043011 A. It flows through every series resistor and splits across the parallel branches.
  5. Series drops are I × R. The parallel group drop is I × 132.000 = 5.677 V, and each branch current is that voltage ÷ its resistance.
Series resistorResistanceVoltage across itCurrent through it
R147.00 Ω2.022 V43.01 mA
R2100.00 Ω4.301 V43.01 mA
Parallel branchResistanceVoltage across itCurrent through it
R1220.00 Ω5.677 V25.81 mA
R2330.00 Ω5.677 V17.20 mA

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The formulas

  • Series: R = R1 + R2 + R3 + … One path, so the same current flows through every resistor and the resistances add.
  • Parallel: 1 ÷ R = 1 ÷ R1 + 1 ÷ R2 + 1 ÷ R3 + … Every branch connects across the same two points, so each sees the full voltage. For two resistors this reduces to R = (R1 × R2) ÷ (R1 + R2).
  • Combination: add the groups. Reduce each parallel bank to one equivalent resistance, then add it to the series resistors in line with it.
  • Ohm's law: I = V ÷ R, V = I × R. Use it on the total to get the total current, then on each part to get its voltage or current.

Two checks catch most mistakes. A parallel total is always smaller than the smallest branch, and a series total is always larger than the largest resistor. The voltage drops in a series string add up to the source voltage, and the branch currents in a parallel bank add up to the total current.

Worked example 1: three resistors in series

100, 220 and 330 Ω on 12 V. R = 100 + 220 + 330 = 650 Ω. I = 12 ÷ 650 = 18.46 mA. The drops are 1.846 V, 4.062 V and 6.092 V, which add to 12 V.

Worked example 2: two resistors in parallel

100 and 220 Ω on 12 V. R = (100 × 220) ÷ (100 + 220) = 22,000 ÷ 320 = 68.75 Ω. Branch currents are 12 ÷ 100 = 120 mA and 12 ÷ 220 = 54.55 mA, so the total is 174.55 mA, and 12 ÷ 68.75 gives the same figure. Three equal 300 Ω resistors in parallel give 300 ÷ 3 = 100 Ω.

Worked example 3: a series group and a parallel group

47 Ω and 100 Ω in series with a bank of 220 Ω and 330 Ω in parallel, on 12 V. The series group is 147 Ω. The bank is (220 × 330) ÷ 550 = 132 Ω. The total is 147 + 132 = 279 Ω and I = 12 ÷ 279 = 43.01 mA. The series drops are 2.022 V and 4.301 V. The bank drop is 43.01 mA × 132 Ω = 5.677 V, so the branches carry 25.81 mA and 17.20 mA, which add to the total.

These are original arithmetic examples, not exam questions.

Where this fits

Series and parallel arithmetic is the base for the Ohm's law work in the Calculations portion. For power, the power formulas and three-phase work, use the Ohm's law calculator. The Calculations portion guide shows where theory sits in the whole portion, and the transformer calculator covers current on each side of a transformer.

Series and parallel resistance questions

How do I find total resistance in series?

Add the resistances. Three resistors of 100, 220 and 330 ohms in series total 650 ohms.

How do I find total resistance in parallel?

Add the reciprocals of the resistances, then take the reciprocal of that sum. For two resistors, multiply them and divide by their sum. 100 and 220 ohms in parallel give 68.75 ohms.

Why is parallel resistance smaller than the smallest resistor?

Each extra branch gives the current another path, so the source can push more current at the same voltage. More current at the same voltage means less total resistance.

How do I get the voltage across each resistor?

Find the total resistance and the total current with Ohm's law, I = V ÷ R. In a series string the same current flows through every resistor, so each drop is I × R. In a parallel bank every branch has the full source voltage instead.

Add series resistances, add the reciprocals for parallel, then apply Ohm's law for the total current and each voltage drop, with every step shown. A study aid; the question in front of you governs.

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Published references

The formulas on this page are standard physics, explained in our own words, and need no code book. The first two sources describe them; the third describes the exam.

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