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    Calculates locally in your browser · Kirchhoff network solve
    Circuit design / passive networks

    Series and parallel resistance

    Solve a three-element resistor network in either topology, including the current, voltage and dissipation in each element. The power column matters as much as the resistance: a correct value in an under-rated resistor still fails.

    TopologySeriesSeries value7900 ΩParallel value599.8 Ω
    Equivalent resistance
    7900.00 Ω
    Sum of the three
    Supply current
    1.519 mA
    From 12.00 V
    Total power
    18.23 mW
    V × I at the terminals
    Voltage division
    1.52 / 3.34 / 7.14 V
    Across the three elements
    Network solution
    Total resistance, supply current and the voltage and power in each element.
    Series
    Network solution
    ElementResistanceCurrentVoltagePower
    R₁1000.0 Ω1.519 mA1.519 V2.31 mW
    R₂2200.0 Ω1.519 mA3.342 V5.08 mW
    R₃4700.0 Ω1.519 mA7.139 V10.84 mW
    Total7900.00 Ω1.519 mA12.000 V18.23 mW

    In series the current is shared and the voltages add; in parallel the voltage is shared and the currents add. The total is always bounded by the smallest element in parallel and by the sum in series.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Power in each element
    Where the dissipation actually lands.
    Power in each element0.0002.565.127.6810.2412.802.31R₁R₁5.08R₂R₂10.84R₃R₃mW
    Equivalent resistance vs identical elements
    Series grows linearly; parallel falls as 1/n.
    Equivalent resistance vs identical elements0.0001344.002688.004032.005376.006720.00112233445566SeriesParallel
    Canonical equations
    The exact relationships implemented by this engine.

    R = R₁ + R₂ + R₃ · current is common

    1/R = Σ1/Rᵢ · voltage is common

    R = R₁R₂/(R₁+R₂) · the product over the sum

    P = V²/R = I²R · check the resistor rating

    engineering methodology

    A clear path from inputs to results.

    Series and parallel networks differ in exactly one way: which quantity is shared. In series the current is common and voltages add; in parallel the voltage is common and currents add. Everything else follows from Ohm’s law.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Enter the three resistances.
    2. Set the supply voltage.
    3. Choose series or parallel.
    4. Read the equivalent resistance and supply current.
    5. Check the power column against each resistor’s wattage rating.

    The formula

    Rseries=∑Ri,1Rparallel=∑1Ri,P=I2RR_{\text{series}} = \sum R_i, \qquad \frac{1}{R_{\text{parallel}}} = \sum \frac{1}{R_i}, \qquad P = I^2R

    Reference: Kirchhoff's circuit laws; standard passive-network analysis.

    Worked example

    1 kΩ, 2.2 kΩ and 4.7 kΩ on 12 V

    Inputs
    • R₁ 1 kΩ, R₂ 2.2 kΩ, R₃ 4.7 kΩ.
    • Supply 12 V.
    Calculation path
    1. 1

      Series: 1,000 + 2,200 + 4,700 = 7,900 Ω; current 12 / 7,900 = 1.519 mA.

    2. 2

      Parallel: 1/(1/1000 + 1/2200 + 1/4700) = 599.8 Ω; current 20.01 mA.

    3. 3

      Series power: 12 × 1.519 mA = 18.2 mW spread across three parts.

    4. 4

      Parallel power: 12 × 20.01 mA = 240 mW, mostly in the smallest resistor.

    Result

    Series 7,900 Ω at 1.52 mA; parallel 600 Ω at 20.01 mA — thirteen times the current.

    Interpreting the result

    The parallel equivalent is always below the smallest resistor, and the series equivalent always above the largest. Those two rules catch most arithmetic errors instantly. The power column matters as much as the resistance: a correct value in a quarter-watt resistor carrying half a watt will fail.

    Assumptions & limitations

    • Ideal resistors at nominal value with no tolerance spread.
    • The source has zero output impedance and the wiring has no resistance.
    • Resistors do not self-heat enough to shift their value.
    • No reactive elements, so the network is purely resistive at all frequencies.

    Frequently asked questions

    Why is the parallel equivalent smaller than every resistor?

    Because each parallel path provides an additional route for current, so total conductance (the reciprocal of resistance) increases. Adding conductance can only lower the equivalent resistance.

    How do I size the resistor wattage?

    Compute I²R for each element, then choose a rating at least twice that for reliability. Resistor power ratings assume free air at 25 °C and derate with temperature.

    How do I compute two in parallel quickly?

    Product over sum: R₁R₂/(R₁+R₂). For 1 kΩ and 2.2 kΩ that is 2.2M/3.2k = 688 Ω.

    What happens if one parallel branch opens?

    The remaining branches keep working and the equivalent resistance rises. That fault tolerance is why parallel wiring is used for lighting circuits.

    Methodology last reviewed 2026-10-18. Results are informational and not financial or engineering advice — see our disclaimer.