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Series and parallel resistors

Combine two to four resistors and observe equivalent resistance, voltage and current distribution, and power in each component.

Workbench / Ideal modelIn your browser
01 / Adjust the inputs

Sliders with a wide range use a logarithmic scale. When zero is allowed, it has its own position. Type a number to set an exact value.

From 0 to 48 V. Constant ideal source.

From 1 Ω to 1 MΩ. Positive, finite values.

From 1 Ω to 1 MΩ. Positive, finite values.

Try an example

+VsR1R2Common current

Labels R1–R4 identify each resistor in the list. The diagram changes with the connection and number of components.

Equivalent resistance300 Ω
Total current40 mA
Total power0.48 W
  • R1: 100 Ω · 4 V · 40 mA · 160 mW
  • R2: 200 Ω · 8 V · 40 mA · 320 mW

Req = 100 + 200 = 300 Ω. I = 12/300 = 40 mA. P = 12 × 0.04 = 0.48 W.

Voltages add to 4 + 8 = 12 V. Current is shared.

The powers add to 0.16 + 0.32 = 0.48 W. This is not a recommendation of component power ratings.

Use it in your classroom

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100 Ω and 200 Ω across 12 V

In series, Req = 300 Ω and current is 12/300 = 0.04 A. The same current flows through both resistors. R1 drops 0.04 × 100 = 4 V; R2 drops 0.04 × 200 = 8 V. The larger resistor has the larger voltage drop.

R1 dissipates 4 × 0.04 = 0.16 W; R2 dissipates 8 × 0.04 = 0.32 W. Total power is 0.48 W, matching the 12 V × 0.04 A supplied by the source.

In parallel, each resistor receives 12 V. Currents are 12/100 = 120 mA and 12/200 = 60 mA: the smaller resistor now carries more current. Total current is 180 mA and Req = 66.6667 Ω. Power rises to 1.44 W and 0.72 W, totaling 2.16 W. Rewiring also changes what each component must withstand.

Solve the network first, then each component

Series: Req = R1 + … + Rn
Parallel: 1/Req = 1/R1 + … + 1/Rn
Itotal = Vs/Req
Series: Ii = Itotal; Vi = Itotal Ri
Parallel: Vi = Vs; Ii = Vs/Ri
Pi = Vi Ii; Ptotal = Vs Itotal = ΣPi

Convert every resistance to ohms first. Series circuits add voltages; parallel circuits add currents. Applying the same rule to both gives physically inconsistent results, so the tool keeps the relevant balance visible.

An ideal network, without mixed connections or short circuits

Two to four resistors from 1 Ω to 1 MΩ, all in series or all in parallel, with a 0–48 V source. Mixed networks, open or zero-ohm resistors, wires, supply current limiting, tolerances and heating are not modeled.

The displayed power is model dissipation, not the component rating needed for a build. Real component selection also requires design margin, environmental conditions and manufacturer specifications. If a real supply cannot maintain its voltage under load, this model no longer describes that circuit.

Watts measure power. Finding energy in joules also requires a duration and steady conditions: E = P × t.

Common mistakes

  • Adding parallel resistances directly.
  • Splitting current equally among branches with different resistances.
  • Assigning the full supply voltage to each series resistor.
  • Comparing total dissipation with the rating printed on just one part.

Quick guide: check before calculating

  1. Predict whether Req should exceed the largest resistor (series) or be smaller than the smallest (parallel).
  2. Try 100 Ω and 200 Ω in both connections. Observe which quantity is shared.
  3. Select four equal 1 kΩ resistors: they give 4 kΩ in series and 250 Ω in parallel.
  4. Compare this with capacitors: the equivalent-value rules are reversed, but charge rules cannot simply be reused for current.

Try it yourself

Two 1 kΩ resistors in parallel at 5 V: equivalent resistance, current and power?

Req = 500 Ω; each branch carries 5 mA, for 10 mA total. Each resistor dissipates 25 mW, giving 50 mW total.

What changes if you double the supply while keeping resistances fixed?

Equivalent resistance stays the same, branch currents and voltages double, and power quadruples because P = V²/R.