Resistors in Series Calculator
Calculate total resistance, current, and individual voltage drops for series circuits using R_total = R₁ + R₂ + ....
Enter up to 5 resistor values and supply voltage. Leave unused fields blank.
What Is Resistors in Series?
In a series circuit, all components share the same current. Total resistance is simply the sum of all individual resistances: R_total = R₁ + R₂ + R₃ + .... This makes series resistor combinations easy to analyze — adding resistors always increases total resistance.
By Ohm's Law, the current through a series circuit is I = V_supply / R_total. This same current I flows through every resistor in the string. The voltage drop across each resistor is Vᵢ = I × Rᵢ. All voltage drops sum to the supply voltage: V₁ + V₂ + ... = V_supply (Kirchhoff's Voltage Law).
Power dissipated in each resistor: Pᵢ = I² × Rᵢ = Vᵢ × I = Vᵢ²/Rᵢ. Resistors with higher R dissipate more power. Total power: P_total = I² × R_total = V_supply × I. In LED circuits, series resistors limit current and 'drop' the excess voltage between supply and LED forward voltage.
Series resistors have many practical applications: voltage dividers (with two resistors), current limiting (LED resistors), pull-up/pull-down resistors in digital circuits, and attenuation networks in audio equipment. Understanding series resistance is fundamental to all circuit analysis.
Formula Reference Table
| Solve For | Formula | Notes |
|---|---|---|
| Total resistance | R_total = R₁ + R₂ + ... + Rₙ | Ohms; always larger than any single Rᵢ |
| Current | I = V_supply / R_total | Same current through all resistors |
| Voltage drop (each) | Vᵢ = I · Rᵢ | Proportional to resistance |
| KVL (Kirchhoff) | V₁ + V₂ + ... = V_supply | Voltage drops sum to supply |
| Power (each) | Pᵢ = I² · Rᵢ | Watts |
| Total power | P = V_supply · I = I² · R_total | Equal to sum of individual powers |
3 Worked Examples
100 Ω, 220 Ω, 470 Ω in series at 12 V.
- R_total = 100 + 220 + 470 = 790 Ω
- I = 12 / 790 = 15.19 mA
- V₁ = 15.19e-3 × 100 = 1.52 V; V₂ = 15.19e-3 × 220 = 3.34 V; V₃ = 15.19e-3 × 470 = 7.14 V
- Check: 1.52 + 3.34 + 7.14 = 12.0 V ✓
LED needs 2.0 V, 20 mA. Supply = 5 V.
- Voltage across resistor = 5 − 2.0 = 3.0 V
- R = V/I = 3.0 / 0.020 = 150 Ω
- Use nearest standard: 150 Ω or 180 Ω (slightly reduces current for safety)
Two 1 kΩ resistors at 9 V. One opens (becomes infinite resistance).
- R_total → ∞; I → 0 mA
- All 9 V appears across the open resistor; 0 V across good one
- This is how open-circuit faults manifest in series circuits
Real-World Applications
Common Mistakes to Avoid
Series resistors always add — total R is always greater than the largest individual value. If your calculation gives a smaller total, recheck.
Voltage drops are proportional to resistance: V₁/V₂ = R₁/R₂. Only equal resistors share voltage equally. Larger resistors have larger drops.
WRONG — in series, current is identical everywhere in the loop. I is the same through R₁, R₂, and R₃.
If one resistor = 0 Ω (shorted), it's ignored. But if R_total ≈ 0 with high voltage, current becomes dangerously high. Shorts cause fuse/breaker trips.
Mix of kΩ and Ω: convert all to Ω first. 1 kΩ = 1,000 Ω. Mixing them causes errors by factors of 1,000.
Frequently Asked Questions
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Formula Explorer connections
Interpretation: This formula links charge, voltage, current, resistance, capacitance, power or circuit time response. Assumption: Confirm DC versus AC conditions, RMS versus peak values, component topology and steady-state versus transient behavior. Ideal components may be assumed.