How do you calculate voltage drop across a resistor?

Apply Ohm's Law: V = I × R. Take the current I (in amps) flowing through the resistor and multiply by its resistance R (in ohms) — the result is the voltage in volts that drops across that resistor. Example: 0.5 A through a 100 Ω resistor drops 50 V. Power dissipated is P = I² × R = 25 W. Use the Resistor mode in the calculator above to compute this with your own values.

How do you calculate voltage drop across resistors in series?

In a series circuit, the same current flows through every resistor, and the voltage drops add up to the source voltage (Kirchhoff's voltage law). Calculate each individual drop with Vi = I × Ri, then sum them: Vtotal = V1 + V2 + … + Vn. Equivalently, Vtotal = I × (R1 + R2 + … + Rn). Use the Series mode in the calculator to add up to six resistors.

How do you calculate voltage drop in a series circuit step by step?

First, find the total series resistance: Rtotal = R1 + R2 + … + Rn. Second, find the current: I = Vsource / Rtotal. Third, calculate the drop across each resistor: Vi = I × Ri. The drops should sum back to Vsource — that is the verification step required by Kirchhoff's voltage law.

How do you calculate total voltage in a circuit?

For a closed series loop, the total voltage equals the source voltage and equals the sum of all drops around the loop: Vsource = V1 + V2 + … + Vn. For parallel branches, the voltage across each branch is the same and equal to the source voltage; the currents through the branches divide instead. The conductor drop V = 2·L·I·ρ/A is just Ohm's Law applied to the wire treated as one more resistance in the loop.

How do you calculate voltage across a capacitor?

Capacitor voltage is not a drop in the conductor sense — it builds up as the capacitor charges. For an RC circuit charging from V0 through resistance R: VC(t) = V0(1 − e−t/RC). For sinusoidal AC, the capacitor presents reactance XC = 1/(2πfC), and the voltage across it is VC = I × XC. For a full RC and reactance calculator, see the Capacitive Reactance Calculator.

Why does my measured voltage drop differ from my calculation?

Common reasons: (1) actual conductor temperature is higher than the 20°C / 75°C base used in the formula — apply temperature correction to ρ; (2) length used in the calculation is shorter than the real cable route; (3) splices and terminations add 0.5–2 V of extra drop on long runs; (4) loose connections are the most frequent culprit and easy to miss. Verify connection torque before suspecting the calculation.

What is the formula for voltage drop in a wire?

For DC and single-phase AC: Vdrop = 2 × L × I × ρ / A, where L is one-way length in metres, I is current in amps, ρ is conductor resistivity (0.0175 Ω·mm²/m for copper, 0.028 for aluminium at 20°C), and A is cross-section in mm². The factor of 2 accounts for the round-trip through the conductor. For three-phase, replace 2 with √3.

Tutorial · Electrical · Ohm's Law · KVL · NEC

How to calculate voltage drop

Three modes in one tutorial: across a resistor, across resistors in series, and along a conductor. Live formula substitution shows your work step by step. NEC 3%/5% check on the conductor mode. Reviewed by a licensed PE.

Use the calculator

Switch between three teaching modes. Each mode shows the formula with your values substituted, so you see the arithmetic at every step rather than just the final number.

CALC.002 Voltage Drop · Tutorial · 3 modes

Current[I]

Resistance[R]

Ohm's Law applied to a single component. Use this when you know the current through a known resistor.

Voltage drop · V_drop

20.00 V

Show your work

V = 2 × 10 = 20.00 V

Power dissipated: 40.00 W
Total resistance: 10.00 Ω

FORMULA · V = I × R OHM'S LAW

Voltage drop — at a glance

Voltage drop formula: For DC and 1-phase AC: V_drop = 2·L·I·ρ/A. For 3-phase, replace 2 with √3.
Voltage drop across a resistor: Apply Ohm's Law: V = I·R. Example: 0.5 A through 100 Ω drops 50 V and dissipates 25 W.
Voltage drop across resistors in series: Same current in every R. Add the drops: V_total = I·(R₁ + R₂ + … + R_n).
Voltage drop along a wire: V_drop = 2·L·I·ρ/A (Cu ρ = 0.0175, Al ρ = 0.028 Ω·mm²/m at 20°C). Factor of 2 covers the round-trip path.
NEC voltage drop limit: 3% on a branch circuit, 5% combined feeder + branch (NEC 210.19(A) Informational Note).
Voltage drop in DC vs 240 V: Drop percentage scales inversely with voltage. A 4 V drop is 33% on 12 V but only 1.7% on 240 V.
3-phase voltage drop formula: V_drop = √3·L·I·ρ·cos φ / A for line-to-line drop on a 3-phase circuit.

The voltage drop formula, in three forms

All three forms below are the same physical law — Ohm's Law — applied to different shapes of resistance. Pick the form that matches your problem.

Eq. 01 — Across a single resistor (Ohm's Law) SI · Ohm 1827

V = I \cdot R

V: voltage drop across resistor, V
I: current through resistor, A
R: resistance, Ω

Eq. 02 — Across resistors in series (KVL) SI · Kirchhoff 1845

V_{total} = \sum_{i=1}^{n} V_i = I \cdot \sum_{i=1}^{n} R_i

V_total: sum of drops in the loop, V
I: series current (same in every R), A
R_i: each resistance in the chain, Ω

Eq. 03 — Along a conductor (DC and 1-phase AC) SI · NEC 2023 Table 9 · IEC 60287

V_{drop} = \frac{2 \cdot L \cdot I \cdot \rho}{A}

V_drop: voltage lost in the wire, V
L: one-way conductor length, m
I: current, A
ρ: resistivity (Cu 0.0175, Al 0.028), Ω·mm²/m
A: cross-sectional area, mm²

Worked example, all three modes

Here is the same arithmetic written out for one realistic case in each mode. Use these numbers as a sanity check against the calculator.

Mode	Inputs	Substitution	Result
Resistor	I = 0.5 A, R = 100 Ω	V = 0.5 × 100	50 V (P = 25 W)
Series	I = 0.05 A; R₁ = 100, R₂ = 200, R₃ = 300 Ω	V_total = 5 + 10 + 15	30 V (R_tot = 600 Ω)
Conductor	30 m of 10 AWG (5.26 mm²) Cu, 20 A, 12 V	V = (2 × 30 × 20 × 0.0175) / 5.26	3.99 V (33%) — fails NEC 5%

The conductor case is the same one used on the main Voltage Drop Calculator — it intentionally fails NEC limits to demonstrate why low-voltage DC runs are punishing: the same 4 V drop is ~33% on 12 V but only 1.7% on 240 V.

How to calculate voltage drop, step by step

Identify what is dropping voltage. A single resistor, a series chain of resistors, or a length of wire. The formula form changes for each — same physics, different shape.
Find the current through it. In a series circuit, the same current flows through every element. From a known voltage and total resistance: I = V_source / R_total. From nameplate or full-load amps for wired equipment.
Find the resistance value. For a component — read the label or measure with a multimeter. For a wire — use cross-section A and resistivity ρ (Cu = 0.0175, Al = 0.028 Ω·mm²/m at 20°C), with the factor of 2 for the round-trip path: R_wire = 2·L·ρ/A.
Apply the formula. V = I × R. For a series circuit add up V across each resistor: V_total = V₁ + V₂ + … + V_n. For a wire, expand directly: V_drop = 2·L·I·ρ/A.
Check against the limit. For a conductor: NEC 3% on a branch circuit, 5% combined feeder + branch (NEC 210.19(A)). For a component: ensure power dissipation P = V·I stays within its rating.
If the drop is too large, fix it. For a wire — increase cross-section (one or two sizes up), shorten the run, or step up the system voltage. For a component — switch to a higher-rated part or reduce current.

Variants and special cases

Voltage drop across a single resistor

This is the simplest application of Ohm's Law: V = I × R. Take the current flowing through the resistor and multiply by its resistance. Power dissipated is P = I² × R, which sets the wattage rating you need: a 0.5 A current through a 100 Ω resistor drops 50 V and dissipates 25 W — which means a ¼-watt resistor will burn up. Always check both V and P when sizing a resistor.

The same approach extends to any single component with a known impedance — including a winding, a heating element, or an inductor at DC steady state. For AC, replace R with the component's impedance Z; for a capacitor, see the FAQ note below on capacitive reactance.

Voltage drop in a series circuit

In a series chain of resistors, the same current flows through every resistor (current is conserved), and the source voltage divides among the resistors in proportion to their resistance values. This is Kirchhoff's voltage law applied to a single loop: the algebraic sum of voltage drops around any closed loop equals the source voltage.

To solve a series circuit: (1) add the resistances to get R_total; (2) divide source voltage by R_total to get the loop current I; (3) multiply I by each individual R_i to get each drop V_i. Verify by summing — V₁ + V₂ + … must equal V_source. If they don't, an arithmetic error has crept in.

Related concepts on this site

The 3% / 5% NEC rule, in one block

For conductor voltage drop only — not relevant to discrete resistors. The National Electrical Code recommends keeping branch-circuit voltage drop below 3% and combined feeder + branch drop below 5%. These are not absolute legal limits but engineering targets enforced through the informational note in Article 210.19(A).

Conductors for branch circuits as defined in 100, sized to prevent a voltage drop exceeding 3 percent at the farthest outlet of power, heating, and lighting loads, or combinations of such loads, and where the maximum total voltage drop on both feeders and branch circuits to the farthest outlet does not exceed 5 percent, will provide reasonable efficiency of operation.

NFPA 70 (NEC) 2023 Edition → Article 210.19(A) Informational Note No. 4

Frequently asked questions

How do you calculate voltage drop across a resistor?: Apply Ohm's Law: V = I × R. Take the current I (in amps) flowing through the resistor and multiply by its resistance R (in ohms) — the result is the voltage in volts that drops across that resistor. Example: 0.5 A through a 100 Ω resistor drops 50 V. Power dissipated is P = I² × R = 25 W. Use the Resistor mode in the calculator above to compute this with your own values.
How do you calculate voltage drop across resistors in series?: In a series circuit, the same current flows through every resistor, and the voltage drops add up to the source voltage (Kirchhoff's voltage law). Calculate each individual drop with V_i = I × R_i, then sum them: V_total = V₁ + V₂ + … + V_n. Equivalently, V_total = I × (R₁ + R₂ + … + R_n). Use the Series mode in the calculator to add up to six resistors.
How do you calculate voltage drop in a series circuit step by step?: First, find the total series resistance: R_total = R₁ + R₂ + … + R_n. Second, find the current: I = V_source / R_total. Third, calculate the drop across each resistor: V_i = I × R_i. The drops should sum back to V_source — that is the verification step required by Kirchhoff's voltage law.
How do you calculate total voltage in a circuit?: For a closed series loop, the total voltage equals the source voltage and equals the sum of all drops around the loop: V_source = V₁ + V₂ + … + V_n. For parallel branches, the voltage across each branch is the same and equal to the source voltage; the currents through the branches divide instead. The conductor drop V = 2·L·I·ρ/A is just Ohm's Law applied to the wire treated as one more resistance in the loop.
How do you calculate voltage across a capacitor?: Capacitor voltage is not a drop in the conductor sense — it builds up as the capacitor charges. For an RC circuit charging from V₀ through resistance R: V_C(t) = V₀(1 − e^−t/RC). For sinusoidal AC, the capacitor presents reactance X_C = 1/(2πfC), and the voltage across it is V_C = I × X_C. For a full RC and reactance calculator, see the Capacitive Reactance Calculator.
Why does my measured voltage drop differ from my calculation?: Common reasons: (1) actual conductor temperature is higher than the 20°C / 75°C base used in the formula — apply temperature correction to ρ; (2) length used in the calculation is shorter than the real cable route; (3) splices and terminations add 0.5–2 V of extra drop on long runs; (4) loose connections are the most frequent culprit and easy to miss. Verify connection torque before suspecting the calculation.
What is the formula for voltage drop in a wire?: For DC and single-phase AC: V_drop = 2 × L × I × ρ / A, where L is one-way length in metres, I is current in amps, ρ is conductor resistivity (0.0175 Ω·mm²/m for copper, 0.028 for aluminium at 20°C), and A is cross-section in mm². The factor of 2 accounts for the round-trip through the conductor. For three-phase, replace 2 with √3.

Sources and methodology

Ohm, G. S. Die galvanische Kette, mathematisch bearbeitet, 1827. Original statement of V = I·R.
Kirchhoff, G. R. Ueber den Durchgang eines elektrischen Stromes durch eine Ebene, Annalen der Physik, 1845. Voltage and current laws.
NFPA. National Electrical Code (NEC) NFPA 70, 2023 Edition. Article 210.19(A); Table 9 (resistance values).
IEC. IEC 60287-1-1: Electric cables — Calculation of the current rating, 2023.
IEEE. IEEE Std 141-1993 (R1999) Recommended Practice for Electric Power Distribution for Industrial Plants (Red Book). Conductor sizing and voltage drop methodology.