Resistor Voltage Drop Calculator
Calculate the voltage drop across a resistor in a series chain, a two-resistor voltage divider, or from Ohm's law directly. Also sizes the series resistor for an LED and rounds it to the nearest E24 value, with power dissipation checked against a standard quarter-watt rating.
| Resistor | Value (Ω) | Drop | Share |
|---|
A Resistor Does Not Have a Voltage Drop
A 470 Ω resistor does not "drop 4.7 volts". It drops whatever the rest of the circuit hands it. Put the same resistor in a different circuit and the number changes completely — because voltage drop is not a property of the component, it is a property of the component in context.
That single idea clears up most of the confusion around this topic. A resistor only develops a potential difference when current flows through it, and how much current flows depends on everything else in the loop. Take a 470 Ω resistor: at 10 mA it drops 4.7 V, at 1 mA it drops 0.47 V, and with nothing connected it drops nothing at all.
So there are really only two questions worth asking: do you already know the current, or do you not? Each has its own method, and the calculator above handles both.
How to Calculate Voltage Drop Across a Resistor
When you know the current: Ohm's law
This is the direct case. Multiply the current through the resistor by its resistance:
Written compactly this is just V = IR, and the potential drop it gives you is what a voltmeter would read across the component's two legs. A 470 Ω resistor carrying 10 mA drops 0.010 × 470 = 4.7 V. Watch the units: current must be in amperes, so 10 mA becomes 0.01 A. Feeding milliamps straight into the formula is the most common arithmetic slip here, and it inflates the answer by a thousand. The same applies at the small end — a microamp is 0.000001 A — and at the large end, where a kilohm is 1,000 Ω and a megohm is 1,000,000 Ω. The calculator above takes the unit from the dropdown so the conversion is handled for you.
When you do not know the current: the voltage divider rule
This is the case most people actually face, and it is where the confusion sits. You know the supply voltage and the resistor values, but not the current. You do not need it.
In a series loop the same current flows everywhere, so each resistor takes a share of the source voltage in exact proportion to its resistance. That gives the voltage divider rule:
No current calculation required. A 470 Ω resistor in series with a 1 kΩ across 12 V takes 12 × 470 ÷ 1470 = 3.84 V, and the 1 kΩ takes the remaining 8.16 V.
If you would rather work it the long way, find the total resistance, divide the supply by it to get the current, then apply Ohm's law to each resistor. You will get the same answer — the divider rule is simply those two steps collapsed into one.
Voltage Drop Across Resistors in Series
Series is the straightforward arrangement: one current path, so the same current flows through every resistor. Total resistance is the plain sum, and the drops add back up to the supply — that is Kirchhoff's voltage law, which says the potential differences around any closed loop must sum to zero.
Take 12 V across 100 Ω, 220 Ω and 470 Ω in series. Total resistance is 790 Ω, so the current is 12 ÷ 790 = 15.19 mA everywhere in the loop:
| Resistor | Value | Voltage drop | Share of supply |
|---|---|---|---|
| R1 | 100 Ω | 1.519 V | 12.7% |
| R2 | 220 Ω | 3.342 V | 27.8% |
| R3 | 470 Ω | 7.139 V | 59.5% |
| Total | 790 Ω | 12.000 V | 100% |
The largest resistor takes the largest share — always, in series. If your calculated drops do not add up to the supply voltage, you have made an arithmetic error somewhere.
Voltage Drop Across Resistors in Parallel
Parallel is where the intuition flips, and it catches people out constantly.
Resistors in parallel all have the same voltage drop. They share the same two nodes, so they see identical potential difference regardless of their values. What differs is the current through each one.
Put a 1 kΩ and a 10 kΩ in parallel across 5 V and both drop exactly 5 V. The 1 kΩ carries 5 mA, the 10 kΩ carries 0.5 mA — ten times less, because current divides in inverse proportion to resistance. That is the current divider, the mirror image of the voltage divider.
Equivalent resistance for a parallel pair comes from the reciprocal sum, 1/Rtotal = 1/R1 + 1/R2, which for two resistors simplifies to the product over the sum. A 1 kΩ with a 10 kΩ gives 909 Ω — always lower than the smallest branch.
In a combination circuit, collapse each parallel group into its equivalent resistance first, then treat the whole thing as a series chain and apply the divider rule.
The Voltage Divider Rule in Practice
Two resistors in series with the output taken across the lower one — the most-used circuit in electronics after the plain resistor itself. The voltage divider formula sets the divider ratio purely from the two resistances:
The 5 V to 3.3 V divider
Far and away the most common real use: level shifting a 5 V signal down for a 3.3 V microcontroller input on an ESP32 or a 3.3 V Arduino. With R1 = 1 kΩ and R2 = 2 kΩ:
Vout = 5 × 2000 ÷ 3000 = 3.333 V, drawing 1.667 mA.
A 1 kΩ / 2 kΩ pair works, but 10 kΩ / 20 kΩ gives the identical ratio at a tenth of the current — better for battery drain, and the quiescent current of a permanently-powered divider is worth thinking about. Push too high, though, and the divider becomes vulnerable to noise and to the loading effect below.
Why a divider collapses under load
The formula assumes nothing is connected to the output. The moment you attach a load, that load resistance sits in parallel with R2, lowering the effective bottom resistance and pulling Vout down.
This is why a voltage divider is a poor power supply and a fine reference voltage. Feeding a high impedance input — a microcontroller ADC or analog input, an op amp buffer — the loading is negligible, because a high input impedance draws almost no current from the node. Driving anything that actually draws current, the output sags as soon as the load changes. Since both resistors share a single junction, the output is simply the node voltage at that point relative to ground.
Rule of thumb: keep the divider current at least ten times the load current, or buffer the output with an op amp. If you need a stable rail rather than a reference, use a regulator instead.
Building a Voltage Divider on a Breadboard
The physical version is four connections, and it is usually the first thing anyone builds on a prototype board. Place R1 and R2 in series across two rows, run a jumper wire from the supply rail to the free end of R1, another from the free end of R2 to ground, and take the output from the junction between them.
Confirm it with a multimeter before wiring anything downstream: black probe on ground, red probe on the junction. To measure the drop across one specific resistor rather than to ground, put a probe on each of its legs — a voltmeter always reads the potential difference between two points, never at a single point.
Swapping R2 for a potentiometer turns the divider into an adjustable reference, with the wiper as the output. That is exactly how a volume control works, and how a thermistor or light-dependent sensor feeds a microcontroller — the sensor is one leg of a divider, and its changing resistance moves the output voltage.
Sizing a Series Resistor for an LED
The most common practical reason to calculate a resistor voltage drop. An LED is a diode, not a resistor — it holds a roughly fixed forward voltage and its current runs away without something to limit it. The series resistor absorbs the difference between the supply and that forward voltage:
A red LED with a 2.0 V forward voltage running at a forward current of 20 mA from a 5 V supply needs (5 − 2.0) ÷ 0.02 A = 150 Ω. The resistor drops 3.0 V and dissipates 60 mW — comfortable for a quarter watt part. Without that current limiting resistor the LED is effectively a short circuit once its forward voltage is exceeded, and it destroys itself in moments.
Two practical notes. Forward voltage varies by colour: red and amber sit near 2.0 V, blue, white and green nearer 3.2 V. And current flows from anode to cathode only, so an LED wired backwards simply stays dark.
Calculated values rarely land on a stock value, so round to the nearest E24 preferred value — the calculator does this and shows the resulting current. Always round up in resistance if you are near the LED's maximum, since a higher resistance means less current.
Power Rating: the Step People Skip
Getting the resistance right is half the job. A resistor also has to survive the heat, and the power it dissipates is:
Our 470 Ω at 10 mA gives a power dissipation of 0.010² × 470 = 47 mW — forty-seven milliwatt, well inside a standard quarter watt (0.25 W) part. But drop 12 V across a 100 Ω and you are at 1.44 W — a quarter watt resistor there will discolour, drift and eventually fail open.
Standard sizes are 1/8 W, 1/4 W, 1/2 W, 1 W and up. Good practice is to derate: pick a part rated at roughly twice the calculated dissipation, since a resistor at its limit runs hot enough to shift its own value and cook whatever sits next to it on the board.
Tolerance matters less than people fear. Stock resistors come in preferred values defined by IEC 60063 — the coarse E12 series at 10% steps, the E24 series at 5%, and E96 for 1% parts. For a divider feeding an ADC, 1% is worth paying for; for an LED current limiter 5% is fine, since a few percent of current makes no visible difference. Where a value really matters, take the tolerance and power rating from the manufacturer's datasheet rather than a generic table — component makers such as KOA Speer and Vishay publish full specifications for every resistor series they build.
Related Calculators
This calculator handles the circuit-theory side — drops across components in a circuit. For conductor and installation work, where the "resistor" is the cable itself, these cover the rest:
Solve V, I, R and P from any two known values.
Cable and conductor drop in AWG or mm², with NEC limits.
Resistor Color Code Calculator
Read a resistor's value and tolerance from its bands.
Size a conductor for current and acceptable drop.
More of these are collected in our engineering calculators hub.
Frequently Asked Questions
How do you calculate voltage drop across a resistor?
Multiply the current through the resistor by its resistance: V = I × R. A 470 Ω resistor carrying 10 mA drops 0.010 × 470 = 4.7 V. Make sure current is in amperes rather than milliamps, since that unit slip is the most common error. If you do not know the current, use the voltage divider rule instead: each resistor in a series chain takes a share of the supply proportional to its resistance.
How do you find voltage drop across a resistor without knowing the current?
Use the voltage divider rule: Vn = Vsupply × Rn ÷ Rtotal. Because the same current flows through every resistor in a series loop, each one takes a share of the supply voltage in exact proportion to its resistance, so the current cancels out of the maths entirely. A 470 Ω resistor in series with a 1 kΩ across 12 V drops 12 × 470 ÷ 1470 = 3.84 V, with no current calculation needed at any point.
Does voltage drop across a resistor?
Yes, whenever current flows through it. The resistor converts a portion of the circuit's electrical energy into heat, and the voltage measured across its two legs is that portion. With no current flowing there is no drop at all, which is why an open circuit shows the full supply voltage across the break and zero across every resistor in the loop.
Do resistors in parallel have the same voltage drop?
Yes. Parallel resistors connect to the same two nodes, so they all see the same potential difference regardless of their values. What differs is current: it divides in inverse proportion to resistance, so a 1 kΩ carries ten times the current of a 10 kΩ beside it. This is the opposite of a series circuit, where current is shared equally and voltage divides.
What is the voltage divider rule?
The voltage divider rule states that in a series circuit, the voltage across any resistor equals the supply voltage multiplied by that resistor's share of the total resistance: Vn = Vsupply × Rn ÷ Rtotal. For the common two-resistor divider with the output taken across the lower resistor, this becomes Vout = Vin × R2 ÷ (R1 + R2). It applies only when no significant load is drawing current from the output.
What resistors do I need for a 5 V to 3.3 V divider?
A 1 kΩ for R1 and 2 kΩ for R2 gives 3.333 V from a 5 V input, drawing 1.667 mA. Using 10 kΩ and 20 kΩ produces the identical voltage at a tenth of the current, which is preferable on battery power. This works for level-shifting a 5 V signal into a 3.3 V microcontroller input, but only for high-impedance inputs — it is not a power supply and will sag under any real load.
Why does my voltage divider output the wrong voltage?
Almost always the loading effect. The divider formula assumes an unloaded output, but whatever you connect sits in parallel with R2, reducing the effective bottom resistance and pulling the output down. Keep the current through the divider at least ten times the load current, use much smaller resistor values, or buffer the output with an op amp. Resistor tolerance is a second, smaller cause — 5% parts can shift the ratio noticeably.
What size resistor do I need for an LED?
Subtract the LED's forward voltage from the supply, then divide by the desired current: R = (Vsupply − Vforward) ÷ Iforward. A red LED at 2.0 V forward running 20 mA from 5 V needs (5 − 2.0) ÷ 0.02 = 150 Ω. Red and amber LEDs sit near 2.0 V forward voltage while blue, green and white are closer to 3.2 V, so always check the datasheet rather than assuming.
How do I measure voltage drop across a resistor with a multimeter?
Set the meter to DC volts and place one probe on each leg of the resistor while the circuit is powered and running. A voltmeter reads the potential difference between two points, so both probes must touch the component you are measuring — probing a single point tells you nothing. Do not break the circuit to insert the meter; that is how you measure current, not voltage.
What power rating resistor do I need?
Calculate the dissipation with P = I² × R or V² ÷ R, then choose a part rated at roughly twice that figure. A 470 Ω resistor at 10 mA dissipates 47 mW, so a standard quarter watt part is fine. Dropping 12 V across 100 Ω dissipates 1.44 W and needs a 2 W or larger resistor. Derating matters because a component running at its rated limit gets hot enough to drift in value and damage nearby parts.
These calculations assume ideal resistors at room temperature in DC circuits. Real components carry tolerance, temperature coefficients and, at higher frequencies, parasitic inductance and capacitance. For circuits where precision matters, check the datasheet values and verify the built circuit with a multimeter rather than relying on calculation alone.