Voltage Calculation in a Series Circuit: Formula & Example

This calculator finds the voltage drop across one resistor in a series circuit from the total supply voltage and the resistor values. Enter the supply voltage and each resistance; it returns the selected resistor’s voltage drop, along with the circuit current and total resistance.

Formula

For resistors connected in series, the voltage across a particular resistor is the supply voltage multiplied by that resistor’s share of the total resistance:

V₁ = Vₛ × R₁ / Rₜ, where Rₜ = R₁ + R₂ for the two resistors represented here.

• V₁ is the voltage drop across resistor 1, in volts (V).
• Vₛ is the source or supply voltage, in volts (V).
• R₁ and R₂ are the resistor values, in ohms (Ω).
• Rₜ is their combined resistance, in ohms (Ω).
• I, the same current through each series component, is Vₛ / Rₜ, in amperes (A).

The calculator’s two-resistor setup illustrates the rule; for more resistors, add every series resistance to find Rₜ, then use Vₛ × R of the chosen resistor / Rₜ. The input values must describe a simple series path and use consistent resistance units.

Worked Example

Using the default inputs, the supply is 12 V, resistor 1 is 100 Ω, and resistor 2 is 200 Ω.

1. Add the resistances: Rₜ = 100 Ω + 200 Ω = 300 Ω.
2. Find the current: I = 12 V / 300 Ω = 0.040 A, or 40 mA.
3. Find the drop across resistor 1: V₁ = 12 V × 100 Ω / 300 Ω = 4.00 V.

The calculator therefore shows 4.00 V across resistor 1, 0.040 A of current, and 300.00 Ω total resistance. The remaining 8.00 V is across the 200 Ω resistor, so the drops add to the 12 V supply.

How Voltage Divides in Series

A series circuit has one path for current. The same current passes through each component, while the source voltage is shared among them. For ideal resistors, Ohm’s law gives V = IR. Since the current I is common, a larger resistance has a proportionally larger voltage drop.

This relationship is called the voltage-divider rule. If one resistor is twice the value of another, its voltage drop is twice as large, provided the circuit contains only those series resistances and the supply is applied across the whole string. The drops together equal the source voltage in a closed circuit, consistent with Kirchhoff’s voltage law.

Voltage is not “used up” by a resistor. Electrical energy is transferred to heat or another form in circuit components; voltage describes a difference in electric potential between two points. A voltage reading is always between two points, such as across a resistor or between the supply terminals.

Using the Calculator and Reading Results

Enter the voltage applied across the entire series string, not a voltage measured across just one component. Enter each resistor’s resistance in ohms. The calculator treats the two values as ideal resistors in series and reports the drop over resistor 1. To calculate the drop over resistor 2, use the same rule with R₂ in the numerator, or swap the resistor entries.

For a longer chain, add all resistor values for total resistance. Then multiply the supply voltage by the chosen resistor’s fraction of that total. For example, a 1 kΩ resistor is 1,000 Ω; a 2.2 kΩ resistor is 2,200 Ω. Convert units before adding: 1 kΩ = 1,000 Ω, and 1 MΩ = 1,000,000 Ω.

The calculator reports volts, amperes, and ohms. Metric and US customary units do not create a separate voltage conversion here: volts, amperes, and ohms are standard electrical units used in both contexts. Keep sufficient precision during intermediate calculations, then round the displayed answer to a sensible number of digits.

Assumptions and Limits

The formula assumes a steady, direct-current circuit with resistors whose values are known, connected end-to-end with no branches between them. It also assumes the entered source voltage is maintained across the full string and that other components do not significantly affect the result. Real resistor values vary within their tolerance, and their resistance can change with temperature.

For a circuit with a load connected to the midpoint of a voltage divider, that load becomes part of the circuit and can change the output voltage. The simple unloaded divider calculation will then be inaccurate unless the load’s resistance is included in an appropriate circuit analysis. Likewise, components such as diodes, lamps, motors, capacitors, and AC impedances may not behave like fixed resistors; do not apply this basic resistor formula without accounting for their behavior.

Zero resistance values are allowed by the calculator, but if both resistors are zero, total resistance is zero and current cannot be computed. In a physical circuit, an ideal zero-ohm path across a voltage source represents a short circuit, not a usable operating condition. A real circuit’s current is limited by source and wiring characteristics, which this calculator does not model.

Common Mistakes and Circuit Checks

• Using only one resistor in the denominator: the denominator is the total series resistance, not just the selected resistor.
• Mixing units: convert kΩ and MΩ to ohms before adding values. A factor-of-1,000 error changes the answer substantially.
• Confusing supply voltage with a component drop: the input is across the complete string. Individual drops are portions of that voltage.
• Assuming equal voltage drops: drops are equal only when the series resistances are equal.
• Rounding too early: retain the full calculated total and current until the final result.

A useful check is that every calculated drop is between zero and the supply voltage for positive resistors, and that all drops sum to the supply voltage. If those checks fail, revisit the wiring description, units, and arithmetic.

Frequently Asked Questions

How do you calculate voltage across a resistor in series?
Multiply the total supply voltage by the resistor’s value, then divide by the sum of all series resistances: Vᵣ = Vₛ × Rᵣ / Rₜ.

Does voltage stay the same across resistors in series?
No. The source voltage is divided among the resistors. Each drop depends on its resistance, and the drops add up to the supply voltage in the ideal series circuit.

How do I find the total voltage in a series circuit?
Add the individual voltage drops, measured with consistent polarity. For a simple closed resistor circuit, their sum equals the applied source voltage.

Is current the same through every resistor in series?
Yes. A series path has no branches, so the same current passes through each component. The current is the source voltage divided by total resistance for an ideal resistive circuit.

What happens if one resistor in a series circuit is disconnected?
The path is open, so current stops throughout the circuit. The supply voltage may then appear across the break or open component, depending on the circuit and measurement setup.

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in vs | Supply voltage | V | 12 | 0 | 1000000
in r1 | Resistor 1 | ohm | 100 | 0 | 1000000000
in r2 | Resistor 2 | ohm | 200 | 0 | 1000000000
out v1 | Voltage across resistor 1 | V | vs * r1 / (r1 + r2) | 2
out i | Series current | A | vs / (r1 + r2) | 3
out rt | Total resistance | ohm | r1 + r2 | 2