How to calculate voltage drop

What is voltage drop?

Voltage, or potential difference, measures the energy each unit of charge carries. As current flows through a resistor, a lamp or a long wire, the charge gives up some of that energy, often as heat. The voltage between the two ends of the component is its voltage drop. A voltmeter connected across the component measures it directly.

The voltage drop formula

This is Ohm's law, which AQA writes as potential difference = current × resistance. To find the drop across one component, use the current through that component and its resistance.

1. Find the current through the component, in amps. Convert milliamps by dividing by 1,000.
2. Find its resistance in ohms. Convert kilohms by multiplying by 1,000.
3. Multiply them to get the drop in volts.

Try one: this question asks you to read a resistance from a current against voltage graph using the same equation.

Voltage drop in a series circuit

AQA's rule for series circuits is that every component carries the same current and the components split the supply's voltage between them. So the drops always add up to the supply voltage. This is Kirchhoff's voltage law: around any closed loop, the voltage gains and drops add to zero (OpenStax).

Worked example 1: two resistors in series

A 12 V cell is connected to a 4 Ω and an 8 Ω resistor in series.

1. Total resistance in series: 4 + 8 = 12 Ω.
2. Current: I = V ÷ R = 12 ÷ 12 = 1 A, the same through both resistors.
3. Drop across the 4 Ω resistor: 1 × 4 = 4 V. Across the 8 Ω resistor: 1 × 8 = 8 V.
4. Check: 4 + 8 = 12 V, the supply voltage.

The bigger resistor takes the bigger share, in proportion to its resistance. Build this circuit and read the voltmeters in our Circuit Builder.

The potential divider shortcut

Because the drops are shared in proportion to resistance, you can skip the current. For two resistors R₁ and R₂ in series across a supply V_in, the drop across R₂ is:

With a 9 V supply, R₁ = 6 kΩ and R₂ = 3 kΩ, the drop across R₂ is 9 × 3 ÷ (6 + 3) = 3 V. This circuit is called a potential divider, and it is how sensors such as light dependent resistors give a voltage a microcontroller can read.

Voltage drop in a parallel circuit

In parallel, AQA notes that the potential difference across each component is the same. Each branch connects straight across the supply, so each branch drops the full supply voltage, whatever its resistance. What differs between branches is the current: lower resistance branches carry more.

Voltage drop inside a battery

Real cells have some internal resistance, r. When current flows, part of the cell's voltage, its emf (ε), is dropped inside the cell itself, so less reaches the circuit. The voltage at the terminals is:

A battery with an emf of 9 V and internal resistance 0.5 Ω supplying 2 A loses 2 × 0.5 = 1 V inside, so its terminal voltage is 9 − 1 = 8 V. This is why a battery's voltage sags when it powers something that draws a big current.

How to calculate voltage drop in a cable

Wires have resistance too, so long cables carrying large currents waste voltage. First find the resistance of the wire from its resistivity (ρ), length (L) and cross-sectional area (A):

Copper's resistivity is 1.68 × 10⁻⁸ Ω m at 20 °C, according to OpenStax University Physics. Remember the current has to go out and back, so the total wire length is twice the cable run.

Worked example 2: a long cable run

A 30 m run of copper cable with 2.5 mm² conductors (2.5 × 10⁻⁶ m²) carries 16 A.

1. Resistance of one conductor: 1.68 × 10⁻⁸ × 30 ÷ (2.5 × 10⁻⁶) = 0.2016 Ω.
2. Out and back: 2 × 0.2016 = 0.4032 Ω.
3. Voltage drop: 16 × 0.4032 = 6.45 V.
4. As a percentage of a 230 V supply: 6.45 ÷ 230 × 100 ≈ 2.8%. On a 120 V supply the same drop would be about 5.4%.

Thicker cable means less resistance. With 4 mm² conductors the drop falls to about 4.0 V, or 1.8% of 230 V. Electricians check cable drop against limits in their wiring rules, such as BS 7671 in the UK or the NEC in the US. Always follow those rules and a qualified electrician for real installations; this example only shows the physics.

Common mistakes

• Using the total resistance for one component. Use that component's own resistance and the current through it.
• Forgetting unit prefixes. 250 mA is 0.25 A and 3.3 kΩ is 3,300 Ω.
• Splitting the voltage in parallel. Each parallel branch gets the full supply voltage.
• Using the one-way cable length. Double it for the return wire.
• Using mm² directly in R = ρL/A. Convert to m²: 1 mm² = 10⁻⁶ m².
• Thinking voltage is used up. Energy is transferred; the drops show where.

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