Ohm’s law
A resistor drops voltage because that is what it is there to do. Give this any two of voltage, current and resistance and it returns the third, along with the power the resistor has to dissipate.
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Fill in the two quantities you know and the third is worked out for you.
Voltage drop across the resistor
9.4V
Choose a resistor rated well above the dissipated power. A factor of two is a common working margin. A resistor run at its rated power gets hot enough to drift in value and to damage what it is mounted on.
Voltage drop across the resistor
V = I × RV = 0.02 A × 470 ΩResult: 9.4 V
Ohm’s law. The drop is set entirely by the current through the resistor and its resistance.
Power dissipated
P = V × IP = 9.4 V × 0.02 AResult: 0.188 W
Equivalent to I² × R and to V² ÷ R. Pick a resistor rated comfortably above this figure.
Series resistors all carry the same current, so each drops voltage in proportion to its resistance. The drops always add back up to the source voltage.
12 V across 100 Ω + 220 Ω + 470 Ω
The idea
Current flowing through a resistance produces a potential difference across it. That is Ohm’s law, and it is the whole of the explanation: V = I × R. Push 20 mA through 470 Ω and 9.4 V appears across the resistor. Push twice the current and you get twice the voltage.
The energy that difference represents does not vanish. It becomes heat, at a rate of P = V × I watts. This is the number people forget, and it is the one that decides whether the resistor survives: 9.4 V at 20 mA is 0.188 W, comfortable for a quarter-watt part but well past a tenth-watt one.
In a series circuit every resistor carries the same current, so each one drops voltage in proportion to its own resistance, and the drops always add back up to the source voltage. That is Kirchhoff’s voltage law, and it is what makes a voltage divider work.
Not the same thing
The drop is the point. You chose the resistance to produce it, to set an LED current, to bias a transistor, to scale a signal. Nothing is wasted that you did not intend to spend, and the only design question is whether the part can shed the heat.
V = I × R
No length. No material. No system multiplier. Two terminals and a number.
The drop is a side effect of getting power somewhere. The conductor has resistance whether you want it or not, the drop is pure loss, and every volt spent on it is a volt the load does not get. So the design question is how to make it smaller.
V_drop = 2 × I × R × L
Depends on length, material, temperature and whether the circuit is DC, single-phase or three-phase. That is what the main calculator is for.
Questions
Use Ohm’s law: V = I × R. Multiply the current through the resistor in amperes by its resistance in ohms. A 470 Ω resistor carrying 20 mA drops 0.02 × 470 = 9.4 V. If you know the voltage and need the current instead, rearrange to I = V ÷ R.
Resistors in series all carry the same current, so each one drops voltage in proportion to its resistance. Add the resistances to get the total, divide the source voltage by that total to get the current, then multiply that current by each resistance. Across 100 Ω, 200 Ω and 300 Ω on a 12 V supply, the current is 20 mA and the drops are 2 V, 4 V and 6 V, which add back to the 12 V you started with.
P = V × I, which is equivalent to I² × R and to V² ÷ R. A 470 Ω resistor dropping 9.4 V at 20 mA dissipates 0.188 W. Choose a resistor with a power rating comfortably above the calculated figure (a factor of two is a common working margin), because a resistor run at its rated power gets hot enough to drift and to damage what it is mounted on.
The physics is identical (both are I × R), but the intent is opposite. A resistor is placed in a circuit to drop voltage; that is its job. A cable drops voltage as an unavoidable side effect of carrying current, and that drop is pure loss. So a resistor calculation has no length, no material and no system multiplier, while a cable calculation needs all three.
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