Series Resistor Calculator

Series components form one unbroken path, so the same current passes through every one of them and the resistances simply add:

$R_{\text{total}} = R_1 + R_2 + \cdots + R_n$

Symbols and units:

• $R_{\text{total}}$ — total series resistance, ohms (Ω)
• $I$ — the single current shared by every element, amperes (A)
• $V_n$ — the drop across element $n$, volts (V)

The current follows from Ohm's law applied to the whole string, and each drop from Ohm's law applied to one element:

$I = \frac{V_{\text{supply}}}{R_{\text{total}}} \qquad V_n = I R_n$

Kirchhoff's voltage law guarantees the drops add back to the supply voltage, which is the check you should run every time.

The operating assumption: DC or steady-state resistive AC, ideal wires, and a source with no significant internal resistance. Reactive parts need impedance instead.

The mistake people make: adding a series total, then using the supply voltage again for a single resistor. Only $IR_n$ belongs to element $n$.

You rarely need the current at all. Because the same $I$ flows everywhere, the supply voltage divides in direct proportion to resistance:

$V_n = V_{\text{supply}} \times \frac{R_n}{R_{\text{total}}}$

The largest resistor takes the largest share. Two equal resistors split the supply exactly in half.

This is also how a series resistor protects an LED. An LED is not ohmic: it holds a roughly fixed forward voltage $V_f$ and the resistor absorbs the rest, so the resistor is sized from the difference:

$R = \frac{V_{\text{supply}} - V_f}{I_f}$

with $I_f$ the wanted forward current in amperes.

Contrast this with parallel, where every branch sees the whole supply voltage and the currents add instead. Same two resistors, opposite behaviour: $2 \times 100\ \Omega$ gives $200\ \Omega$ in series but $50\ \Omega$ in parallel.

The mistake people make: sizing the LED resistor from the supply voltage rather than from $V_{\text{supply}} - V_f$, which lets far too much current through.

• Adding series resistors as reciprocals — reciprocals belong to parallel. In series you just add.
• Using the supply voltage in $V = IR$ for one element — use that element's own current and resistance.
• Mixing prefixes — $2.2$ kΩ is $2200$ Ω; adding $470 + 1 + 2.2$ is meaningless.
• Forgetting to check the sum — the individual drops must add back to the supply voltage. If they do not, the arithmetic is wrong.
• Rounding the current too early — a current rounded to two figures propagates into every drop. Carry extra digits and round only the final answers.
• Ignoring the power in each resistor — $P_n = I^2 R_n$, and the biggest resistor in a series string dissipates the most.
• Assuming an LED behaves like a resistor — its forward voltage is nearly constant, so it needs the difference formula, not a divider.