Resistors in series

Old-style decorative light strings had a maddening flaw: if one bulb's filament snapped, the entire string went dark, every other bulb included. That single fact is a clue to how the bulbs were wired — end to end, in one unbroken loop, with no branching path for current to take. This is a series combination, the simplest way to connect resistors, and it follows directly from the Ohm's law you already know. Put resistors one after another in a single loop and two things happen at once: the same current is forced through every one of them, while the battery's total push (potential difference) gets shared out, split unevenly across each resistor in proportion to how much it resists. Add up those shares and a strikingly simple rule falls out — the equivalent resistance of a series combination is just the plain sum of the individual resistances. This article builds that rule from Ohm's law, works two full numericals with it, and flags the mix-ups (current vs voltage, series vs parallel) that cost the most marks.

The core insight

In a series combination the current is the same through every resistor, so the equivalent resistance is simply the sum of the individual resistances, $R_s = R_1 + R_2 + R_3 + \dots$

The core idea

One loop, one current. Resistors in series are joined end-to-end so charge has exactly one path to follow through all of them. Because charge cannot pile up or vanish anywhere along a single loop, the same current $I$ must flow through every resistor and through the battery itself — this is the one fact that defines a series combination and everything else follows from it.

The voltage splits, current doesn't. Each resistor still obeys Ohm's law individually, so a bigger resistor takes a bigger "share" of the battery's potential difference even while the same current passes through all of them. For resistors $R_1$ , $R_2$ , $R_3$ in series across a battery of potential difference $V$ , with individual potential differences $V_1$ , $V_2$ , $V_3$ , the total splits as their sum:

V = V_1 + V_2 + V_3

where: $V$ is the potential difference supplied by the battery, and $V_1$ , $V_2$ , $V_3$ are the potential differences across $R_1$ , $R_2$ , $R_3$ respectively — the drops add up to account for the whole push the battery supplies.

One number replaces the whole chain. For any external circuit, a single equivalent resistance $R_s$ is defined as the one resistor that would draw the same current from the same battery. Substituting Ohm's law into the voltage-sum equation (worked out in full below) gives the defining formula of this topic:

R_s = R_1 + R_2 + R_3 + \dots

where: $R_s$ is the equivalent (total) resistance of the series combination and $R_1, R_2, R_3, \dots$ are the individual resistances being added — in series, resistance only ever grows, so $R_s$ is always larger than the biggest individual resistor.

Figure

Three resistors R1 R2 R3 joined in a single loop with one battery showing the same current I flowing through all three and separate voltage drops V1 V2 V3 adding up to the battery voltage V

Common traps

Assuming voltage is the same across every resistor in series — it is current that stays the same in series; voltage divides across the resistors unequally, in proportion to each resistance.
Forgetting that $R_s$ in a series combination is always larger than any individual resistor — a numerical answer smaller than one of the given resistors is a sure sign of an arithmetic slip.
Adding potential differences without first confirming the resistors are genuinely in series, that is, a single loop with one current path — this rule does not apply to branching, parallel circuits.
Solving multi-resistor problems by guessing values instead of the systematic route: find $R_s$ , then $I$ from the whole circuit, then apply $V = IR$ to each resistor individually.

Derivation ADeriving Rs equals R1 plus R2 plus R3 from Ohm's law

The goal is to show that the equivalent-resistance rule used throughout this article, $R_s = R_1 + R_2 + R_3$ , is not a separate law of its own but a direct, two-line consequence of Ohm's law applied to each resistor plus the one defining fact of a series circuit — that the same current flows through all of them — so a student never needs to memorise it as an isolated formula.

Step 1. Start from the defining feature of a series combination: a single current $I$ flows through every resistor, since there is only one path. Apply Ohm's law separately to each resistor using this shared current: $V_1 = IR_1$ , $V_2 = IR_2$ , $V_3 = IR_3$ .

Step 2. Apply Ohm's law to the combination as a whole, treating it as one equivalent resistor $R_s$ carrying the same current $I$ from the same battery of potential difference $V$ : $V = IR_s$ .

Step 3. Use the fact that the total potential difference is the sum of the individual drops, $V = V_1 + V_2 + V_3$ , and substitute every term from Steps 1 and 2: $IR_s = IR_1 + IR_2 + IR_3$ .

Step 4. Divide both sides by the common current $I$ (valid since $I$ is the same nonzero current throughout the loop), leaving $R_s = R_1 + R_2 + R_3$ . Because every term on the right is a positive resistance, $R_s$ must be larger than any single $R_1$ , $R_2$ , or $R_3$ on its own — the equivalent resistance of a series chain only ever grows as more resistors are added.

Worked example

Three resistors of 2 ohm, 3 ohm, and 5 ohm are connected in series across a 20 V battery. Find (a) the equivalent resistance, (b) the current in the circuit, and (c) the potential difference across the 5 ohm resistor.

Given: $R_1 = 2\ \Omega$ , $R_2 = 3\ \Omega$ , $R_3 = 5\ \Omega$ , $V = 20\ \text{V}$ .

Step 1. Find the equivalent resistance by simply adding the three resistances: $R_s = R_1 + R_2 + R_3 = 2 + 3 + 5 = 10\ \Omega$ .

Step 2. Find the current using Ohm's law on the whole combination, since this current is the same everywhere in the loop: $I = \dfrac{V}{R_s} = \dfrac{20\ \text{V}}{10\ \Omega} = 2\ \text{A}$ .

Step 3. Find the potential difference across the 5 ohm resistor using the same current $I = 2$ A found in Step 2, since series circuits carry equal current through every resistor: $V_3 = IR_3 = 2\ \text{A} \times 5\ \Omega = 10\ \text{V}$ .

Answer: equivalent resistance = 10 ohm, current = 2 A, and the potential difference across the 5 ohm resistor = 10 V.

Competitive stretch

Two resistors are connected in series across a 12 V battery, and a current of 0.5 A flows in the circuit. If one resistor is 10 ohm, the equivalent resistance is $R_s = V/I = 12/0.5 = 24\ \Omega$ , so the second resistor is $R_2 = R_s - R_1 = 24 - 10 = 14\ \Omega$ . The same two-step logic — find $R_s$ from $V$ and $I$ , then subtract known resistances — solves any series problem where only the total effect is given.

Ask the tutor: Have a multi-resistor series problem you're stuck on — ask the AI tutor to work through it with you step by step.

Why does the same current flow through every resistor in a series circuit?

Because a series circuit is a single loop with no branch points, charge has exactly one path to travel and nowhere to divert. Since charge cannot be created, destroyed, or pile up anywhere along that loop, the rate at which it flows — the current — must be identical at every point in the loop, whether measured before, between, or after any resistor. This is the one defining feature of series wiring, and every other result in this topic, including $R_s = R_1 + R_2 + R_3$ , is built on it.

If the current is the same everywhere, why do different resistors have different voltages across them?

Ohm's law, $V = IR$ , links voltage and resistance for a fixed current: with $I$ the same for every resistor, a larger resistance forces a larger voltage drop across it, and a smaller resistance produces a smaller drop. Think of the battery's total potential difference as a fixed amount of "push" that gets shared out among the resistors in proportion to how strongly each one resists the current — a bigger obstacle needs a bigger share of the push to force the same current through it.

Why does one broken bulb turn off an entire series-connected string of lights?

In a series circuit every component shares the single loop that current must travel through. If one bulb's filament breaks, that point in the loop becomes an open gap, and since there is no alternate path around it, the current everywhere in the loop drops to zero — not just at the broken bulb, but in every other bulb on the same string too. This is exactly why decorative series-wired light strings go completely dark from a single failed bulb, and it is one practical reason household circuits use parallel wiring instead, so one failed appliance does not switch off every other one.

Resistors in series

The core idea

Common traps

Worked example

Competitive stretch

Before this

Where this leads