How do ideal resistors combine in series?
Quick answer
For ideal resistors connected in one series path with no branches, add their resistances: Rₛ = R₁ + R₂ + … . The same current flows through each resistor. Find it from the circuit’s total voltage divided by Rₛ, then find each resistor’s voltage drop using Vᵢ = IRᵢ. The drops add up to the total source voltage. [S1]
The question, the details, the next step.
What does “resistors in series” mean, and what rule should I use to combine them?
Resistors are in series when they are connected one after another along a single path, with no branch that lets current split. In the ideal series model, the same current passes through every resistor. Their equivalent resistance—the single resistance that can replace the group for the circuit calculation—is the sum:
Rₛ = R₁ + R₂ + …
For example, three ideal resistors of 2 Ω, 4 Ω, and 6 Ω in one unbranched path have an equivalent resistance of 12 Ω. This addition rule applies to the stated series connection; it is not the rule for resistors on separate parallel branches. [S1]
How do I use that equivalent resistance to find the current and each voltage drop?
First use the total resistance to find the current from the stated circuit voltage: I = Vₛ/Rₛ. Because the path has no branches, that current is the same through every resistor. Then apply Ohm’s law to each one: Vᵢ = IRᵢ. Each resistor’s drop depends on its resistance, and all the individual drops add to the source voltage. [S1]
Worked example: Assume a 12 V source connected across ideal 2 Ω, 4 Ω, and 6 Ω resistors in series, with no branches. The total is 2 + 4 + 6 = 12 Ω, so the current is 12 V ÷ 12 Ω = 1 A. The drops are 1 A × 2 Ω = 2 V, 1 A × 4 Ω = 4 V, and 1 A × 6 Ω = 6 V. Check: 2 V + 4 V + 6 V = 12 V.
Does the voltage split evenly between the resistors?
Not generally. In this ideal model, each drop is I times that resistor’s resistance, so resistors with larger resistance have larger drops when the shared current is the same. Equal resistors in series have equal drops; unequal ones do not. In the example, the 6 Ω resistor gets 6 V, while the 2 Ω resistor gets 2 V. The source establishes the total voltage, and the individual drops sum to it. [S1]
Can you give me a reusable way to set up and check a series-resistor calculation?
Use this template, filling in values from the circuit you are actually analyzing:
- Stated source voltage: [Vₛ]
- Series resistances: [R₁], [R₂], … [Rₙ]
- Connection check: one path, no branches
- Equivalent resistance: Rₛ = [R₁] + [R₂] + … = [total]
- Shared current: I = [Vₛ] ÷ [Rₛ]
- Drops: V₁ = I[R₁], V₂ = I[R₂], …
- Check: V₁ + V₂ + … = [Vₛ]
Keep units consistent: resistance in ohms, current in amperes, and voltage in volts. If there is a branch, or the connection is uncertain, do not use the series rule until the circuit path has been checked. This is an idealized calculation; the example does not establish the behavior of a real component or a particular physical circuit.
A practical checklist
Confirm the resistors form one path with no branches.
Write down the source voltage and each resistance with units.
Add the series resistances to get Rₛ.
Calculate the shared current using I = Vₛ/Rₛ.
Calculate each drop with Vᵢ = IRᵢ and check that the drops sum to Vₛ.
A few more useful answers
What if one series resistor has a much larger resistance than the others?
The same current still passes through each resistor in the ideal unbranched series model. The larger resistor has a proportionally larger voltage drop, since Vᵢ = IRᵢ. [S1]
Can I add resistances if the circuit has a branch?
Not as one series group across that branch: the series rule assumes one path with no branching. Identify the actual connections before choosing a combination rule. [S1]
How can I check my arithmetic?
Calculate each drop as current times resistance, then add the drops. For the stated ideal series model, their sum should equal the source voltage. [S1]
Sources
References used for this guide.
- [S1] 21.1 Resistors in Series and Parallel — College Physics 2e openstax.org · Checked