In a series circuit there is one path, so the current is the same everywhere and the p.d. is shared between the components.
In a parallel circuit the p.d. across each branch is the same and the current splits between branches.
This lesson extends using R = V ÷ I with units to circuits with more than one component, inside potential difference, resistance and circuits.
What are the rules?
| Series | Parallel | |
|---|---|---|
| Current | same in every component | splits: I total = I₁ + I₂ |
| p.d. | shared: V total = V₁ + V₂ | same across each branch |
| Combined resistance | R = R₁ + R₂ | smaller than the smallest branch |
For two parallel resistors the combined value comes from 1/R = 1/R₁ + 1/R₂, where your syllabus requires it. You can always find it as supply p.d. ÷ total current.
How do I work through a question?
- Sketch the circuit and mark what you know.
- Decide series or parallel for each part.
- In a series part, find the current once, then the p.d. across each resistor.
- In a parallel part, use the supply p.d. on every branch and find each branch current.
- Add or compare to check: branch currents sum to the total, series p.d.s sum to the supply.
Worked example
Resistors of 6.0 Ω and 12 Ω are connected in parallel across a 12 V supply. (Invented example data.) Find the current in each branch, the total current and the combined resistance.
Step 1, p.d. on each branch: both branches have 12 V.
Step 2, branch currents: I₁ = 12 ÷ 6.0 = 2.0 A and I₂ = 12 ÷ 12 = 1.0 A.
Step 3, total current: 2.0 + 1.0 = 3.0 A.
Step 4, combined resistance: R = V ÷ I = 12 ÷ 3.0 = 4.0 Ω.
Step 5, check: 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4, so R = 4.0 Ω. It is smaller than 6.0 Ω, as it must be.
The mistake to watch for
A common slip is to add the resistances in a parallel circuit.
Mistaken answer: R = 6.0 + 12 = 18 Ω, so I = 12 ÷ 18 = 0.67 A
The student treated the branches as one path. That gives a total current smaller than the current in one branch alone, which cannot be right.
The correction is to ask “does the battery see one path or two?” With two paths the total current must be larger than either branch current. If your combined resistance is larger than the smallest branch, stop and recheck.
Check yourself
Try these, then open each answer.
1. A 10 Ω and a 20 Ω resistor are in series with a 9.0 V battery. Find the current.
Show answer
R = 10 + 20 = 30 Ω. I = 9.0 ÷ 30 = 0.30 A. The same current flows through both.
2. A 20 Ω and a 30 Ω resistor are in parallel across 6.0 V. Find the current in each branch and the total current.
Show answer
I₁ = 6.0 ÷ 20 = 0.30 A. I₂ = 6.0 ÷ 30 = 0.20 A. Total = 0.30 + 0.20 = 0.50 A.
3. Two identical 6.0 Ω resistors are in parallel. What is the combined resistance?
Show answer
1/R = 1/6 + 1/6 = 2/6, so R = 6 ÷ 2 = 3.0 Ω. Identical resistors in parallel give half the value of one.
Where this leads next
Next, see how a component’s behaviour shows up on a line in a current-voltage graph. The circuit reasoning simulator is a good place to test a parallel prediction, and the practice set mixes both circuit types.
Students who solve each branch correctly but lose the thread in a longer circuit can gain from a teacher redrawing it with them. That is part of our online one-to-one Physics tuition.