Series & parallel circuits
Ohm's law in action: watch the current respond as you change voltage, resistance and wiring.
Just need the number? Ohm's Law and Circuits
Current vs voltage
Readouts
What's happening
Current is voltage divided by total resistance. In series the resistances add, so each added resistor slows the whole loop.
In parallel each resistor offers its own path, total resistance drops, and the battery drives more current. The moving dots pace with the actual current; the graph shows the I–V line whose slope is 1/R.
- Ohm's lawI = V / R
- Resistors in seriesR = R₁ + R₂
- Resistors in parallel1/R = 1/R₁ + 1/R₂
On the syllabus
- GCSE Physics AQA 4.2.2, Edexcel, OCRseries and parallel circuits, Ohm's law and power
- A-Level Physicsinternal resistance and terminal voltage
Challenge
Predict first: with R₁ = R₂, is the parallel current twice, four times, or half the series current at the same voltage? Work it out, then flip the toggle.
Four times the series current: series gives I = V/2R₁, parallel gives I = 2V/R₁, a 4× ratio.
Learn this properly
Lessons from GCSE Combined and Separate Science on this topic: worked steps, then exam-style questions with new numbers every time.
- Potential difference and resistanceElectric circuitsFree
- Series circuitsElectric circuitsPlus
- Parallel circuitsElectric circuitsPlus
See the whole courseThe full course, topic tests and mock papers come with Plus. See Plus
FAQ
How do you add resistors in series and in parallel?
In series they add, R = R₁ + R₂. In parallel the reciprocals add, 1/R = 1/R₁ + 1/R₂, so the total is smaller than the smallest resistor.
Why do parallel circuits draw more current?
Each branch is another path for charge, so the total resistance drops. With the same battery voltage, I = V/R goes up.
What is internal resistance?
A real battery has some resistance inside it. When current flows, part of the EMF is used up inside, so the terminal voltage is V = ε − Ir.
What resistor do I add in parallel to get a given total?
Rearrange 1/R = 1/R₁ + 1/R₂ to R₂ = R₁R/(R₁ − R). To bring a 20 Ω resistor down to 12 Ω, add R₂ = 20 × 12 / (20 − 12) = 30 Ω in parallel. The target must be below 20 Ω, because parallel always lowers the total. Pick Parallel R₂ for a total under Solve for to see each step.