How to solve circuit problems
What stays the same and what splits in each circuit type, a worked parallel example, and the resistance mistake almost everyone makes.
What the examiner is testing
Whether you know which quantity stays the same and which one splits. Get that right and the arithmetic is easy; get it backwards and every number is wrong.
$$ V = IR $$
The one table worth memorising
| Series | Parallel | |
|---|---|---|
| Current | Same everywhere | Splits between branches |
| Voltage | Splits between components | Same across each branch |
| Total resistance | \( R_1 + R_2 \) | Less than the smallest branch |
The pattern is symmetric: whichever quantity stays the same in series splits in parallel, and vice versa.
Worked example
A 12 V supply is connected to a 4 Ω resistor and a 6 Ω resistor in parallel. Find the current through each and the total current.
Step 1 — in parallel, the voltage across each branch is the full supply voltage. Both resistors get 12 V. This is the step people skip.
Step 2 — current through each branch, using \( I = V/R \):
$$ I_1 = \frac{12}{4} = 3.0 \text{ A} \qquad I_2 = \frac{12}{6} = 2.0 \text{ A} $$
Step 3 — total current is the sum of the branches.
$$ I_{\text{total}} = 3.0 + 2.0 = 5.0 \text{ A} $$
Step 4 — total resistance, from the supply's point of view:
$$ R_{\text{total}} = \frac{V}{I} = \frac{12}{5.0} = 2.4 \ \Omega $$
Check: 2.4 Ω is less than 4 Ω, the smaller of the two resistors. In parallel it always must be. If your total comes out bigger than either resistor, you have added them as though they were in series.
The same components in series
Now the resistances add: \( R = 4 + 6 = 10 \ \Omega \), so the current is \( 12/10 = 1.2 \text{ A} \) — the same through both. The voltages split in proportion: 4.8 V across the 4 Ω and 7.2 V across the 6 Ω, adding to the 12 V supply.
Notice the current dropped from 5.0 A to 1.2 A with exactly the same components. That is the whole point of the topic.
The three mistakes that lose marks
1. Adding parallel resistances like series ones. The total in parallel is always smaller than the smallest branch. Use that as your check.
2. Splitting the voltage in a parallel circuit. Each branch gets the full supply voltage.
3. Forgetting that voltages in series must sum to the supply. If yours do not add up to the source, something is wrong — this is a free check on the whole question.
30-second recap
Series: current the same, voltage splits, resistances add. Parallel: voltage the same, current splits, total resistance drops below the smallest branch. Use \( V = IR \) on one branch at a time, then check that your parallel total is smaller than the smallest resistor and that series voltages add to the supply.