In a series circuit, the current is the same everywhere, so a larger resistor has a larger potential difference across it. In parallel, each branch has the same potential difference as the supply.
This lesson follows predicting what an open branch changes and belongs to electrical reasoning from diagrams and measurements.
Worked example: a series pair
An original circuit: a 12 V supply is connected to a 10 Ω resistor R1 and a 20 Ω resistor R2 in series. A voltmeter is across each resistor.
- Total resistance = 10 + 20 = 30 Ω.
- Current = 12 ÷ 30 = 0.40 A, the same through both.
- V1 = 0.40 × 10 = 4 V and V2 = 0.40 × 20 = 8 V.
- Check: 4 + 8 = 12 V, the supply.
R2 has twice the resistance, so its voltmeter reads twice as much. You can state the comparison without a calculation: same current, larger R, larger V.
What if R2 is replaced by a larger resistor?
Replace R2 with a 50 Ω resistor, keeping the same connections.
- Total resistance = 60 Ω, so the current falls to 12 ÷ 60 = 0.20 A.
- V1 = 0.20 × 10 = 2 V and V2 = 0.20 × 50 = 10 V.
The voltmeter across R1 now reads less, even though R1 has not changed. A larger resistor elsewhere in the series path reduces the current, so V across R1 falls too.
The mistake that costs marks
The common slip is to assume that two resistors in series share the supply equally, so each voltmeter reads 6 V.
| Claim | Why it fails |
|---|---|
| “Two resistors share 12 V equally, so 6 V each” | Sharing depends on resistance, not on the number of components |
| “Voltmeter across R1 reads 12 V” | That is true only if R1 is the whole circuit |
| Correct: “same current, V = IR, so larger R has larger V” | Uses the quantity that is shared in series |
Equal sharing is correct only when the resistors are equal, which is a special case.
Check yourself
A 9 V supply is connected across a 30 Ω resistor and a 60 Ω resistor in series. Find the reading across each resistor.
Answer
Total resistance = 90 Ω, so the current = 9 ÷ 90 = 0.10 A.
The 30 Ω resistor reads 0.10 × 30 = 3 V. The 60 Ω resistor reads 0.10 × 60 = 6 V.
Check: 3 + 6 = 9 V.
What to study next
Next, look at how a lamp’s own resistance changes in explaining a non-linear current-voltage graph. Then test the four lessons in the section practice set. The units and significant figure checker helps with units and rounding.
For a teacher to go through voltmeter questions with you, see online one-to-one Physics tuition or try the one-hour trial class (from RM50).