Current, potential difference and resistance
Current is the flow of charge, potential difference is the push that drives it and resistance is what opposes it. Two equations and four rules for series and parallel circuits cover most of the topic.
Part 1 of 3: Learn it
In short
- Charge = current × time (Q = I t) and potential difference = current × resistance (V = I R).
- In series, the current is the same everywhere and the potential differences add up.
- In parallel, each branch has the same potential difference and the branch currents add up.
Where this is in your specification
Spec points: AQA 4.2.1 and 4.2.2 (Combined Trilogy 6.2.1 and 6.2.2), Edexcel Topic 10, OCR P3.1 and P3.2
| Board | Topic: Electric circuits |
|---|---|
| AQA 8463 | 4.2.1 and 4.2.2 (8463); 6.2.1 and 6.2.2 (8464) |
| Edexcel 1PH0 | Topic 10 |
| OCR J249 | P3.1 and P3.2 |
The quantities
| Quantity | Symbol | Unit | Measured with |
|---|---|---|---|
| Current | I | ampere (A) | ammeter, in series |
| Potential difference | V | volt (V) | voltmeter, in parallel |
| Resistance | R | ohm (Ω) | from V ÷ I |
| Charge | Q | coulomb (C) |
Two equations
Time must be in seconds. Rearranged: I = V ÷ R and R = V ÷ I.
Series and parallel
| Series | Parallel | |
|---|---|---|
| Current | the same through every component | splits between branches; branch currents add up to the total |
| Potential difference | shared between components; adds up to the supply | the same across every branch |
| Total resistance | R total = R₁ + R₂ | less than the smallest single resistor |
Adding a resistor in series increases the total resistance. Adding one in parallel decreases it, because it gives the charge another path.
I-V characteristics
- Fixed resistor (at constant temperature): a straight line through the origin. Current is directly proportional to potential difference, so the resistance stays the same (an ohmic conductor).
- Filament lamp: a curve that gets flatter. As the filament heats up its resistance increases.
- Diode: current flows in one direction only. Turned the other way round, almost no current gets through because its resistance is huge.
What is the resistance of a lamp with 3.0 V across it and 0.2 A through it?
Show the answer
R = 3.0 ÷ 0.2 = 15 Ω.
Part 2 of 3: See it worked
Worked examples
Example 1
A 12 Ω resistor is connected to a 6.0 V battery. Find the current, and the charge that flows in 2 minutes.
- I = V ÷ R = 6.0 ÷ 12 = 0.50 A
- t = 2 × 60 = 120 s
- Q = I t = 0.50 × 120 = 60 C
Answer: 0.50 A and 60 C.
Example 2
A 4 Ω and a 6 Ω resistor are in series with a 5 V supply. Find the current and the potential difference across the 6 Ω resistor.
- R total = 4 + 6 = 10 Ω
- I = 5 ÷ 10 = 0.5 A (the same through both)
- V across 6 Ω = 0.5 × 6 = 3 V
Answer: 0.5 A, and 3 V across the 6 Ω resistor (the other 2 V is across the 4 Ω).
Common mistakes
- Connecting a voltmeter in series or an ammeter in parallel.
- Using minutes instead of seconds in Q = I t.
- Adding parallel resistances as if they were in series.
- Calling a filament lamp ohmic. Its resistance changes as it heats up.
Two identical lamps are in parallel across a 9 V battery. What is the potential difference across each?
Show the answer
9 V: every parallel branch has the full supply potential difference.
Part 3 of 3: Test yourself
Check yourself
Answer each one in your head or on paper first, then open it to check.
What is the resistance of a lamp with 3.0 V across it and 0.2 A through it?
R = 3.0 ÷ 0.2 = 15 Ω.
Two identical lamps are in parallel across a 9 V battery. What is the potential difference across each?
9 V: every parallel branch has the full supply potential difference.
What happens to an LDR's resistance in bright light?
It decreases.
Jobs that use this
Each link opens the job profile on the National Careers Service (England). In the rest of the UK: My World of Work (Scotland), Careers Wales, nidirect careers (Northern Ireland).
Ces fiches sont en anglais car elles suivent les programmes d'examen britanniques.
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