This note covers how components are connected in series and in parallel, and how current, p.d. and resistance behave in each arrangement. It also introduces the diode, LDR and thermistor, and the potential divider. Almost every circuit question uses these rules.
Key ideas
Why this matters. A circuit diagram is a map of how charge can flow. Reading it correctly tells you where the current splits, where the p.d. is shared and which parts will keep working if something fails.
Circuit components. You should recognise and draw the standard symbols for: cell, battery, switch, fixed resistor, variable resistor, lamp, fuse, ammeter, voltmeter, diode, LED, thermistor, LDR, and a connecting wire.
Series circuits. Components are joined end to end in a single loop.
- The current is the same at every point.
- The supply p.d. is shared between the components: Vtotal = V1 + V2 + …
- Resistances add: Rtotal = R1 + R2 + …
- If one component breaks, the whole circuit stops.
- Adding more resistors in series makes the total resistance larger, so the current falls. Lamps in series with the same cell are dimmer than a single lamp.
Parallel circuits. Components are on separate branches between the same two points.
- The p.d. across each branch is the same as the supply p.d.
- The currents in the branches add to give the current from the supply: Itotal = I1 + I2 + …
- The combined resistance is smaller than the smallest branch resistance. Adding another branch gives the current an extra path.
- If one branch breaks, the others keep working. This is why homes use parallel wiring.
- Each branch can be switched on and off separately, and each appliance receives the full mains p.d.
- A bigger resistance in a branch means a smaller current in that branch. The branch with the smallest resistance carries the largest current.
For extended work, the combined resistance of resistors in parallel is found from 1/Rtotal = 1/R1 + 1/R2 + …
Measuring in circuits. An ammeter goes in series with the component, so all of the current passes through it. A voltmeter goes in parallel with the component, across its two ends. An ideal ammeter has very low resistance and an ideal voltmeter has very high resistance, so neither changes the circuit much.
Special components.
- Diode: allows current in one direction only. Its resistance is very large in the other direction. An LED is a diode that gives out light.
- Thermistor (NTC type): its resistance decreases as temperature rises. Used in temperature sensors.
- Light-dependent resistor (LDR): its resistance decreases as light intensity increases. Used in light sensors.
- Variable resistor: its resistance can be adjusted by hand, for example to change the brightness of a lamp.
Potential divider (extended). Two resistors in series across a supply share the p.d. in proportion to their resistances. The output is taken across one of them. If a thermistor or LDR is one of the pair, the output p.d. changes as the temperature or light level changes. This can switch a device on or off.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| R = R1 + R2 + … | Total resistance of resistors in series. |
| 1/R = 1/R1 + 1/R2 | Total resistance of two resistors in parallel (extended). |
| V = IR | Apply to the whole circuit or to any single component. |
| Vout = R2 ÷ (R1 + R2) × Vin | Potential divider output across R2 (extended). |
| Series | Same current. P.d.s add. |
| Parallel | Same p.d. Currents add. |
Worked examples
Example 1 Resistors of 4.0 Ω, 6.0 Ω and 10 Ω are connected in series to a 12 V supply. Find the current and the p.d. across the 6.0 Ω resistor.
Total resistance: 4.0 + 6.0 + 10 = 20 Ω.
Current: I = V/R = 12 ÷ 20 = 0.60 A. This is the same through each resistor.
P.d. across the 6.0 Ω resistor: V = IR = 0.60 × 6.0 = 3.6 V.
Answer: 0.60 A and 3.6 V
Example 2 (extended) A 6.0 Ω resistor and a 12 Ω resistor are connected in parallel across a 12 V supply. Find the current in each resistor, the total current and the combined resistance.
Each branch has the full 12 V across it.
Current in 6.0 Ω: 12 ÷ 6.0 = 2.0 A. Current in 12 Ω: 12 ÷ 12 = 1.0 A.
Total current: 2.0 + 1.0 = 3.0 A.
Combined resistance: R = V/I = 12 ÷ 3.0 = 4.0 Ω. Check: 1/6.0 + 1/12 = 3/12 = 1/4.0. This is smaller than 6.0 Ω, as expected.
Answer: 2.0 A, 1.0 A, 3.0 A, 4.0 Ω
Example 3 (extended) A 4.0 Ω resistor is in series with a parallel pair made of 6.0 Ω and 3.0 Ω. The supply is 12 V. Find the p.d. across the parallel pair.
Parallel pair: 1/R = 1/6.0 + 1/3.0 = 3/6.0, so R = 2.0 Ω.
Total resistance: 4.0 + 2.0 = 6.0 Ω.
Supply current: I = 12 ÷ 6.0 = 2.0 A. This flows through the 4.0 Ω resistor and then through the pair.
P.d. across the pair: V = IR = 2.0 × 2.0 = 4.0 V.
Answer: 4.0 V
Common errors
- Adding resistances directly in a parallel circuit. The combined value must be smaller than the smallest branch.
- Forgetting to take the reciprocal at the end of 1/R calculations.
- Saying the current is shared in a series circuit. The p.d. is shared. The current is the same.
- Saying that adding a lamp in parallel makes the other lamps dimmer. Each branch still has the full supply p.d.
- Mixing up the thermistor and LDR. Both have lower resistance when the temperature or light level goes up, but they respond to different things.
Exam tip: In a mixed circuit, replace the parallel part by its single equivalent resistance first. Then treat the circuit as a simple series circuit, and go back to the parallel part for the branch currents.
Check yourself
- (Extended) A 2.0 kΩ resistor and a 1.0 kΩ resistor are in series across a 9.0 V supply. What is the p.d. across the 1.0 kΩ resistor?
Answer
Vout = 1.0 ÷ (2.0 + 1.0) × 9.0 = 3.0 V. - (Extended) Two 5.0 Ω resistors are connected in parallel. What is their combined resistance?
Answer
1/R = 1/5.0 + 1/5.0 = 2/5.0, so R = 2.5 Ω. - Two lamps are in series with a cell and an ammeter reads 0.30 A just after the cell. What does a second ammeter read between the two lamps?
Answer
0.30 A. In a series circuit the current is the same everywhere. - (Extended) An LDR is in series with a fixed resistor across a battery. A voltmeter is connected across the fixed resistor. What happens to its reading when the LDR is moved into brighter light?
Answer
The LDR resistance falls, so the circuit current rises and more of the supply p.d. is across the fixed resistor. The voltmeter reading increases.