4.3.2 Series and parallel circuits
- Syllabus
- 0625–2026–2027
- Topic
- 4.3.2
- Level
- —
Learning objectives
4.3.2.1Current at every point in a series• Know: the current at every point in a series circuit is the same4.3.2.2How to construct and use series and• Know how to construct and use series and parallel circuits4.3.2.3Combined e.m.f. of several sources in• Calculate the combined e.m.f. of several sources in series4.3.2.4Combined resistance of two or more• Calculate the combined resistance of two or more resistors in series4.3.2.5That, for a parallel circuit, the• State that, for a parallel circuit, the current from the source is larger than the current in each branch4.3.2.6Combined resistance of two resistors• State: the combined resistance of two resistors in parallel is less than that of either resistor by itself4.3.2.7Advantages of connecting lamps in• State the advantages of connecting lamps in parallel in a lighting circuit4.3.2.8Circuit rules: (a) current into a• Recall/use circuit rules: (a) current into a junction equals current out (b) total p.d. across series components equals the sum of their p.d.s (c) p.d. across parallel branches is the same4.3.2.9The sum of the currents into a• Explain that the sum of the currents into a junction is the same as the sum of the currents out of the junction4.3.2.10Combined resistance of two resistors• Calculate the combined resistance of two resistors in parallel
Follow current around a series circuit
In a series circuit there is only one continuous path, so the current is the same at every point in that path. An ammeter placed before or after any component therefore gives the same reading.
Current is the rate of flow of charge. Charge does not get used up by a lamp or resistor, and there is no junction where the flow can split, so the same amount of charge passes every point each second.
Potential difference can be shared between series components, but current is not. Do not add the current readings from different positions in the same single path.
Construct and use series and parallel circuits
Series components lie on one path; parallel components lie on separate branches connected between the same two junctions. Build a circuit by identifying the junctions first, then placing each component on the intended path or branch.
| Feature | Series | Parallel |
|---|---|---|
| paths for current | one | two or more branches |
| control | one break stops the whole path | a switch can control one branch independently |
| component failure | breaks the only path | other complete branches can still work |
| typical use | components that must carry the same current | devices that need the same supply p.d. and independent operation |
Before switching on, trace every complete path from one supply terminal to the other, check for an accidental zero-resistance path across the supply, put ammeters in series, and put voltmeters in parallel across the component being measured.
Combine the e.m.f. of sources in series
Sources connected in series add their e.m.f.s when they are oriented to drive current in the same direction. Treat each e.m.f. as signed: add aiding sources and subtract a source that is reversed.
E_{\text{total}} = E_1 + E_2 + \cdots
A 1.5 V cell in series with a 6.0 V battery, both aiding, gives Eexttotal=1.5+6.0=7.5 V. If the 1.5 V cell is reversed, the combined e.m.f. is 6.0−1.5=4.5 V.
Check the long-line positive terminals before choosing plus or minus. Adding the printed values without checking orientation can give the wrong combined e.m.f.
Add resistances in series
Series resistors carry the same current, and each contributes opposition along the only path. Their combined resistance is the sum of the individual resistances.
R_{\text{series}} = R_1 + R_2 + \cdots
For 12 Ω, 8 Ω and 5 Ω in series, Rextseries=12+8+5=25 Ω. The result must be greater than every individual resistance because another series resistor adds opposition.
Only add resistances directly when the same current must pass through them one after another with no junction between them.
Compare source and branch currents
In a parallel circuit, current from the source reaches a junction and divides between the branches. The source current is the sum of the branch currents, so it is larger than the current in any one conducting branch.
I_{\text{source}} = I_1 + I_2 + \cdots
If two branch currents are 0.30 A and 0.20 A, the source current is 0.30+0.20=0.50 A. The currents need not divide equally unless the branches have the same resistance.
Recognise the resistance of a parallel pair
The combined resistance of two resistors in parallel is less than the resistance of either resistor on its own.
Adding a parallel resistor creates another path for charge. At the same p.d., more total current can flow from the source; since R=V/I, a larger total current means a smaller combined resistance.
For example, a 3 Ω resistor in parallel with a 6 Ω resistor must have a combined resistance below 3 Ω. A result between 3 Ω and 6 Ω, or their sum 9 Ω, cannot be correct.
Why lighting circuits use parallel lamps
Lamps are connected in parallel so each lamp is directly across the supply and each branch can operate independently.
| Advantage | Reason |
|---|---|
| normal operating p.d. | each lamp receives the full supply p.d. |
| independent failure | if one lamp breaks, other complete branches remain lit |
| independent control | a switch in one branch can turn that lamp on or off without opening the others |
Parallel operation usually draws a larger total current as more lamps are added. The advantage is not that the supply transfers less power or that fewer wires are needed.
Use the three circuit rules
Solve a series–parallel circuit by identifying junctions and branches before using current or potential-difference values.
| Situation | Rule |
|---|---|
| junction | total current in = total current out |
| components in one series path | their p.d.s add to the supply p.d. |
| parallel branches between the same two junctions | each branch has the same p.d. |
A 12 V supply feeds two parallel branches. Both branches have 12 V across them. If one branch takes 0.40 A and the other 0.25 A, the source current is 0.65 A. Within a branch containing two series components with p.d.s 7 V and 5 V, the p.d.s add to 12 V.
Add p.d.s along one series route, not across separate parallel branches. Add currents at a junction, not readings taken at different points of the same unbranched path.
Explain current conservation at a junction
At a junction, the sum of the currents entering equals the sum of the currents leaving.
Current measures charge passing per second, and electric charge is conserved. In a steady circuit, charge cannot be created, destroyed or continually accumulate at the junction. Therefore every coulomb arriving each second must leave through one of the outgoing branches.
If 1.2 A enters and currents of 0.45 A and 0.30 A leave along two branches, the remaining outgoing current is 1.2−0.45−0.30=0.45 A.
The rule concerns the signed total in each direction; it does not require every branch current to be equal.
Calculate two resistors in parallel
For two resistors in parallel, the reciprocal of the combined resistance equals the sum of the reciprocals. The same result can be written as a product-over-sum shortcut.
\frac{1}{R_{\text{parallel}}}=\frac{1}{R_1}+\frac{1}{R_2}\qquad\text{so}\qquad R_{\text{parallel}}=\frac{R_1R_2}{R_1+R_2}
For 3.0 Ω and 6.0 Ω in parallel, R=(3.0imes6.0)/(3.0+6.0)=18/9.0=2.0 Ω. The answer passes the check because 2.0 Ω is below both 3.0 Ω and 6.0 Ω.
Reduce a mixed circuit in stages: first find each clear series or parallel group, redraw it as one equivalent resistor, then combine the next group. Keep units in ohms throughout.