4.3.7Electrons, Waves and Photons
Circuits with Multiple Sources of e.m.f
5 min read 5 parts Lesson 33 of 62 Unit quiz
Section 1 of 5Cells in Series
A battery is simply several cells connected so their e.m.f.s combine in a chosen way.
Aiding
- When cells point the same way, the total e.m.f. is the sum of their individual e.m.f.s.
Opposing
- If one cell is reversed, its e.m.f. is subtracted, so the total can be much smaller.
Internal resistance
- The internal resistances of cells in series also add, which increases the energy lost inside the battery.
Purpose
- Series connection gives a higher p.d. for devices that need more than a single cell can supply.
Section 2 of 5Cells in Parallel
Cells in Parallel
Identical cells
- The e.m.f. of the combination equals that of one cell, because the p.d. across each is the same.
Current sharing
- Each cell supplies a share of the load current, so each is drained more slowly.
Internal resistance
- The combined internal resistance is smaller than that of one cell, so less energy is wasted inside.
Use
- Parallel cells are chosen for longer life or larger current, not for higher voltage.
| Arrangement | Total e.m.f. | Main benefit |
|---|---|---|
| Series (aiding) | Sum of e.m.f.s | Higher p.d. |
| Parallel (identical) | Same as one cell | Longer-lasting supply |
Section 3 of 5Loops With Several Sources
1
Choose a direction around the loop and treat e.m.f.s acting with that direction as positive.
2
Treat an e.m.f. acting against the chosen direction as negative, and similarly for currents.
3
Write the second law as the signed sum of e.m.f.s equals the sum of IR terms around the loop.
4
Solve for the unknown current and interpret a negative answer as flowing opposite to the chosen direction.
5
Confirm the energy sense: the stronger source drives current through the weaker one in the opposite direction.
Section 4 of 5Worked Example
Question
- A 6.0 V cell and a 2.0 V cell are connected in series in opposition with a 4.0 Ω resistor.
- Ignoring internal resistance, find the current.
- Net e.m.f. = 6.0 − 2.0 = 4.0 V, since the cells oppose each other.
- Apply Σε = ΣIR4.0 = I × 4.0.
- I = 4.0 / 4.0 = 1.0 A in the direction set by the 6.0 V cell.
Answer Current = 1.0 A.
Section 5 of 5Exam Checklist & Tips
4.3.7 – Key Points for the Exam
- Battery Two or more cells connected together.
- Series cells Cells connected end to end so their e.m.f.s add or subtract.
- Parallel cells Cells connected across the same two points, giving one cell's e.m.f. with lower internal resistance.
- Opposing e.m.f. A source whose polarity works against the driving current.
Common Mistakes
- Adding e.m.f.s for cells in parallel; they do not add.
- Ignoring the polarity of a reversed cell and adding all e.m.f.s.
- Forgetting that internal resistances also combine and affect the current.
- Dropping the sign of a current or e.m.f. in a multi-source loop.