4.2.2Series and parallel
There are two ways of joining electrical components, in series and in parallel. Some circuits include both series and parallel parts.
For components connected in series:
• there is the same current through each component • the total potential difference of the power supply is shared between the components • the total resistance of two components is the sum of the resistance of each component:
Rtotal = R1 + R2 resistance, R, in ohms, Ω
For components connected in parallel:
• the potential difference across each component is the same • the total current through the whole circuit is the sum of the currents through the separate components • the total resistance of two resistors is less than the resistance of the smallest individual resistor.
You should be able to describe the difference between series and parallel circuits, and use circuit diagrams to construct and check series and parallel circuits that include a variety of common circuit components.
- Series circuit
- a circuit in which components are joined one after another, so the same current passes through each, the supply potential difference is shared between them, and the total resistance is the sum of the separate resistances
- Parallel circuit
- a circuit in which components are joined across the same two points, so each has the same potential difference across it and the total current is the sum of the currents through the branches
4.2.2Why the total resistance changes
You should be able to explain qualitatively why adding resistors in series increases the total resistance, whilst adding resistors in parallel decreases the total resistance.
In series, the charge has to pass through every resistor in turn. Each one opposes the flow, so adding another adds its opposition to the total — the current has more to get through and the total resistance rises.
In parallel, each new resistor provides another separate path for the charge to take. More paths means more charge can flow in total for the same potential difference, so the total current rises and the total resistance falls.
That is why the parallel total is less than the smallest individual resistor: adding a path can only ever make it easier for charge to flow, never harder — even if the new path is a poor one.
Common mistake
In parallel the total resistance is not the average of the two, and it is not somewhere between them. It is always less than the smaller one. If your answer for two 10 Ω resistors in parallel is more than 5 Ω, something is wrong.
4.2.2Calculations in series circuits
You should be able to calculate the currents, potential differences and resistances in dc series circuits, and to solve problems for circuits which include resistors in series using the concept of equivalent resistance.
You are not required to calculate the total resistance of two resistors joined in parallel — so a parallel calculation will only ever ask for currents and potential differences, using the two rules above.
Equivalent resistance means replacing several series resistors with a single one of the same total value, so that V = I R can be applied to the whole circuit at once.
Worked example
A 6.0 V supply is connected in series with a 4.0 Ω and an 8.0 Ω resistor. Calculate the current, and the potential difference across each resistor.
- Find the equivalent resistance: Rtotal = R1 + R2 = 4.0 + 8.0 = 12.0 Ω.
- Apply V = I R to the whole circuit: I = V ÷ R = 6.0 ÷ 12.0 = 0.50 A.
- In series the current is the same through each component, so 0.50 A flows through both.
- Across the 4.0 Ω resistor: V = I R = 0.50 × 4.0 = 2.0 V.
- Across the 8.0 Ω resistor: V = 0.50 × 8.0 = 4.0 V.
- Check: 2.0 + 4.0 = 6.0 V, which is the supply — the potential difference is shared, as it must be.
Answer: 0.50 A throughout; 2.0 V across the 4.0 Ω resistor and 4.0 V across the 8.0 Ω resistor.
Worked example
Two lamps are connected in parallel across a 12 V supply. One draws 0.30 A and the other 0.50 A. State the potential difference across each lamp and calculate the total current from the supply.
- In parallel the potential difference across each component is the same as the supply.
- So each lamp has 12 V across it.
- The total current through the whole circuit is the sum of the currents through the separate components.
- 0.30 + 0.50 = 0.80 A.
Answer: 12 V across each lamp, and a total supply current of 0.80 A.
You should also be able to explain the design and use of dc series circuits for measurement and testing purposes.
A series circuit is used for testing because the single current makes the readings unambiguous: one ammeter anywhere in the loop measures the current through everything, and adding a variable resistor in series controls that current for every component at once. That is exactly the arrangement used in required practicals 3 and 4.
Common mistake
Check that the potential differences across series components add up to the supply. It is a free check on the whole calculation, and it catches an arithmetic slip immediately.
Before you practise
- Series: same current through each component; the supply pd is shared; Rtotal = R1 + R2.
- Parallel: same potential difference across each component; the currents in the branches add to give the total.
- The total resistance of two resistors in parallel is less than the smaller of the two.
- Adding resistors in series increases total resistance because the charge must pass through each in turn.
- Adding resistors in parallel decreases total resistance because each one provides another path for charge.
- You are not required to calculate the total resistance of two resistors in parallel.
- Use equivalent resistance to reduce a series circuit to one resistor, then apply V = I R.
- Check your answer: series potential differences must add up to the supply.