Physics · 4 Electricity and magnetism

4.3.2 Series and parallel circuits

Before you start

$R = V / I$ and its rearrangements (4.2.4).
E.m.f. and p.d., and that charge carries energy from source to components (4.2.3).
Circuit symbols and the difference between a junction and a crossing (4.3.1).

Syllabus objectives covered (10)

Know that the current at every point in a series circuit is the same
Know how to construct and use series and parallel circuits
Calculate the combined e.m.f. of several sources in series
Calculate the combined resistance of two or more resistors in series
State that, for a parallel circuit, the current from the source is larger than the current in each branch
State that the combined resistance of two resistors in parallel is less than that of either resistor by itself
State the advantages of connecting lamps in parallel in a lighting circuit
Recall and use in calculations, the fact that: (a) the sum of the currents entering a junction in a parallel circuit is equal to the sum of the currents that leave the junction (b) the total p.d. across the components in a series circuit is equal to the sum of the individual p.d.s across each component (c) the p.d. across an arrangement of parallel resistances is the same as the p.d. across one branch in the arrangement of the parallel resistances
Explain that the sum of the currents into a junction is the same as the sum of the currents out of the junction
Calculate the combined resistance of two resistors in parallel

Part 1 of 5

The two ways to connect things

In a series circuit the components are joined end to end in a single loop. There is one path, so all the charge must go through every component.
In a parallel circuit the components are on separate branches between the same two points. The charge divides between the branches and rejoins afterwards.
Everything in this subtopic follows from that one difference: one path or several. Work out which you are looking at before doing anything else.
In a real circuit, look for a junction dot where the wire splits. If nothing splits, it is series.
One practical consequence, which matters in 4.4 as well: in a series circuit, if one component fails or is removed, the single path is broken and everything stops. In parallel, the other branches carry on.

Part 2 of 5

Series circuits

Current is the same at every point. There is only one path, so every coulomb that leaves the cell passes through every component in turn.
This is often stated backwards by students as 'the current gets used up'. It does not — charge is not consumed. Only energy is transferred.
P.d.s add up. The total p.d. across the components equals the e.m.f. of the source:
$V_{total} = V_{1} + V_{2} + V_{3} + \dots$
That is conservation of energy: each coulomb picks up a fixed amount of energy at the cell and gives all of it away as it passes through the components.
E.m.f.s add up too, if the cells face the same way. Three 1.5 V cells in series give 4.5 V.
If a cell is reversed, its e.m.f. subtracts. Two 1.5 V cells opposing each other give zero. Check the direction of every cell symbol before adding.
Resistances add up:
$R_{total} = R_{1} + R_{2} + R_{3} + \dots$
The total is always larger than any single resistor, because the charge has to fight its way through all of them one after another.

Part 3 of 5

Parallel circuits

P.d. is the same across every branch. Each branch is connected between the same two points, so each experiences the same p.d.
That is the mirror image of the series rule. In series the current is shared and the p.d. divides; in parallel the p.d. is shared and the current divides.
Currents add up at a junction. The sum of the currents entering a junction equals the sum leaving it.
$I_{total} = I_{1} + I_{2} + I_{3} + \dots$
The reason is that charge is conserved: charge cannot pile up at a junction or vanish there, so whatever arrives each second must leave each second.
It follows that the current from the source is larger than the current in any individual branch, since it is the sum of them all.
Total resistance falls. Adding a resistor in parallel gives the current an extra route, so more current flows for the same p.d. — which means less resistance overall.
So the combined resistance of two resistors in parallel is less than either one on its own. This is genuinely surprising the first time, and it is a favourite one-mark question.
One path against two. In series the current is common; in parallel the p.d. is.

Part 4 of 5

Calculating parallel resistance

For two resistors in parallel, the reciprocals add:
$\frac{1}{R_{total}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
The step people forget is the last one. This equation gives you $1 / R$ , not $R$ — you must take the reciprocal at the end.
For $6 Ω$ and $3 Ω$ : $1 / R = 1 / 6 + 1 / 3 = 1 / 6 + 2 / 6 = 3 / 6 = 1 / 2$ , so $R = 2 Ω$ .
Notice $2 Ω$ is less than both $6$ and $3$ , as it must be. That is your check on every parallel calculation — if the answer is not smaller than the smallest resistor, you have gone wrong.
A useful special case: two equal resistors in parallel give exactly half of one of them. Two $10 Ω$ resistors give $5 Ω$ .
For exactly two resistors you may also use the product-over-sum form, which avoids the reciprocal slip:
$R_{total} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
That shortcut works for two resistors only. For three or more, go back to the reciprocals.

Part 5 of 5

Why house lights are in parallel

Three advantages, and you should be able to give all three.
Each lamp gets the full supply p.d., so every lamp is equally bright. In series they would share the p.d. and all be dim.
Each can be switched independently, because each branch can have its own switch.
If one lamp fails, the others stay on, because the other branches are still complete paths.
The last one is the clinching argument. A series lighting circuit fails completely when a single bulb blows — which is exactly the behaviour of old fairy lights, and exactly why nobody wires a house that way.

Worked examples

Example 1. A $4.0 Ω$ and a $8.0 Ω$ resistor are connected in series to a $6.0$ V battery. Calculate the current in the circuit and the p.d. across each resistor.

Show the working

Find the total resistance first. In series, resistances add.
$R = 4.0 + 8.0 = 12 Ω$
Now find the current using the total resistance and the supply p.d.
$I = \frac{V}{R} = \frac{6.0}{12} = 0.50 A$
In series the current is the same everywhere, so $0.50$ A flows through both resistors.
Find each p.d. from $V = I R$ using that shared current.
$V_{1} = 0.50 \times 4.0 = 2.0 V V_{2} = 0.50 \times 8.0 = 4.0 V$
Check: $2.0 + 4.0 = 6.0$ V, the supply p.d. The p.d.s add up in series, so the answer is consistent.

Answer: Current $= 0.50$ A throughout. P.d. across the $4.0 Ω$ resistor $= 2.0$ V; across the $8.0 Ω$ resistor $= 4.0$ V.

Example 2. A $12 Ω$ resistor and a $4.0 Ω$ resistor are connected in parallel. Calculate their combined resistance.

Show the working

Use the reciprocal equation for parallel resistors.
$\frac{1}{R} = \frac{1}{12} + \frac{1}{4.0}$
Put them over a common denominator.
$\frac{1}{R} = \frac{1}{12} + \frac{3}{12} = \frac{4}{12}$
Take the reciprocal — the step it is easy to stop just short of.
$R = \frac{12}{4} = 3.0 Ω$
Check: $3.0 Ω$ is smaller than both $12 Ω$ and $4.0 Ω$ , as a parallel combination must be.

Answer: $R = 3.0 Ω$ .

Example 3. A $6.0$ V supply is connected to two resistors in parallel, of $10 Ω$ and $15 Ω$ . Calculate the current in each branch and the total current from the supply.

Show the working

Start from the parallel rule for p.d.: each branch has the full $6.0$ V across it, because both are connected between the same two points.
Now treat each branch separately with $I = V / R$ .
$I_{1} = \frac{6.0}{10} = 0.60 A$
And the second branch.
$I_{2} = \frac{6.0}{15} = 0.40 A$
The currents add at the junction, so the total from the supply is the sum.
$I = 0.60 + 0.40 = 1.0 A$
Two checks. The total exceeds each branch current, as it must. And $R = 6.0 / 1.0 = 6.0 Ω$ , which is less than both $10$ and $15$ — consistent with a parallel combination.

Answer: $0.60$ A in the $10 Ω$ branch, $0.40$ A in the $15 Ω$ branch, and $1.0$ A from the supply.

Where people go wrong

Mistake: Saying the current 'gets used up' as it goes round a series circuit.

Instead: The current is the same at every point. Energy is transferred, not charge.

Charge is conserved. If current were lost, charge would have to be piling up somewhere.

Mistake: Stopping the parallel calculation at $1 / R$ .

Instead: Take the reciprocal to get $R$ .

The equation gives the reciprocal of the resistance. Forgetting the final step is the most common error in the subtopic — and the answer that results is always too small to be sensible.

Mistake: Expecting parallel resistance to be the sum, or the average.

Instead: It is always less than the smallest of the resistors.

Each extra branch is an extra route for charge, so more current flows for the same p.d., meaning less resistance.

Mistake: Adding e.m.f.s without checking the cells face the same way.

Instead: Cells facing the same way add; a reversed cell subtracts.

A cell opposing the others pushes charge the other way, cancelling part of the total e.m.f.

Mistake: Using the total resistance to find the p.d. across one component in series.

Instead: Use the current (which is common) and that component's own resistance.

In series it is the current that is shared, not the p.d. Use $V = I R$ with the individual $R$ .

Mistake: Using the product-over-sum shortcut for three parallel resistors.

Instead: That form works for two resistors only; use reciprocals for three or more.

It is an algebraic simplification of the two-resistor case, not a general rule.

Check yourself

In a series circuit with three lamps, the current through the first is $0.30$ A. What is the current through the third?
Answer
$0.30$ A — the current is the same at every point in a series circuit.
Four $1.5$ V cells are connected in series, all facing the same way. What is the combined e.m.f.? What if one is reversed?
Answer
$6.0$ V. With one reversed, it subtracts instead of adding, giving $4.5 - 1.5 = 3.0$ V.
Calculate the combined resistance of $5 Ω$ , $10 Ω$ and $15 Ω$ in series.
Answer
$5 + 10 + 15 = 30 Ω$ .
Calculate the combined resistance of two $8.0 Ω$ resistors in parallel.
Answer
$4.0 Ω$ — two equal resistors in parallel give half the value of one.
A current of $1.2$ A enters a junction and splits into two branches. One branch carries $0.5$ A. What does the other carry, and why?
Answer
$0.7$ A. The currents leaving a junction must sum to the current entering it, because charge is conserved.
Two resistors are in parallel across a $9.0$ V supply. What is the p.d. across each?
Answer
$9.0$ V across each — parallel branches are connected between the same two points.
Give three advantages of wiring house lamps in parallel.
Answer
Each lamp gets the full supply p.d. so all are equally bright; each can be switched independently; and if one fails the others stay on.
Explain why adding a second resistor in parallel decreases the total resistance.
Answer
It provides an additional path for the current, so a larger total current flows for the same p.d., which means the combined resistance is lower.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A $6.0$ V battery is connected in series with a $20 Ω$ resistor and a lamp. The p.d. across the lamp is $2.0$ V.

(a) Determine the p.d. across the resistor. 2 marks

Mark scheme

C1 uses the fact that p.d.s in series sum to the e.m.f.
A1 4.0 V; unit required

(b) Calculate the current in the circuit. 2 marks

Mark scheme

C1 I = V / R applied to the resistor; ecf from (a)
A1 0.20 A; unit required

(c) Determine the resistance of the lamp. 2 marks

Mark scheme

C1 R = V / I with 2.0 V and the current from (b); ecf
A1 10 Ω; unit required

(d) State what happens to the current in the resistor if the lamp is removed from the circuit, and give a reason. 2 marks No calculator

Mark scheme

B1 the current becomes zero
B1 the single path in the series circuit is broken, so no current can flow anywhere

Question 2. A $12$ V supply is connected to a $30 Ω$ resistor and a $60 Ω$ resistor in parallel.

(a) State the p.d. across the $60 Ω$ resistor. 1 mark No calculator

Mark scheme

B1 12 V

(b) Calculate the current in each resistor and the total current from the supply. 3 marks

Mark scheme

C1 I = V / R applied to each branch separately
A1 0.40 A and 0.20 A
A1 total 0.60 A; unit required

(c) Calculate the combined resistance of the two resistors. 2 marks

Mark scheme

C1 1/R = 1/30 + 1/60, or R = V / I total used
A1 20 Ω; unit required

(d) Explain why the combined resistance is less than $30 Ω$ . 2 marks No calculator

Mark scheme

B1 the second resistor provides an additional path for the current
B1 so a larger total current flows for the same p.d., meaning a lower resistance, owtte

Question 3. The lamps in a house are connected in parallel across the mains supply.

(a) State two advantages of connecting the lamps in parallel rather than in series. 2 marks No calculator

Mark scheme

B1 each lamp receives the full supply p.d., so all are equally bright
B1 any one from: each lamp can be switched independently / if one lamp fails the others still work

(b) Explain why the total current drawn from the supply increases as more lamps are switched on. 2 marks No calculator

Mark scheme

B1 each additional lamp adds another parallel branch, reducing the total resistance
B1 the currents in the branches add at the junction, so the supply current is larger

Words to know

Series: Components connected end to end in a single loop, so the same current passes through each.
Parallel: Components on separate branches between the same two points, so each has the same p.d. across it.
Junction: A point where a circuit divides or rejoins. The currents entering equal the currents leaving.
Combined resistance: The single resistance equivalent to a group of resistors. Resistances add in series; in parallel the reciprocals add.

All Physics lessons Next: 4.3.3 Action and use of circuit components