Series and parallel circuits — deriving the rules — Physics

Before this

Ohm's law and resistance
Electric potential and potential energy
Electric current — charge flowing · coming soon

This leads to

More resistors, less resistance?

A 10 Ω resistor is connected to a 12 V battery. Now you add a second resistor of 1000 Ω, a hundred times bigger. Does the total resistance of the circuit go up or down?

Most people say "up, of course". You have added a large obstacle. If you connect the two resistors one after the other, that is right: the total becomes 1010 Ω, and the current falls from 1.2 A to about 0.012 A. But connect the 1000 Ω resistor side by side with the 10 Ω one, and the total resistance falls, to about 9.90 Ω. The battery now pushes out slightly more current than before.

So adding a resistor can lower the resistance. If you have only memorised "add them in series, add the reciprocals in parallel", this result has no explanation. This lecture builds both rules from two facts you already know. Once you see where the rules come from, results like this one stop being surprising. The last lecture promised exactly this: to use resistance to derive the rules for series and parallel circuits.

The two facts underneath every circuit

Fact 1: charge is conserved. Charge is not created or destroyed in a circuit. It also does not pile up in a wire. If more charge flowed into a piece of wire each second than flowed out, charge would build up there. That charge would push back on the charge arriving, until the flow in and the flow out became equal. So in a steady circuit, the current into any point equals the current out of it.

Fact 2: energy is conserved. The last two lectures gave every point in a circuit a potential. A battery lifts each coulomb of charge to a higher potential. The charge then transfers that energy in the components as it "falls" back to the start. A coulomb that goes once round the circuit ends where it began, at the same potential. So the potential it gains in the battery equals the potential it loses in the rest of the circuit. A connecting wire with negligible resistance has the same potential all along it, because no energy is transferred in it.

Everything in this lecture comes from these two facts. The next lecture gives them their usual names, as two laws.

Series: one path for the charge

Resistors are in series when they are connected one after another, so that the charge has only one path.

The current. There is nowhere for charge to branch off. By Fact 1, every coulomb that passes through the first resistor must pass through the second. So the current is the same at every point in a series circuit. A resistor does not "use up" current. It transfers energy, and the charge carries on.

The p.d. Follow the potential round the circuit below. The battery lifts each coulomb to 12 V. It falls by 8 V in the 4 Ω resistor and by another 4 V in the 2 Ω resistor, back to 0 V.

The potential at four points (blue), measured from the battery's negative terminal. The p.d. across the 4 Ω resistor is 12 V − 4 V = 8 V, and across the 2 Ω resistor it is 4 V − 0 V = 4 V. The current, 2 A, is the same everywhere.

By Fact 2, the falls add up to the rise. So in series, the p.d.s across the components add up to the p.d. of the battery: $V = V_{1} + V_{2}$ .

The total resistance. Now use the definition of resistance on each resistor. The current $I$ is the same in both, so $V_{1} = I R_{1}$ and $V_{2} = I R_{2}$ :

V = I R_{1} + I R_{2} = I (R_{1} + R_{2})

A single resistor that draws the same current $I$ from the same p.d. $V$ would have $V = I R$ . Compare the two lines, and $R = R_{1} + R_{2}$ .

Resistors in series

The current is the same in each resistor. The p.d.s add up: $V = V_{1} + V_{2} + \dots$

The total resistance is the sum: $R = R_{1} + R_{2} + \dots$

Check it on the figure: $4 + 2 = 6$ Ω, and $12 \div 6 = 2$ A. The p.d.s are $2 \times 4 = 8$ V and $2 \times 2 = 4$ V. They add up to 12 V. ✓ Notice one more thing. The bigger resistor takes the bigger share of the p.d. In series, the p.d. is shared in the same ratio as the resistances.

Parallel: several paths between the same two points

Resistors are in parallel when each one connects across the same two points, so the charge can choose between several paths. The step-through below builds the rule one move at a time. Read the reason for each step before you go on.

Step through — the rule for resistors in parallel

Two resistors, R₁ and R₂, are both connected between points A and B. Press Next to add one step.

Each step gives one line and the reason it is allowed.

1. Both resistors join A to B. A connecting wire has the same potential all along it, so each end of each resistor is at the potential of A or of B.
2. So both resistors have the same p.d., V. The p.d. is the difference between the potentials of A and B, and that is the same for both.
3. I₁ = V ÷ R₁ and I₂ = V ÷ R₂. The definition of resistance, used on each resistor.
4. The total current is I = I₁ + I₂. Charge is conserved: the charge arriving at A each second must leave along the two branches.
5. So I = V ÷ R₁ + V ÷ R₂. Put step 3 into step 4.
6. One resistor R that draws the same current from the same p.d. has I = V ÷ R. This is what "the total resistance" means.
7. V ÷ R = V ÷ R₁ + V ÷ R₂, so 1 ÷ R = 1 ÷ R₁ + 1 ÷ R₂. Divide both sides by V.

Result: 1/R = 1/R₁ + 1/R₂. For two resistors, this gives R = R₁R₂ ÷ (R₁ + R₂).

Resistors in parallel

The p.d. is the same across each branch. The currents add up: $I = I_{1} + I_{2} + \dots$

The total resistance is given by $\frac{1}{R} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots$ . For two resistors, $R = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$ .

The two-resistor form comes from adding the fractions: $\frac{1}{R _{1}} + \frac{1}{R _{2}} = \frac{R _{1} + R _{2}}{R _{1} R _{2}}$ , then turning the fraction upside down. Some books call it "product over sum". It works for two resistors only.

That is it. That is the whole idea. Series and parallel are two answers to one question: what is shared? In series, the charge has one path, so the current is shared by every resistor and the p.d. is split. In parallel, every branch has the same two ends, so the p.d. is shared and the current is split.

Why a parallel resistor lowers the total

Now return to the hook. Put the numbers into the parallel rule:

\frac{1}{R} = \frac{1}{10} + \frac{1}{1000} = 0.101

R = \frac{1}{0.101} \approx 9.90 Ω

Here is the physical reason. The 10 Ω resistor still has 12 V across it, so it still carries 1.2 A. The new branch adds its own current, $12 \div 1000 = 0.012$ A. So the battery now supplies 1.212 A. More current from the same p.d. means less total resistance. A new path never blocks the old ones. It can only add current.

This gives a quick check for every parallel answer: the total is always smaller than the smallest resistor. In the same way, a series total is always bigger than the biggest resistor. Explore both below.

Explore — adding a second resistor

R₁ is 10 Ω. Choose how R₂ is connected, then change R₂ and watch the total resistance and the current from a 12 V battery.

Connection:

R₂: 10 Ω

This table shows R₁ = 10 Ω combined with different values of R₂, and the current each combination draws from a 12 V battery (3 significant figures).

R₂ (Ω)	Series total (Ω)	Series current (A)	Parallel total (Ω)	Parallel current (A)
1	11	1.09	0.909	13.2
10	20	0.600	5.00	2.40
100	110	0.109	9.09	1.32
1000	1010	0.0119	9.90	1.21

Every series total is bigger than both resistors. Every parallel total is smaller than both. Two equal resistors in parallel give half of one.

You should have seen the two rules pull in opposite directions. In series, the total always sits above the bigger resistor. In parallel, it always sits below the smaller one. When R₂ is huge, the parallel total is almost exactly R₁, because almost no current takes the huge path. When R₂ is tiny, it carries most of the current, and the total is close to R₂.

Stop and think

A torch has two identical lamps in parallel across its battery. One lamp breaks. Does the other lamp become brighter, dimmer, or stay the same? Assume the battery keeps the same p.d.

Show the answer

It stays the same. The working lamp still has the full battery p.d. across it, so by the definition of resistance it still carries the same current. The battery supplies less current in total, because one path has gone. Many students expect the remaining lamp to get "the current the broken lamp was using". That is the series picture, and it does not apply here: each parallel branch takes the current that its own p.d. and resistance decide. (A real battery's p.d. rises a little when less current is drawn, for a reason the lecture on internal resistance explains.)

Worked examples

Worked Example 1 — Three resistors in series

Resistors of 6 Ω, 3 Ω and 2 Ω are connected in series to a 12 V battery. Find the total resistance, the current, and the p.d. across each resistor.

Think first. One path, so one current. Find the total first.

$R = 6 + 3 + 2 = 11$ Ω, so $I = 12 \div 11 \approx 1.09$ A, the same in every resistor.

Keep the exact current, $\frac{12}{11}$ A, for the p.d.s: $V_{1} = \frac{12}{11} \times 6 \approx 6.55$ V, $V_{2} = \frac{12}{11} \times 3 \approx 3.27$ V, $V_{3} = \frac{12}{11} \times 2 \approx 2.18$ V.

Check. $6.55 + 3.27 + 2.18 = 12.00$ V. ✓ The 6 Ω resistor takes half of the 12 V, a little more than 6 V, because it is a little more than half of the 11 Ω total. ✓

Worked Example 2 — Two resistors in parallel

Resistors of 6 Ω and 3 Ω are connected in parallel to a 12 V battery. Find the current in each, the total current and the total resistance.

Think first. Each branch has the full 12 V across it, so each current comes straight from the definition.

$I_{1} = 12 \div 6 = 2$ A and $I_{2} = 12 \div 3 = 4$ A. The total current is $2 + 4 = 6$ A, so $R = 12 \div 6 = 2$ Ω.

Check. By the rule: $\frac{1}{6} + \frac{1}{3} = \frac{3}{6} = \frac{1}{2}$ , so $R = 2$ Ω. ✓ By product over sum: $\frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2$ Ω. ✓ And 2 Ω is smaller than 3 Ω, the smaller resistor. ✓ The smaller resistor carries the larger current.

Worked Example 3 — A mixed circuit

A 4 Ω resistor is in series with a parallel pair of 6 Ω and 3 Ω. The whole circuit is connected to a 12 V battery. Find the current in every resistor.

Think first. The parallel pair can be replaced by one resistor. Then the circuit is a simple series circuit.

The pair: from Worked Example 2, 6 Ω and 3 Ω in parallel make 2 Ω. The total is $4 + 2 = 6$ Ω, so the battery current is $12 \div 6 = 2$ A. All of it passes through the 4 Ω resistor, so the p.d. there is $2 \times 4 = 8$ V.

The pair gets the rest of the battery's p.d.: $12 - 8 = 4$ V. Notice that this is not 12 V. The parallel branches share a p.d. with each other, but not with the battery. So $I_{6} = 4 \div 6 \approx 0.667$ A and $I_{3} = 4 \div 3 \approx 1.33$ A.

Check. $0.667 + 1.333 = 2.00$ A, the current that left the 4 Ω resistor. ✓ $8 + 4 = 12$ V round the circuit. ✓

Worked Example 4 — When "parallel first" is wrong

Two branches are connected across a 12 V battery. Branch A has a 2 Ω and a 4 Ω resistor in series. Branch B has a single 3 Ω resistor. Find the total resistance and the battery current.

Branch A (tinted) holds two resistors in series. It must be combined first, to 6 Ω, before it can be put in parallel with branch B.

Think first. There is no simple parallel pair here. The 3 Ω resistor is in parallel with the whole of branch A, not with the 2 Ω or the 4 Ω alone. So combine the innermost group first: the series pair inside branch A.

Branch A: $2 + 4 = 6$ Ω. Then 6 Ω in parallel with 3 Ω gives 2 Ω, as in Worked Example 2. So the battery current is $12 \div 2 = 6$ A.

Check. Each branch has 12 V across it. Branch A carries $12 \div 6 = 2$ A, and branch B carries $12 \div 3 = 4$ A, which add up to 6 A. ✓ In branch A, the 2 A gives $2 \times 2 = 4$ V and $2 \times 4 = 8$ V, which add up to the branch's 12 V. ✓

So the method is not "parallel first". It is "combine the innermost group first, then work outwards", where a group is a set of resistors that are all in series or all in parallel with each other.

Three traps

Trap 1 — "The first resistor uses up some of the current"

In series, the current is the same everywhere. A resistor transfers energy from the charge, which is why the potential drops across it. The charge itself carries on, at the same rate. If an ammeter before the first resistor reads 2 A, an ammeter after the last one also reads 2 A.

Trap 2 — Forgetting to turn the fraction over

For 6 Ω and 3 Ω in parallel, $\frac{1}{6} + \frac{1}{3} = 0.5$ . That is $\frac{1}{R}$ , not $R$ . The answer is $R = 1 \div 0.5 = 2$ Ω. The "smaller than the smallest" check catches this at once when the numbers are bigger. For 60 Ω and 30 Ω, a student who stops at $\frac{1}{20} = 0.05$ Ω has an answer far too small. The real answer, 20 Ω, is below 30 Ω, as it must be.

Trap 3 — "Parallel branches always get the battery's p.d."

Parallel branches have the same p.d. as each other. They have the battery's p.d. only when nothing else is in series with them. In Worked Example 3, the pair had only 4 V, because the 4 Ω resistor took 8 V first. Always find the p.d. across the parallel group before you use it.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Resistors of 4 Ω and 12 Ω are connected in series to an 8.0 V supply. Find the current and the p.d. across each resistor.

Hint 1
One path, so one current. Start with the total resistance.

Hint 2
$R = 4 + 12 = 16$ Ω. Then $I = V \div R$ .

Full answer
$I = 8.0 \div 16 = 0.50$ A. $V_{4} = 0.50 \times 4 = 2.0$ V and $V_{12} = 0.50 \times 12 = 6.0$ V. Check: $2.0 + 6.0 = 8.0$ V. ✓ The 12 Ω resistor is three times the 4 Ω one, and it takes three times the p.d. ✓
The same two resistors, 4 Ω and 12 Ω, are now connected in parallel to a 6.0 V supply. Find the current in each, the total current and the total resistance.

Hint 1
What is the p.d. across each branch?

Hint 2
Each has 6.0 V across it. Find each current from the definition, then add.

Full answer
$I_{4} = 6.0 \div 4 = 1.5$ A and $I_{12} = 6.0 \div 12 = 0.50$ A, so $I = 2.0$ A and $R = 6.0 \div 2.0 = 3.0$ Ω. Check: $\frac{4 \times 12}{4 + 12} = \frac{48}{16} = 3$ Ω. ✓ 3 Ω is below 4 Ω, the smaller resistor. ✓
You have three identical 30 Ω resistors. Find every total resistance you can make using all three of them.

Hint 1
There are four arrangements: all in series, all in parallel, and two mixed ones.

Hint 2
For the mixed ones, either put two in parallel and the third in series with them, or put two in series and the third in parallel with them.

Full answer
All in series: $30 + 30 + 30 = 90$ Ω. All in parallel: $\frac{1}{R} = \frac{3}{30}$ , so $R = 10$ Ω. Two in parallel make 15 Ω, plus 30 Ω in series: 45 Ω. Two in series make 60 Ω, in parallel with 30 Ω: $\frac{60 \times 30}{60 + 30} = 20$ Ω. Check: every total lies between 10 Ω (all paths side by side) and 90 Ω (one long path). ✓ In general, $n$ identical resistors $R$ in parallel give $R \div n$ .
A stretch. A 12 V battery is connected to a 6.0 Ω resistor in series with a parallel pair: a 12 Ω resistor and an unknown resistor $R$ . The battery current is 1.5 A. Find $R$ .

Hint 1
Work backwards. What total resistance gives 1.5 A from 12 V? What must the parallel pair be?

Hint 2
The total is 8.0 Ω, so the pair is 2.0 Ω. Now find the p.d. across the pair and the current in the 12 Ω branch.

Full answer
Total $= 12 \div 1.5 = 8.0$ Ω, so the pair is $8.0 - 6.0 = 2.0$ Ω. The 6.0 Ω resistor takes $1.5 \times 6.0 = 9.0$ V, so the pair has $12 - 9.0 = 3.0$ V. The 12 Ω branch carries $3.0 \div 12 = 0.25$ A, so $R$ carries $1.5 - 0.25 = 1.25$ A, and $R = 3.0 \div 1.25 = 2.4$ Ω. Check with the rule: $\frac{1}{2.4} + \frac{1}{12} = 0.4167 + 0.0833 = 0.5$ , so the pair is 2.0 Ω. ✓ 2.0 Ω is below 2.4 Ω, the smaller resistor. ✓

The bigger idea

The wiring in a home is a set of parallel branches. Each lamp, fan and fridge is connected across the same two supply wires, so each one gets the full mains p.d. and works whatever the others are doing. Switching one off removes a path. It does not break the others. Strings of small decorative lights are often wired the other way, in series, which is why one failed bulb in an old string can make the whole string go dark.

Parallel wiring has a cost. Every appliance you switch on adds its own current, and all of that current flows in the supply wires together. Too much current would overheat those wires. This is why each circuit in a building has a fuse or a circuit breaker, which cuts off the supply when the total current is too large.

Look back at what you used in this lecture: the definition of resistance, charge that cannot pile up, and energy that is not lost. Those two conservation facts solve circuits far more tangled than series and parallel, where no groups can be combined at all. The next lecture writes them as two laws.

Key concepts in this lesson

Charge is conserved, so the current into any point of a circuit equals the current out of it.
Energy is conserved, so the potential a coulomb gains in the battery equals the potential it loses around the rest of the circuit.
In series, the current is the same in each resistor, the p.d.s add up, and the total resistance is $R_{1} + R_{2} + \dots$ .
In parallel, the p.d. is the same across each branch, the currents add up, and $\frac{1}{R} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots$ , so the total is smaller than the smallest resistor.
To solve a mixed circuit, combine the innermost group first and work outwards, then find the p.d. across each group before you use it.