Series and Parallel Circuits

Every circuit you will encounter at A-Level can be broken down into components connected in series, in parallel, or a combination of both. Mastering the rules for each arrangement is essential for circuit analysis.

Series Circuits

In a series circuit, all components are connected end to end in a single loop. There is only one path for the current to flow through.

Three Key Rules for Series Circuits

1. Current is the same through every component.

There is only one path, so the same charge flows through each component every second. If 0.50 A flows through the battery, then 0.50 A flows through every resistor, lamp, or other component in the loop.

2. The total potential difference is shared between the components.

The sum of the potential differences across all components equals the total p.d. of the supply:

V_total = V₁ + V₂ + V₃ + ...

The p.d. across each component is proportional to its resistance (since V = IR and I is the same for all). A larger resistance gets a larger share of the total voltage.

3. The total resistance is the sum of the individual resistances.

R_total = R₁ + R₂ + R₃ + ...

Adding more resistors in series always increases the total resistance.

Worked Example 1

Three resistors of 10 Ω, 20 Ω, and 30 Ω are connected in series to a 12 V battery. Calculate (a) the total resistance, (b) the current, and (c) the p.d. across the 20 Ω resistor.

Solution:

(a) R_total = 10 + 20 + 30 = 60 Ω
(b) I = V/R = 12/60 = 0.20 A
(c) V₂₀ = IR = 0.20 × 20 = 4.0 V

Parallel Circuits

In a parallel circuit, components are connected across each other, creating multiple paths for current to flow.

Three Key Rules for Parallel Circuits

1. The potential difference is the same across each branch.

Each branch is connected directly across the supply, so every branch has the same p.d.

2. The total current is the sum of the currents in each branch.

Current splits at a junction and recombines when the branches meet again:

I_total = I₁ + I₂ + I₃ + ...

More current flows through the branch with less resistance (since I = V/R and V is the same for all branches).

3. The total resistance is found using the reciprocal formula.

1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + ...

For two resistors in parallel, there is a useful shortcut:

R_total = (R₁ × R₂) / (R₁ + R₂)

Adding more resistors in parallel always decreases the total resistance (because you are providing additional paths for current).

Worked Example 2

Two resistors, 6.0 Ω and 12 Ω, are connected in parallel across a 12 V supply. Calculate (a) the total resistance, (b) the current through each resistor, and (c) the total current from the supply.

Solution:

(a) R_total = (6.0 × 12) / (6.0 + 12) = 72/18 = 4.0 Ω
(b) I₆ = V/R = 12/6.0 = 2.0 A; I₁₂ = V/R = 12/12 = 1.0 A
(c) I_total = 2.0 + 1.0 = 3.0 A (Check: I = V/R_total = 12/4.0 = 3.0 A ✓)

Combined (Series-Parallel) Circuits

Most real circuits contain both series and parallel combinations. The strategy for solving these is:

flowchart TD
    A[Start: Read the circuit] --> B[Identify series and parallel sections]
    B --> C[Calculate equivalent resistance for each parallel group]
    C --> D[Redraw simplified circuit]
    D --> E[Add remaining series resistances for R_total]
    E --> F[Find total current: I = V / R_total]
    F --> G[Find p.d. across each section using V = IR]
    G --> H[Find branch currents in parallel sections using I = V/R]
    H --> I[Check: Do voltages sum to EMF? Do currents at junctions balance?]

Worked Example 3

A circuit consists of a 12 V battery connected to a 4.0 Ω resistor in series with a parallel combination of 6.0 Ω and 3.0 Ω resistors.

Solution:

Step 1: Find the parallel combination. R_parallel = (6.0 × 3.0) / (6.0 + 3.0) = 18/9.0 = 2.0 Ω
Step 2: Find total resistance. R_total = 4.0 + 2.0 = 6.0 Ω
Step 3: Find total current. I = V/R = 12/6.0 = 2.0 A
Step 4: Find p.d. across each section. V across 4.0 Ω = IR = 2.0 × 4.0 = 8.0 V V across parallel combination = IR = 2.0 × 2.0 = 4.0 V (check: 8.0 + 4.0 = 12 V ✓)
Step 5: Find current in each parallel branch. I₆ = 4.0/6.0 = 0.67 A I₃ = 4.0/3.0 = 1.33 A (check: 0.67 + 1.33 = 2.0 A ✓)

Worked Example 4: Three-Level Nested Circuit

A 24 V battery (negligible internal resistance) is connected to the following: a 2.0 Ω resistor in series with a parallel combination of [8.0 Ω] and [4.0 Ω in series with 12 Ω]. Calculate the current from the battery and the p.d. across each component.

Solution:

First: the second parallel branch is 4.0 + 12 = 16 Ω in series
Parallel combination: R_p = (8.0 × 16) / (8.0 + 16) = 128/24 = 5.33 Ω
Total: R_total = 2.0 + 5.33 = 7.33 Ω
Total current: I = 24/7.33 = 3.27 A
V across 2.0 Ω = 3.27 × 2.0 = 6.55 V
V across parallel section = 3.27 × 5.33 = 17.45 V (check: 6.55 + 17.45 = 24 V ✓)
Current through 8.0 Ω branch = 17.45/8.0 = 2.18 A
Current through (4.0 + 12) branch = 17.45/16 = 1.09 A (check: 2.18 + 1.09 = 3.27 A ✓)
P.d. across 4.0 Ω = 1.09 × 4.0 = 4.36 V
P.d. across 12 Ω = 1.09 × 12 = 13.09 V (check: 4.36 + 13.09 = 17.45 V ✓)

Identical Resistors: Quick Shortcuts

For n identical resistors, each of resistance R:

In series: R_total = nR
In parallel: R_total = R/n

These shortcuts are very useful in exams. For example, three 12 Ω resistors in parallel have a combined resistance of 12/3 = 4.0 Ω.

Quick Reference: Parallel Resistance Results

R₁	R₂	R_total (parallel)	Relationship
R	R	R/2	Two equal resistors
R	2R	2R/3	1:2 ratio
R	3R	3R/4	1:3 ratio
R	10R	10R/11 ≈ 0.91R	One much larger
R	100R	100R/101 ≈ R	Very unequal: R_total ≈ smaller R

This table reveals a useful pattern: when one parallel resistor is much larger than the other, the combination is approximately equal to the smaller resistor. The large resistor has almost no effect because very little current flows through it.

Worked Example: Current Splitting at a Junction

In a parallel circuit, it is often useful to find how the total current divides between branches without first calculating the p.d. The current divider rule states:

For two resistors R₁ and R₂ in parallel carrying total current I_total:

Current through R₁: I₁ = I_total × R₂ / (R₁ + R₂)
Current through R₂: I₂ = I_total × R₁ / (R₁ + R₂)

Notice the "cross-over" — the current through R₁ depends on R₂, and vice versa. More current flows through the smaller resistance.

Example: A 2.0 A current splits between 3.0 Ω and 6.0 Ω in parallel.

I through 3.0 Ω = 2.0 × 6.0/(3.0 + 6.0) = 12/9.0 = 1.33 A
I through 6.0 Ω = 2.0 × 3.0/(3.0 + 6.0) = 6.0/9.0 = 0.67 A
Check: 1.33 + 0.67 = 2.0 A ✓

The 3.0 Ω branch (half the resistance) carries twice the current — as expected.

Power Distribution in Series vs Parallel

Understanding how power distributes differently in series and parallel is a common exam topic.

In series (same current I through all):

P = I²R, so the largest resistance dissipates the most power

In parallel (same voltage V across all):

P = V²/R, so the smallest resistance dissipates the most power

Worked Example 5: Power in a Combined Circuit

A 12 V battery is connected to 4.0 Ω in series with a parallel combination of 6.0 Ω and 3.0 Ω. Find the power dissipated in each resistor.

Solution (using values from Worked Example 3):

P in 4.0 Ω: P = I²R = (2.0)² × 4.0 = 16 W (or P = VI = 8.0 × 2.0 = 16 W)
P in 6.0 Ω: P = V²/R = (4.0)²/6.0 = 2.7 W
P in 3.0 Ω: P = V²/R = (4.0)²/3.0 = 5.3 W
Total power = 16 + 2.7 + 5.3 = 24 W. Check: P = VI = 12 × 2.0 = 24 W ✓

Note: within the parallel section, the 3.0 Ω (smaller R) dissipates more power — as expected for parallel components with the same p.d.

Common Mistakes to Avoid

Mistake	Why It Happens	How to Avoid It
Using parallel formula for series (or vice versa)	Rushing, not checking circuit structure	Count the paths: one path = series, multiple = parallel
Forgetting the final reciprocal in 1/R_total	Mechanical error	Always check: is R_total < smallest R? (for parallel)
Assuming current splits equally	Only true for equal resistances	Use I = V/R for each branch
Adding a parallel R increases R_total	Intuition says "more = more"	More paths = less resistance: R_total always decreases
Forgetting internal resistance	Not reading the question carefully	If EMF and r are given, total R = R_external + r

A-Level Deep Dive: Series and Parallel Circuits

Spec mapping

Edexcel 9PH0 specification, Topic 3 — Electric circuits, sub-topic 3.4 (series and parallel networks) establishes the two combination rules $R_\text{series} = R_1 + R_2 + \dots$ and $1/R_\text{parallel} = 1/R_1 + 1/R_2 + \dots$ , together with the conservation principles that current is the same through every component in a series loop and that potential difference is the same across every branch of a parallel network (refer to the official Pearson Edexcel specification document for exact wording). Although series/parallel analysis is the fourth sub-topic of Topic 3, it is the workhorse skill used everywhere else in 9PH0-01 and 9PH0-03: emf with internal resistance (3.5) is a series problem with the cell as one element; potential dividers in Topic 3 are series problems with a tap point; the Wheatstone-style balanced networks that appear in synoptic questions are nested series/parallel structures. The Edexcel formula booklet lists both combination formulae, but candidates are expected to apply them via Kirchhoff's voltage and current laws, not merely substitute.

Worked example with full mark scheme

Question (8 marks):

A 24 V battery of negligible internal resistance is connected to the following network: a $3.0\ \Omega$ resistor $R_1$ in series with a parallel combination of two branches — branch X is a single $8.0\ \Omega$ resistor, and branch Y is a $2.0\ \Omega$ resistor in series with a $6.0\ \Omega$ resistor.

(a) Show that the total resistance of the network is $7.0\ \Omega$ . (3)

(b) Calculate the current drawn from the battery. (1)

(d) State and justify which of the two parallel branches dissipates more power. (1)

Solution with mark scheme:

(a) Step 1 — combine the series pair inside branch Y.

$R_Y = 2.0 + 6.0 = 8.0\ \Omega$

M1 — applying the series rule $R = R_1 + R_2$ inside the branch.

Step 2 — combine the two parallel branches.

Branch X is $8.0\ \Omega$ and branch Y is now $8.0\ \Omega$ . For two equal resistors in parallel, $R_p = R/2$ :

$R_p = \dfrac{8.0 \times 8.0}{8.0 + 8.0} = \dfrac{64}{16} = 4.0\ \Omega$

M1 — correct application of the parallel formula $R_p = R_1 R_2 / (R_1 + R_2)$ (or the reciprocal form).

Step 3 — add the remaining series resistor.

$R_\text{total} = R_1 + R_p = 3.0 + 4.0 = 7.0\ \Omega$

A1 — printed answer obtained, with at least one extra significant figure shown in the working (here the answer is exact, so 7.0 Ω is sufficient).

(b) Step 1 — apply $I = V/R$ to the whole network.

$I = \dfrac{V}{R_\text{total}} = \dfrac{24}{7.0} \approx 3.43\ \text{A}$

A1 — $I \approx 3.4\ \text{A}$ to 2 s.f. Common error: candidates use $24/4.0$ here (forgetting the series $3.0\ \Omega$ ) and quote 6.0 A. That is a single substitution slip but it propagates through every later step.

The same current $I = 3.43\ \text{A}$ flows through $R_1$ first, dropping $V_1 = IR_1 = 3.43 \times 3.0 = 10.3\ \text{V}$ across it. By Kirchhoff's voltage law, the p.d. across the parallel section is the remainder:

$V_p = 24 - 10.3 = 13.7\ \text{V}$

M1 — correct use of $V_p = \varepsilon - IR_1$ , equivalently $V_p = IR_p = 3.43 \times 4.0 = 13.7\ \text{V}$ . Either route earns the mark.

Step 2 — apply $I = V/R$ to each branch using the same p.d.

$I_X = \dfrac{V_p}{R_X} = \dfrac{13.7}{8.0} \approx 1.71\ \text{A}$ $I_Y = \dfrac{V_p}{R_Y} = \dfrac{13.7}{8.0} \approx 1.71\ \text{A}$

M1 — recognising that the same p.d. $V_p$ acts across each branch (a Topic 3.4 fundamental).

A1 — both branch currents $\approx 1.7\ \text{A}$ , with the consistency check $I_X + I_Y = 3.43\ \text{A}$ matching part (b).

(d) B1 — the two branches dissipate equal power because they have equal resistance and the same p.d. across them: $P = V^2/R$ gives $P_X = P_Y = 13.7^2 / 8.0 \approx 23\ \text{W}$ . Many candidates assume that "the branch with more resistors" must dissipate more power; the correct argument is in terms of total branch resistance, which is identical here.

Total: 8 marks (M1 M1 A1 + A1 + M1 M1 A1 + B1).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A potential divider is constructed from a fixed $1.5\ \text{k}\Omega$ resistor in series with a thermistor whose resistance varies between $500\ \Omega$ at 80 °C and $5.0\ \text{k}\Omega$ at 0 °C. The supply p.d. is $6.0\ \text{V}$ and the output is taken across the thermistor.

(a) Calculate the output p.d. when the thermistor is at 80 °C. (2)

(b) Calculate the output p.d. when the thermistor is at 0 °C. (2)

(c) Explain how the output p.d. would change if the fixed $1.5\ \text{k}\Omega$ resistor were replaced with a $15\ \text{k}\Omega$ resistor. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1b) — using the potential-divider relation $V_\text{out} = V_\text{in} \times R_\text{th} / (R_\text{th} + R_\text{fixed})$ with consistent units.
A1 (AO1.1b) — $V_\text{out} = 6.0 \times 500 / (500 + 1500) = 6.0 \times 0.25 = 1.5\ \text{V}$ .

(b)

M1 (AO1.1b) — substituting $R_\text{th} = 5000\ \Omega$ .
A1 (AO1.1b) — $V_\text{out} = 6.0 \times 5000 / (5000 + 1500) = 6.0 \times 0.769 \approx 4.6\ \text{V}$ .

(c)

M1 (AO2.1) — recognising that increasing the fixed resistor increases the denominator at every temperature, so the fraction $R_\text{th} / (R_\text{th} + R_\text{fixed})$ decreases, meaning the output p.d. decreases at every temperature.
A1 (AO2.5) — concluding with quantitative reasoning: at 80 °C the new output is $6.0 \times 500/15500 \approx 0.19\ \text{V}$ , and at 0 °C it is $6.0 \times 5000/20000 = 1.5\ \text{V}$ , so the range of output narrows toward zero — making the divider a worse temperature sensor.

Total: 6 marks split AO1 = 4, AO2 = 2. This question is a Topic 3.4 application dressed in sensor clothing — every step is a series-circuit calculation with a temperature-dependent element. AO2 marks reward physical reasoning about why the choice of fixed resistor matters for sensor sensitivity, not just numerical substitution.

Synoptic links

Connects to:

Series Circuits

Three Key Rules for Series Circuits

Worked Example 1

Parallel Circuits

Three Key Rules for Parallel Circuits

Worked Example 2

Combined (Series-Parallel) Circuits

Worked Example 3

Worked Example 4: Three-Level Nested Circuit

Identical Resistors: Quick Shortcuts

Quick Reference: Parallel Resistance Results

Worked Example: Current Splitting at a Junction

Power Distribution in Series vs Parallel

Worked Example 5: Power in a Combined Circuit

Common Mistakes to Avoid

A-Level Deep Dive: Series and Parallel Circuits

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Synoptic links

More in Physics