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Understanding how components behave in series and parallel combinations is essential for analysing electrical circuits. This lesson covers the rules for current and voltage distribution, how to calculate total resistance for different configurations, and strategies for solving combined circuit problems.
Spec mapping: This lesson maps to AQA 7408 specification section 3.5.1.5 — Circuits, covering the rules for current and voltage in series and parallel arrangements, the derivation of the series and parallel resistance combination formulae, and the systematic reduction of mixed series-parallel networks to a single equivalent resistance. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- 3.5.1.7 — Kirchhoff's laws: the series and parallel rules are special cases of Kirchhoff's two laws — series current invariance follows from KCL applied to a node with one in and one out; parallel voltage equality follows from KVL applied around a loop containing only the two parallel branches. Lesson 9 generalises both.
- 3.5.1.6 — EMF and internal resistance: internal resistance is conceptually a hidden series resistor inside the source. The series rule R_total = R + r is the bridge between this lesson and Lesson 7's treatment of real cells.
- 3.5.1.5 — Potential dividers: the voltage-sharing rule in series circuits (V_i = I R_i, so V_i/V_total = R_i/R_total) is exactly the potential-divider equation. Lesson 8 uses this every time.
In a series circuit, components are connected end-to-end in a single loop. There is only one path for the current to flow.
| Property | Rule |
|---|---|
| Current | Same through all components: I_total = I₁ = I₂ = I₃ |
| Voltage | Shared between components: V_total = V₁ + V₂ + V₃ |
| Resistance | Adds up: R_total = R₁ + R₂ + R₃ |
Why current is the same: There is only one path, so all the charge that flows through one component must flow through every other component. If charge accumulated anywhere, conservation of charge would be violated.
Why voltage adds: The total energy transferred per coulomb (the total p.d.) equals the sum of the energy transferred per coulomb across each component. This follows from conservation of energy.
Three resistors of 100 Ω, 220 Ω, and 330 Ω are connected in series to a 12 V supply. Calculate (a) the total resistance, (b) the current, and (c) the p.d. across each resistor.
Solution:
(a) R_total = 100 + 220 + 330 = 650 Ω
(b) I = V/R_total = 12 / 650 = 0.01846 A = 18.5 mA
(c) V₁ = IR₁ = 0.01846 × 100 = 1.85 V V₂ = IR₂ = 0.01846 × 220 = 4.06 V V₃ = IR₃ = 0.01846 × 330 = 6.09 V
Check: V₁ + V₂ + V₃ = 1.85 + 4.06 + 6.09 = 12.0 V ✓
Exam Tip: In a series circuit, the largest resistor has the largest p.d. across it. The p.d. is shared in the same ratio as the resistances: V₁:V₂:V₃ = R₁:R₂:R₃.
In a parallel circuit, components are connected across the same two points. There are multiple paths for current.
| Property | Rule |
|---|---|
| Current | Shared between branches: I_total = I₁ + I₂ + I₃ |
| Voltage | Same across all branches: V_total = V₁ = V₂ = V₃ |
| Resistance | Reciprocal formula: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃ |
Why voltage is the same: All parallel branches are connected across the same two points, so the p.d. across each branch is the same.
Why current splits: At a junction, charge is conserved. The total current entering the junction equals the total current leaving it (Kirchhoff's first law).
For resistors in parallel, V is the same across each:
Total current: I_total = I₁ + I₂ + I₃ = V/R₁ + V/R₂ + V/R₃ = V(1/R₁ + 1/R₂ + 1/R₃)
Since I_total = V/R_total:
1/R_total = 1/R₁ + 1/R₂ + 1/R₃
For exactly two resistors in parallel, the formula simplifies to:
Rtotal=R1+R2R1R2
This "product over sum" formula is very useful and saves time in calculations.
Two resistors of 60 Ω and 40 Ω are connected in parallel across a 12 V supply. Calculate (a) the total resistance, (b) the total current, and (c) the current through each resistor.
Solution:
(a) Using the product-over-sum formula:
R_total = (60 × 40) / (60 + 40) = 2400 / 100 = 24 Ω
Check using reciprocal: 1/R = 1/60 + 1/40 = 2/120 + 3/120 = 5/120, so R = 120/5 = 24 Ω ✓
(b) I_total = V/R_total = 12 / 24 = 0.50 A
(c) I₁ = V/R₁ = 12 / 60 = 0.20 A I₂ = V/R₂ = 12 / 40 = 0.30 A
Check: I₁ + I₂ = 0.20 + 0.30 = 0.50 A = I_total ✓
Key Fact: The total resistance of parallel resistors is always LESS than the smallest individual resistance. Adding a resistor in parallel always decreases the total resistance (because you are providing an additional path for current).
Three 120 Ω resistors are connected in parallel. Calculate the total resistance.
Solution:
1/R_total = 1/120 + 1/120 + 1/120 = 3/120 = 1/40
R_total = 40 Ω
General rule: n identical resistors R in parallel give R_total = R/n.
Many circuits contain both series and parallel combinations. The strategy is to simplify step by step.
A circuit consists of a 10 Ω resistor in series with a parallel combination of 20 Ω and 30 Ω. The supply voltage is 10 V. Find (a) the total resistance, (b) the total current, (c) the p.d. across each section, and (d) the current through each parallel resistor.
Solution:
(a) First, find the parallel combination: R_parallel = (20 × 30) / (20 + 30) = 600 / 50 = 12 Ω
Total resistance: R_total = 10 + 12 = 22 Ω
(b) Total current: I = V/R_total = 10 / 22 = 0.4545 A = 0.455 A
(c) p.d. across the 10 Ω resistor: V₁ = IR₁ = 0.4545 × 10 = 4.55 V
p.d. across the parallel combination: V₂ = IR_parallel = 0.4545 × 12 = 5.45 V
Check: 4.55 + 5.45 = 10.0 V ✓
(d) Both parallel resistors have the same p.d. (5.45 V) across them:
I₂₀ = V₂/20 = 5.45 / 20 = 0.273 A
I₃₀ = V₂/30 = 5.45 / 30 = 0.182 A
Check: 0.273 + 0.182 = 0.455 A = I_total ✓
In the following circuit, R₁ = 8 Ω is in series with a parallel combination of R₂ = 6 Ω and the series combination of R₃ = 4 Ω and R₄ = 12 Ω. Supply voltage = 24 V.
Solution:
Step 1: R₃ and R₄ are in series: R₃₄ = 4 + 12 = 16 Ω
Step 2: R₂ and R₃₄ are in parallel: R₂₃₄ = (6 × 16) / (6 + 16) = 96 / 22 = 4.364 Ω
Step 3: R₁ is in series with R₂₃₄: R_total = 8 + 4.364 = 12.36 Ω
Total current: I = 24 / 12.36 = 1.94 A
p.d. across R₁: V₁ = 1.94 × 8 = 15.52 V
p.d. across parallel section: V_par = 1.94 × 4.364 = 8.47 V
Check: 15.52 + 8.47 = 23.99 ≈ 24.0 V ✓ (rounding)
Current through R₂: I₂ = 8.47 / 6 = 1.41 A
Current through R₃₄ branch: I₃₄ = 8.47 / 16 = 0.529 A
Check: 1.41 + 0.529 = 1.94 A ✓
Exam Tip: Always check your answers by verifying that (1) currents at junctions add up correctly and (2) voltages around loops add up to the supply voltage. This catches arithmetic errors.
| Configuration | Total Resistance |
|---|---|
| n resistors R in series | R_total = nR |
| n resistors R in parallel | R_total = R/n |
| 2 in series, then in parallel with 2 in series | R_total = R |
| Action | Effect on total R | Effect on total I (fixed V supply) |
|---|---|---|
| Add resistor in series | R increases | I decreases |
| Add resistor in parallel | R decreases | I increases |
| Remove a parallel branch | R increases | I decreases |
A 12.0 V battery (assume zero internal resistance for this example) drives a circuit consisting of a 4.0 Ω resistor R₁ in series with a parallel pair: R₂ = 6.0 Ω and R₃ = 3.0 Ω. Calculate: (a) the equivalent resistance of the parallel pair, (b) the total resistance of the circuit, (c) the total current drawn from the battery, (d) the p.d. across R₁ and across the parallel section, (e) the current through R₂ and through R₃. Verify the results using both Kirchhoff's current law (at the junction) and Kirchhoff's voltage law (around the loop).
Solution — Part (a): Parallel pair equivalent.
For two resistors in parallel, the product-over-sum shortcut applies:
R_par = (R₂ × R₃) / (R₂ + R₃) = (6.0 × 3.0) / (6.0 + 3.0) = 18 / 9 = 2.0 Ω
Sanity check: the parallel equivalent must be smaller than the smaller of the two branches (3.0 Ω). 2.0 Ω satisfies this.
Part (b): Total circuit resistance.
R₁ is in series with R_par, so:
R_total = R₁ + R_par = 4.0 + 2.0 = 6.0 Ω
Part (c): Total current.
I_total = V / R_total = 12.0 / 6.0 = 2.0 A
This is the current through the battery, the connecting wires, and the series resistor R₁.
Part (d): Voltages across the two sections.
p.d. across R₁: V₁ = I_total × R₁ = 2.0 × 4.0 = 8.0 V
p.d. across the parallel section: V_par = I_total × R_par = 2.0 × 2.0 = 4.0 V
Note both resistors in the parallel pair share the same 4.0 V — that is the defining property of a parallel connection.
KVL check (Kirchhoff's voltage law): V₁ + V_par = 8.0 + 4.0 = 12.0 V = ε ✓
Part (e): Branch currents in the parallel pair.
I₂ = V_par / R₂ = 4.0 / 6.0 = 0.667 A
I₃ = V_par / R₃ = 4.0 / 3.0 = 1.333 A
KCL check (Kirchhoff's current law): I₂ + I₃ = 0.667 + 1.333 = 2.000 A = I_total ✓
Interpretation: Notice that the smaller-resistance branch (R₃ = 3.0 Ω) carries the larger current (1.33 A vs 0.67 A). The ratio of currents in a parallel pair is the inverse of the ratio of resistances: I₂ / I₃ = R₃ / R₂ = 3/6 = 1/2, consistent with 0.667 / 1.333 ≈ 0.5. This "current-divider" relationship is the parallel-circuit dual of the potential-divider relationship for series circuits.
Note: The combined KVL + KCL check at the end is the gold-standard way to validate a circuit calculation. If either check fails, the algebra contains an error somewhere. AQA examiners explicitly reward candidates who show the check rather than leaving the final answers unverified.
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