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Real circuits contain many components arranged in combinations of series and parallel. To analyse them efficiently we need compact rules for finding the total resistance of a set of components, and for distributing current and voltage between them.
This lesson derives the series and parallel resistance formulas from first principles (Kirchhoff's first law and the definition of pd), gives worked examples of mixed circuits, and previews the potential divider. It is OCR A-Level Physics A Module 4.3.1.
Two or more components are in series if the same current flows through each of them in turn. There is no junction between them — whatever goes in one end comes out the other, and must then enter the next component.
flowchart LR
A[+] --> R1[R1] --> R2[R2] --> R3[R3] --> B[-]
By Kirchhoff's first law at any junction in a series chain, the current is the same everywhere:
I = I₁ = I₂ = I₃ = …
The pds across each resistor add up to the supply voltage (this is really Kirchhoff's second law, which we will make rigorous in Lesson 11):
V_total = V₁ + V₂ + V₃ + …
Applying V = IR to each component with the same current I:
But also V_total = I × R_total, so:
R_total = R₁ + R₂ + R₃ + …
Series resistances add directly. Always.
Three resistors of 100 Ω, 220 Ω and 470 Ω are connected in series across a 9.0 V battery (internal resistance negligible). Find the total resistance, the current, and the pd across each resistor.
Check: V₁ + V₂ + V₃ = 1.14 + 2.51 + 5.36 = 9.01 V ≈ 9.0 V ✓
Notice that the largest resistor drops the largest voltage and the smallest drops the smallest. This is the principle behind the potential divider, which we will exploit in Lesson 12.
Two or more components are in parallel if they share the same pair of nodes — i.e. the pd across each of them is the same.
flowchart LR
A[+] --> J1((Node))
J1 --> R1[R1] --> J2((Node))
J1 --> R2[R2] --> J2
J1 --> R3[R3] --> J2
J2 --> B[-]
The pd across each parallel branch is identical:
V = V₁ = V₂ = V₃ = …
By Kirchhoff's first law at the top junction, the total current entering equals the sum of the branch currents:
I_total = I₁ + I₂ + I₃ + …
Applying I = V/R to each branch with the same V:
But I_total = V/R_total, so:
1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + …
Reciprocals of parallel resistances add, not the resistances themselves.
For exactly two parallel resistors:
R_total = (R₁ × R₂) / (R₁ + R₂)
(derived by combining the two terms 1/R₁ + 1/R₂ over a common denominator and inverting). Only valid for two resistors — for three or more you must use the reciprocal formula.
Three resistors of 100 Ω, 220 Ω and 470 Ω are connected in parallel across a 9.0 V supply. Find the total resistance, the total current, and the current in each branch.
Check: I₁ + I₂ + I₃ = 0.090 + 0.041 + 0.019 = 0.150 A ✓
Observations:
Two rules of thumb for parallel resistors:
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