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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.
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 ✓
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