Resistors in series
When resistors are joined end to end so that there is only one path for current, they are in series.
- The same current I flows through each resistor.
- The battery's p.d. is shared: V = V₁ + V₂ + V₃. The largest resistor gets the largest share.
- If one resistor (or bulb) breaks, the whole circuit stops.
Derivation: equivalent resistance in series
By Ohm's law, V₁ = IR₁, V₂ = IR₂, V₃ = IR₃. The total is V = V₁ + V₂ + V₃ = I(R₁ + R₂ + R₃). If one resistor Rs is to replace all three with the same I, then V = IRs. Comparing:
Rs = R₁ + R₂ + R₃
So Rs is always larger than the largest single resistor.
Resistors in parallel
When all resistors are joined between the same two points, each on its own branch, they are in parallel.
- The voltage across each resistor is the same, equal to V.
- The total current splits: I = I₁ + I₂ + I₃. The smallest resistor takes the largest current.
- If one branch breaks, the others keep working.
Derivation: equivalent resistance in parallel
Each branch has the same V, so I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃. Total I = V(1/R₁ + 1/R₂ + 1/R₃). For one equivalent resistor Rp, I = V/Rp. Comparing:
1/Rp = 1/R₁ + 1/R₂ + 1/R₃
For two resistors, Rp = R₁R₂ / (R₁ + R₂). For n equal resistors R, series gives nR and parallel gives R/n. Rp is always smaller than the smallest resistor, because adding a branch gives charge an extra path.
Why homes use parallel wiring
In series, appliances would share the 220 V, each would get the same current even though they need different currents, and one failure would switch everything off. In parallel every appliance gets the full 220 V, draws the current it needs, and has its own switch. For mixed circuits, first simplify each parallel group, then add the series parts, then use Ohm's law.
Key formulas and definitions
- Series: R_s = R₁ + R₂ + R₃; same I; V = V₁ + V₂ + V₃
- Parallel: 1/R_p = 1/R₁ + 1/R₂ + 1/R₃; same V; I = I₁ + I₂ + I₃
- Two in parallel: R_p = R₁R₂ / (R₁ + R₂)
- n equal resistors R: series nR, parallel R/n
Worked examples
1. Find the total resistance of 3 Ω, 5 Ω and 7 Ω in series.
R_s = 3 + 5 + 7 = 15 Ω.
2. Find the equivalent of 6 Ω and 3 Ω in parallel.
R_p = 6 × 3 / (6 + 3) = 18/9 = 2 Ω.
3. 2 Ω, 4 Ω and 6 Ω are in series with a 12 V battery. Find the current and the p.d. across each.
R_s = 12 Ω, I = 12/12 = 1 A. V₁ = 2 V, V₂ = 4 V, V₃ = 6 V (total 12 V).
4. The same three resistors are in parallel across 12 V. Find each current, the total current and R_p.
I₁ = 12/2 = 6 A, I₂ = 12/4 = 3 A, I₃ = 12/6 = 2 A. Total 11 A. R_p = 12/11 ≈ 1.09 Ω.
5. Four 20 Ω resistors: find the equivalent resistance when all are in series and when all are in parallel.
Series: 4 × 20 = 80 Ω. Parallel: 20/4 = 5 Ω. Ratio 16 : 1.
6. A 4 Ω resistor is in series with a parallel pair of 6 Ω and 12 Ω, across 16 V. Find the total current and the current in the 6 Ω resistor.
Parallel pair: 6 × 12 / 18 = 4 Ω. Total R = 4 + 4 = 8 Ω, I = 16/8 = 2 A. p.d. across the pair = 2 × 4 = 8 V, so the 6 Ω current = 8/6 ≈ 1.33 A.
7. How can you join three 6 Ω resistors to get (a) 9 Ω (b) 4 Ω?
(a) Two in parallel (3 Ω) plus one in series: 3 + 6 = 9 Ω. (b) Two in series (12 Ω) in parallel with one: 12 × 6 / 18 = 4 Ω.
Common mistakes
- Adding resistances directly in parallel. Add the reciprocals, then flip the answer.
- Forgetting the last step: after finding 1/R_p = 1/2, R_p is 2 Ω, not 0.5 Ω.
- Saying the current is the same in parallel branches. The voltage is the same; the currents differ.
- Thinking a parallel combination has more resistance than each part. It is always less than the smallest one.