11 min readConcept sheetSection A MCQ (1 mark) and true-or-false parts; Section B short answer (3 marks) numerical on equivalent resistance, or long answer (5 marks) with the derivation; also a NEET favourite

Resistors in series and in parallel

In series the same current flows: Rs = R₁ + R₂ + R₃. In parallel the same potential difference acts: 1/Rp = 1/R₁ + 1/R₂ + 1/R₃.

SeriesRs = R₁ + R₂ + R₃; same current through each resistor

Parallel1/Rp = 1/R₁ + 1/R₂ + 1/R₃; same potential difference across each

Two in parallelRp = R₁R₂/(R₁ + R₂)

Key resultIn parallel, the total is smaller than the smallest resistor

Series vs parallel

SeriesResistors are joined end to end, one after another. There is only one path, so the same current flows through each. The potential differences add up. If one part breaks, the whole circuit stops.

ParallelResistors are joined between the same two points. Each has the same potential difference. The currents in the branches add up to the main current. If one branch breaks, the others keep working.

Quick check: n equal resistors R give nR in series and R/n in parallel.

Derivation: resistors in series

1
Draw R₁, R₂ and R₃ joined end to end between points P and Q, with a cell, a key and an ammeter in the same loop.
2
There is one path, so the same current i flows through all three resistors.
3
By Ohm's law, the p.d. across each is V₁ = iR₁, V₂ = iR₂ and V₃ = iR₃.
4
The total p.d. between P and Q is the sum: V = V₁ + V₂ + V₃ = i(R₁ + R₂ + R₃).
5
If Rs is the equivalent resistance, then V = iRs.
6★
Comparing the two: Rs = R₁ + R₂ + R₃. So the equivalent resistance in series is the sum of the separate resistances.

★ key step

Derivation: resistors in parallel

1
Draw R₁, R₂ and R₃ as three separate branches between points P and Q. Join P and Q to a cell, a key and an ammeter in the main line.
2
Each resistor is joined across P and Q, so each has the same p.d. V.
3
The main current divides: i = i₁ + i₂ + i₃.
4
By Ohm's law, i₁ = V/R₁, i₂ = V/R₂ and i₃ = V/R₃. If Rp is the equivalent resistance, i = V/Rp.
5
Substitute: V/Rp = V/R₁ + V/R₂ + V/R₃. Divide every term by V.
6★
So 1/Rp = 1/R₁ + 1/R₂ + 1/R₃. The reciprocal of the equivalent resistance is the sum of the reciprocals.

★ key step

Why homes are wired in parallel

Each appliance gets the full supply voltage and takes the current it needs.
Each appliance has its own switch, so the fan can run while the light is off.
If one appliance fails, the others keep working.
Adding appliances in parallel lowers the total resistance, so the main current rises. Too many on one socket can overload it.
The small bulbs of some Deepawali light strings are in series, so one fused bulb can darken the whole string.
A fuse is joined in series with the circuit it protects, so all the current passes through it.

Series and parallel

Problem

(a) Resistors of 2 Ω, 3 Ω and 5 Ω are joined in series to a 12 V battery. Find the total resistance, the current and the p.d. across each. (b) Resistors of 3 Ω and 6 Ω are joined in parallel to a 6 V battery. Find the total resistance, the main current and the current in each.

Working

(a) Rs = 2 Ω + 3 Ω + 5 Ω = 10 Ω.
i = V/Rs = 12 V ÷ 10 Ω = 1.2 A, the same in each resistor.
V₁ = 1.2 A × 2 Ω = 2.4 V; V₂ = 1.2 A × 3 Ω = 3.6 V; V₃ = 1.2 A × 5 Ω = 6.0 V. Check: 2.4 + 3.6 + 6.0 = 12 V.
(b) 1/Rp = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2, so Rp = 2 Ω.
Main current i = V/Rp = 6 V ÷ 2 Ω = 3 A.
i₁ = 6 V ÷ 3 Ω = 2 A; i₂ = 6 V ÷ 6 Ω = 1 A. Check: 2 A + 1 A = 3 A.

Answer

(a) 10 Ω; 1.2 A; 2.4 V, 3.6 V and 6.0 V. (b) 2 Ω; 3 A; 2 A in the 3 Ω and 1 A in the 6 Ω.

A mixed circuit

Problem

A 4 Ω resistor is joined in series with a parallel pair of 6 Ω and 12 Ω, across a 12 V battery. Find the total resistance, the main current, and the current in the 6 Ω resistor.

Working

Parallel pair: Rp = (6 × 12)/(6 + 12) = 72/18 = 4 Ω.
Total: R = 4 Ω + 4 Ω = 8 Ω.
Main current i = V/R = 12 V ÷ 8 Ω = 1.5 A.
P.d. across the 4 Ω resistor = 1.5 A × 4 Ω = 6 V, so the parallel pair gets 12 V − 6 V = 6 V.
Current in the 6 Ω resistor = 6 V ÷ 6 Ω = 1 A (and 0.5 A in the 12 Ω). Check: 1 A + 0.5 A = 1.5 A.

Answer

Total resistance 8 Ω; main current 1.5 A; 1 A in the 6 Ω resistor

In real life

In your home the fan, tube light, fridge and TV are all in parallel, so switching off the TV does not stop the fridge. A multimeter used by an electrician reads the resistance of such a combination directly.

Exam angle

2 marks: “Why are household appliances not joined in series? Give two reasons.” In series the supply voltage is shared, so no appliance gets its full voltage. If one appliance fails or is switched off, all the others stop too.

3 marks: “You have three resistors of 4 Ω each. How would you join them to get (a) 12 Ω, (b) 6 Ω, (c) 4/3 Ω?” (a) All three in series: 4 + 4 + 4 = 12 Ω. (b) Two in parallel (2 Ω) with the third in series: 2 + 4 = 6 Ω. (c) All three in parallel: 1/Rp = 3/4, so Rp = 4/3 Ω.

5 marks: “Derive the formula for the equivalent resistance of three resistors joined in parallel.” Same p.d. V across each; i = i₁ + i₂ + i₃; i₁ = V/R₁ and so on; i = V/Rp; so V/Rp = V/R₁ + V/R₂ + V/R₃, giving 1/Rp = 1/R₁ + 1/R₂ + 1/R₃. Draw the circuit.

Remember it

Series: Same current, Sum the resistances. Parallel: same Potential, add the reciprocals, and the answer is smaller than the smallest.