Electricity • Topic 2 of 3

Resistance, Resistivity & Combinations

Why do some wires resist current more than others? The resistance of a conductor depends on four things, and understanding them helps us design everything from heater coils to power-transmission cables.

  • Length ($l$): Resistance is directly proportional to length — a longer wire resists more, so $R \propto l$.
  • Area of cross-section ($A$): Resistance is inversely proportional to thickness — a thicker wire offers an easier path, so $R \propto \frac{1}{A}$.
  • Material: Copper and silver conduct far better than nichrome or iron.
  • Temperature: For most metals, resistance increases as temperature rises.

Combining the length and area dependence gives the key relation $R=\rho\frac{l}{A}$, where $\rho$ (rho) is the resistivity of the material. Resistivity is a property of the material itself and does not depend on the wire's shape or size. Its SI unit is the ohm-metre ($\Omega\ \text{m}$). Metals have very low resistivity ($\approx 10^{-8}\ \Omega\ \text{m}$), while insulators like rubber have enormous resistivity. Alloys such as nichrome have higher resistivity than pure metals and do not oxidise easily at high temperature, which is why heating elements are made of them.

Resistors in series: When resistors are joined end to end, the same current flows through each, and the potential difference is shared. The combined resistance simply adds up: $R_s=R_1+R_2+R_3+\dots$ The series resistance is always larger than the biggest individual resistor.

Resistors in parallel: When resistors are connected across the same two points, the same potential difference appears across each, while the current divides between them. The reciprocals add: $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\dots$ The parallel resistance is always smaller than the smallest individual resistor. This is why household appliances are wired in parallel — each gets the full 220 V and can be switched independently.

Series versus parallel resistor networksSERIESR1R2R3Rs = R1 + R2 + R3 (same current)PARALLELR1R2R31/Rp = 1/R1 + 1/R2 + 1/R3
Solved Examples
1
Worked Example
A copper wire of length 2 m and cross-sectional area $4 \times 10^{-6}\ \text{m}^2$ has resistivity $1.6 \times 10^{-8}\ \Omega\ \text{m}$. Find its resistance.
Solution
  1. Step 1: Use $R=\rho\frac{l}{A}$.
  2. Step 2: Substitute: $R=1.6 \times 10^{-8} \times \frac{2}{4 \times 10^{-6}}$.
  3. Step 3: Simplify the fraction: $\frac{2}{4 \times 10^{-6}}=5 \times 10^{5}$.
  4. Step 4: Multiply: $R=1.6 \times 10^{-8} \times 5 \times 10^{5}=8 \times 10^{-3}\ \Omega$.

Answer: $R=8 \times 10^{-3}\ \Omega=0.008\ \Omega$

2
Worked Example
A wire of resistance R is stretched so that its length is doubled (volume unchanged). What is its new resistance?
Solution
  1. Step 1: $R=\rho\frac{l}{A}$. If length doubles to $2l$ and volume is constant, area halves to $\frac{A}{2}$.
  2. Step 2: New resistance $R'=\rho\frac{2l}{A/2}=\rho\frac{4l}{A}$.
  3. Step 3: So $R'=4R$.

Answer: $R'=4R$ (resistance becomes four times)

3
Worked Example
Three resistors of $2\ \Omega$, $3\ \Omega$ and $5\ \Omega$ are connected in series. Find the equivalent resistance.
Solution
  1. Step 1: In series, $R_s=R_1+R_2+R_3$.
  2. Step 2: Substitute: $R_s=2+3+5$.
  3. Step 3: Compute: $R_s=10\ \Omega$.

Answer: $R_s=10\ \Omega$

4
Worked Example
Two resistors of $6\ \Omega$ and $3\ \Omega$ are connected in parallel. Find the equivalent resistance.
Solution
  1. Step 1: Use $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}$.
  2. Step 2: Substitute: $\frac{1}{R_p}=\frac{1}{6}+\frac{1}{3}=\frac{1}{6}+\frac{2}{6}=\frac{3}{6}=\frac{1}{2}$.
  3. Step 3: Invert: $R_p=2\ \Omega$.

Answer: $R_p=2\ \Omega$

5
Worked Example
How can three resistors of $3\ \Omega$ each be combined to give an equivalent resistance of $1\ \Omega$?
Solution
  1. Step 1: Connect all three in parallel.
  2. Step 2: $\frac{1}{R_p}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3}=1$.
  3. Step 3: So $R_p=1\ \Omega$.

Answer: Connect the three $3\ \Omega$ resistors in parallel to get $1\ \Omega$.

6
Worked Example
A $4\ \Omega$ and a $12\ \Omega$ resistor are connected in parallel, and this combination is joined in series with a $5\ \Omega$ resistor. Find the total resistance.
Solution
  1. Step 1: First the parallel part: $\frac{1}{R_p}=\frac{1}{4}+\frac{1}{12}=\frac{3}{12}+\frac{1}{12}=\frac{4}{12}=\frac{1}{3}$, so $R_p=3\ \Omega$.
  2. Step 2: Now add the series resistor: $R=R_p+5=3+5$.
  3. Step 3: Compute: $R=8\ \Omega$.

Answer: $R=8\ \Omega$

Key Points

  • Resistance depends on length, area, material and temperature: $R \propto l$ and $R \propto \frac{1}{A}$.
  • $R=\rho\frac{l}{A}$; resistivity $\rho$ is a material property with SI unit ohm-metre.
  • In series: same current, $R_s=R_1+R_2+\dots$ (larger than any single resistor).
  • In parallel: same voltage, $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\dots$ (smaller than the smallest).
  • Household appliances are wired in parallel so each receives full voltage and works independently.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The resistance of a wire is doubled if you:
Explanation: $R \propto l$, so doubling the length doubles the resistance (area unchanged).
Q2.The SI unit of resistivity is:
Explanation: From $R=\rho\frac{l}{A}$, $\rho$ has units of ohm-metre ($\Omega\ \text{m}$).
Q3.Two $10\ \Omega$ resistors in parallel give:
Explanation: $\frac{1}{R_p}=\frac{1}{10}+\frac{1}{10}=\frac{1}{5}$, so $R_p=5\ \Omega$.
Q4.Which combination gives the largest equivalent resistance from three equal resistors?
Explanation: Series adds resistances, giving the maximum total.
Practice more MCQs → 📝 Assignment · 30 marks