This unit covers the relationships between voltage, current and resistance, introduces the material property resistivity, and shows how to analyse simple DC circuits using series‑parallel rules, Kirchhoff’s laws and the voltage‑divider concept. All formulas are presented with SI units the first time they appear.
| Symbol | Name | Definition | SI Unit |
|---|---|---|---|
| \(V\) | Potential difference (voltage) | Energy per unit charge between two points | V (volt) |
| \(I\) | Current | Rate of charge flow | A (ampere) |
| \(R\) | Resistance | Opposition to current in a component | Ω (ohm) |
| \(\rho\) | Resistivity | Intrinsic property of a material; independent of shape | Ω·m (ohm‑metre) |
| \(L\) | Length of a conductor | Physical length of the material | m (metre) |
| \(A\) | Cross‑sectional area | Area through which current flows | m² (square metre) |
| \(\alpha\) | Temperature coefficient of resistivity | Fractional change in \(\rho\) per kelvin for a metal | K⁻¹ |
| \(r\) | Internal resistance of a source | Resistance “seen” inside a battery or cell | Ω |
| \(\mathcal{E}\) | Electromotive force (emf) | Ideal terminal voltage of a source when no current flows | V |
\[
V = I R
\]
\[
I = \frac{V}{R},\qquad R = \frac{V}{I}
\]
\[
R = \rho \frac{L}{A}
\]
where \(L\) is the length and \(A\) the cross‑sectional area.
\[
\rho(T) = \rho{0}\bigl[1+\alpha\,(T-T{0})\bigr]
\]
\[
V = \mathcal{E} - I r
\]
\[
P = IV = I^{2}R = \frac{V^{2}}{R}
\]
| Configuration | Equivalent Resistance |
|---|---|
| Series: \(R{1},R{2},\dots,R_{n}\) | \(R{\text{eq}} = R{1}+R{2}+ \dots + R{n}\) |
| Parallel: \(R{1},R{2},\dots,R_{n}\) | \(\displaystyle\frac{1}{R{\text{eq}}}= \frac{1}{R{1}}+ \frac{1}{R{2}}+ \dots + \frac{1}{R{n}}\) |
Kirchhoff’s Current Law (KCL)
The algebraic sum of currents meeting at a junction is zero:
\[
\sum I{\text{in}} = \sum I{\text{out}}
\]
Kirchhoff’s Voltage Law (KVL)
The algebraic sum of potential differences round any closed loop is zero:
\[
\sum V = 0
\]
Two resistors \(R{1}\) and \(R{2}\) in series across a supply voltage \(V{s}\) give a voltage across \(R{2}\) of
\[
V{R{2}} = V{s}\,\frac{R{2}}{R{1}+R{2}}
\]
The same expression applies for the voltage across \(R_{1}\) by swapping the indices.
A copper wire is 2.0 m long with a cross‑sectional area of \(0.5\;\text{mm}^{2}\). Find its resistance.
\[
R = (1.68\times10^{-8})\frac{2.0}{0.5\times10^{-6}}
= 6.72\times10^{-2}\;\Omega.
\]
A 12 V battery supplies three resistors: \(R{1}=4\;\Omega\) (series) and a parallel branch containing \(R{2}=6\;\Omega\) and \(R_{3}=12\;\Omega\). Find the total current.
\[
\frac{1}{R{23}} = \frac{1}{6} + \frac{1}{12}= \frac{3}{12}\;\Rightarrow\;R{23}=4\;\Omega.
\]
\[
R{\text{tot}} = R{1}+R_{23}=4+4=8\;\Omega.
\]
\[
I = \frac{V{s}}{R{\text{tot}}}= \frac{12}{8}=1.5\;\text{A}.
\]
A nichrome wire has \(\rho_{0}=1.10\times10^{-6}\;\Omega\!\cdot\!\text{m}\) at \(20^{\circ}\text{C}\) and \(\alpha =1.7\times10^{-4}\;\text{K}^{-1}\). Find \(\rho\) at \(100^{\circ}\text{C}\).
\[
\rho(100^{\circ}\text{C}) = 1.10\times10^{-6}\bigl[1+1.7\times10^{-4}(100-20)\bigr]
=1.10\times10^{-6}\bigl[1+0.0136\bigr]
\approx 1.115\times10^{-6}\;\Omega\!\cdot\!\text{m}.
\]
A 9 V battery has \(\mathcal{E}=9\;\text{V}\) and internal resistance \(r=0.5\;\Omega\). It supplies two series resistors \(R{1}=2\;\Omega\) and \(R{2}=3\;\Omega\).
Reduce a 15 V supply to 5 V using two resistors in series. Find the required ratio \(R{2}:R{1}\).
\[
5 = 15\;\frac{R{2}}{R{1}+R{2}}\;\Longrightarrow\;\frac{R{2}}{R{1}+R{2}}=\frac{1}{3}
\]
\[
\Rightarrow\;R{2}= \frac{1}{2}R{1}\quad\text{or}\quad R{1}=2R{2}.
\]
Any pair satisfying this ratio (e.g., \(R{1}=20\;\Omega,\;R{2}=10\;\Omega\)) will give the desired 5 V output.
| Material | Resistivity \(\rho\) (Ω·m) | Typical Use |
|---|---|---|
| Copper | 1.68 × 10⁻⁸ | Electrical wiring |
| Aluminium | 2.82 × 10⁻⁸ | Power transmission lines |
| Silver | 1.59 × 10⁻⁸ | High‑frequency contacts |
| Constantan (Ni‑Cu alloy) | 4.9 × 10⁻⁷ | Thermocouples |
| Glass (dry) | ≈ 10¹⁴ | Insulators |
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