Basic Electricity

Voltage, current, and charge

Voltage (V), or potential difference, measured in volts (V), is the electrical “pressure” that drives charge around a circuit. Formally, one volt is the potential difference across a component when it converts 1 joule of electrical energy into other forms for each coulomb of charge passing through it:

$V = \frac{E}{Q}$

where E is energy in joules (J) and Q is charge in coulombs (C).

Current (I), measured in amperes (A), is the rate of flow of electric charge. One ampere equals one coulomb per second:

$I = \frac{Q}{t}$

In metallic conductors, current is carried by free electrons drifting through the metal lattice. The electron drift speed is remarkably slow (less than 1 mm/s), but all electrons begin drifting simultaneously when a voltage is applied, so current appears instantaneously.

Charge (Q), measured in coulombs (C). One electron carries a charge of approximately -1.6 × 10^⁻19 C. So one coulomb is about 6.25 × 10^18 electrons. The sign convention matters: electrons are negative charge carriers, but we use conventional current which flows from the positive terminal to the negative terminal of a supply. This is opposite to the physical direction of electron flow. The convention was established before electrons were discovered, when current was imagined as positive charges flowing out of the positive battery terminal. For circuit analysis, conventional current is used throughout and the distinction rarely matters in practice: stick with conventional current (positive to negative).

Energy: The electromotive force (emf) of a battery is 1 volt if it supplies 1 joule of energy to each coulomb of charge passing through it. Similarly, a component has a pd of 1 volt across it if it converts 1 joule to heat/light/other forms per coulomb.

Ohm’s Law

Resistance (R), measured in ohms (Ω), is the opposition to current flow. For an ohmic (linear) conductor, the voltage across it is proportional to the current through it:

$V = IR$

This is Ohm’s Law. Rearranged forms:

$I = \frac{V}{R}, \quad R = \frac{V}{I}$

A component with a specific, known resistance is called a resistor. Resistors are made from materials like carbon or metal film and are available in standard values.

Worked example: A 5V supply is connected across a 100 Ω resistor. What current flows?

$I = \frac{V}{R} = \frac{5}{100} = 0.05\text{ A} = 50\text{ mA}$

(1 mA = 10^⁻3 A; 1 kΩ = 10^3 Ω.)

Power

Power (P) is the rate of energy conversion, measured in watts (W). One watt is one joule per second. For a resistor:

$P = IV$

Substituting Ohm’s Law gives two alternative forms:

$P = I^2 R \quad\text{and}\quad P = \frac{V^2}{R}$

Worked example: For the 100 Ω resistor with 5V across it, what power is dissipated?

$P = IV = 0.05 \times 5 = 0.25\text{ W}$

Or using P = V²/R: P = 5²/100 = 25/100 = 0.25 W.

Or using P = I²R: P = (0.05)² × 100 = 0.0025 × 100 = 0.25 W.

Series and parallel resistance

Series resistance

When resistors are connected end-to-end (in series), the same current flows through each. The total resistance is the sum:

$R_{\text{series}} = R_1 + R_2 + R_3 + \ldots$

Proof: The total voltage across the chain is V = V₁ + V₂ + V₃ = IR₁ + IR₂ + IR₃ = I(R₁ + R₂ + R₃), so the equivalent single resistance is R₁ + R₂ + R₃.

Parallel resistance

When resistors are connected across the same two nodes (in parallel), they share the same voltage. The total (equivalent) resistance is given by:

$\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots$

Proof: The total current splits, so I = I₁ + I₂ + I₃ = V/R₁ + V/R₂ + V/R₃ = V(1/R₁ + 1/R₂ + 1/R₃), giving 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃.

For two resistors in parallel, a useful shortcut:

$R_{\text{parallel}} = \frac{R_1 R_2}{R_1 + R_2}$

Worked example: 3 kΩ and 6 kΩ in parallel:

$R_{\text{parallel}} = \frac{3000 \times 6000}{3000 + 6000} = \frac{18 \times 10^6}{9000} = 2000\;\Omega = 2\text{ k}\Omega$

Relevance to digital circuits

Gates consume power: each switching event dissipates energy in the transistors and interconnect. Wires and PCB tracks have non-zero resistance, causing voltage drops (IR drops) across long traces. Inputs draw small currents, and outputs must source or sink sufficient current to drive connected loads (fan-out). For the Tripos, you need to be able to analyse simple resistor networks using Ohm’s Law and to calculate power in digital circuits, particularly in the n-MOS inverter stage (see /modules/digital-electronics/12-transistors-and-cmos-logic/02-nmos-inverter/).