Current and Circuits

Electric current is the rate at which charge flows past a point in a circuit, given by:

$$I = \frac{\Delta q}{\Delta t}$$

Current is measured in amperes, where $1\ \text{A} = 1\ \text{C s}^{-1}$. In metals the moving charges are free (delocalised) electrons; in electrolytes both positive and negative ions can move.

Conventional current is drawn from the positive terminal of a cell, around the circuit, to the negative terminal, even though electrons drift the opposite way. A current of $0.50\ \text{A}$ means $0.50\ \text{C}$ of charge, about $3.1 \times 10^{18}$ electrons, passes each second. Cells and batteries drive direct current (dc), in which the charge flow is in one direction only.

The microscopic model: at the microscopic scale current is quantified by:

$$I = nAvq$$

• $n$ is the charge-carrier density of the conductor
• $A$ is its cross-sectional area
• $v$ is the drift speed
• $q$ is the charge on each carrier

Conduction electrons are accelerated by the electric field set up by the potential difference, but they constantly collide with the positive ions of the crystal lattice, so their net migration is a slow drift. These collisions transfer kinetic energy to the lattice ions, which vibrate with greater amplitude; macroscopically the conductor warms up and its resistivity increases.

You should be able to define current, relate it to electron flow, and use $I = nAvq$ to compare drift speeds in different conductors.

Potential difference (pd) between two points is the energy transferred per unit charge as charge moves between them:

$$V = \frac{W}{q}$$

Here $1\ \text{V} = 1\ \text{J C}^{-1}$. Across a lamp, a pd of $6.0\ \text{V}$ means each coulomb of charge transfers $6.0\ \text{J}$ of electrical energy to light and heat.

Electromotive force (emf, $\varepsilon$) describes the source instead: it is the total energy given to each coulomb by the cell as chemical (or other) energy is converted into electrical energy.

Real cells also have internal resistance $r$, so some energy is dissipated inside the cell itself. When current flows the terminal pd is less than the emf, and the loop equation applies:

$$\varepsilon = I(R + r)$$

This is why a battery's terminal voltage sags when it supplies a large current, for example a car battery while the starter motor runs.

You must define emf and pd in terms of energy per unit charge, distinguish them clearly, and use $\varepsilon = I(R + r)$ in circuit calculations.

Resistance measures how strongly a component opposes current, defined in ohms ($\Omega$) by:

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

For a uniform conductor, resistance grows with length and falls with cross-sectional area:

$$R = \frac{\rho L}{A}$$

Here the resistivity $\rho$ (units $\Omega\ \text{m}$) depends only on the material and its temperature. Doubling the length doubles $R$; doubling the area halves it.

Copper's resistivity (about $1.7 \times 10^{-8}\ \Omega\ \text{m}$) is roughly sixty times smaller than nichrome's, which is why copper is used for connecting wires and nichrome for heating elements.

Ohm's law states that, at constant temperature, the current through a metallic conductor is directly proportional to the pd across it — in other words, $R$ stays constant.

You should be able to:

• rearrange $R = \frac{\rho L}{A}$ for any variable
• convert areas given in $\text{mm}^2$ to $\text{m}^2$ (multiply by $10^{-6}$)
• describe an experiment in which $R$ is plotted against $L$ for a wire so that the gradient equals $\frac{\rho}{A}$

An $I$–$V$ characteristic shows how the current through a component varies with the pd across it. An ohmic conductor, such as a metal wire held at constant temperature, gives a straight line through the origin: its resistance is constant.

A filament lamp is non-ohmic: as the pd rises, the filament gets hotter, the lattice ions vibrate more vigorously, and the resistivity increases, so the $I$–$V$ curve bends over with a decreasing gradient. The curve is symmetric through the origin because reversing the pd simply reverses the current.

A diode is also non-ohmic: it conducts only when forward-biased above a small threshold pd and passes almost no current in reverse.

Two skills matter here: sketching and interpreting these characteristic curves, and remembering that for any device the resistance at a chosen point is found from that point's values:

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

It is not the gradient of the curve, except for an ohmic conductor where the two happen to coincide.

Components transfer electrical energy at a rate given by the power:

$$P = VI$$

Each coulomb carries $V$ joules, and $I$ coulombs arrive per second. Combining this with $R = \frac{V}{I}$ gives two equivalent forms, $P = I^2 R$ and $P = \frac{V^2}{R}$.

Choose the form that matches what is common in the circuit: in series every component carries the same current, so $P = I^2 R$ compares them directly; in parallel every branch has the same pd, so $P = \frac{V^2}{R}$ is quickest.

The energy transferred in time $t$ is $E = Pt$. For example, a $1.2\ \text{kW}$ kettle on a $240\ \text{V}$ supply draws $I = \frac{1200}{240} = 5.0\ \text{A}$ and transfers $1.2 \times 10^{5}\ \text{J}$ in $100\ \text{s}$.

You should be able to calculate the power dissipated in an individual resistor within a network, and the total energy delivered by a source over a stated time.

In a series connection:

• the same current passes through every component
• the pds across the components add up to the supply pd
• the resistances add:

$$R_s = R_1 + R_2 + \dots$$

In parallel, each branch has the same pd and the branch currents add at the junctions (conservation of charge). The combined resistance obeys:

$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$$

It is always smaller than the smallest branch, e.g. two $10\ \Omega$ resistors in parallel give $5.0\ \Omega$.

Mixed networks are reduced step by step: collapse each parallel group into one equivalent resistor, then add along the series chain.

Two resistors in series also act as a potential divider, splitting the supply pd in the ratio of their resistances; this is how variable resistors (potentiometers) provide an adjustable output voltage, as in a volume control.

You must be able to reduce a network to a single resistance, track the current through and pd across each element using conservation of charge and energy, and check answers against the rule that adding parallel branches always lowers the total resistance.