Current and circuits

The **rate of flow of charge** — the charge passing a point each second. Unit: ampere (A).

The **energy given to each coulomb of charge** as it passes through a component. Unit: volt (V).

The **coulomb (C)**.

The **ampere (A)** — one ampere is one coulomb of charge per second.

$I = \dfrac{\Delta q}{\Delta t}$ — charge ÷ time. **Given** in the data booklet.

$V = \dfrac{W}{q}$ — energy ÷ charge. **Given** in the data booklet.

**1 joule of energy given to every 1 coulomb of charge** (1 V = 1 J C⁻¹).

$\Delta q = I \times \Delta t$ — current × time.

$W = V \times q$ — voltage × charge.

**Through** it — an ammeter goes in series (in the line).

**Across** it — a voltmeter goes in parallel.

I = Δq/Δt = 0.80 ÷ 5.0 = 0.16 A.

How hard it is to push current through a component: $R = \dfrac{V}{I}$ (voltage across it ÷ current through it). Unit: the **ohm (Ω)**.

The voltage across a component equals the current through it times its resistance: $V = IR$. Given in the data booklet as R = V ÷ I.

The **ohm (Ω)**.

**R = V ÷ I** at a point on the graph. For a straight line through the origin, R is the same at every point.

A **straight line through the origin** — current is proportional to voltage, so R is constant.

A **curve** — R = V ÷ I changes from point to point, so the resistance is not constant.

As the current increases the filament gets **hotter**, and a hotter metal wire has a **higher resistance**, so the I–V graph curves over.

$R = \dfrac{\rho L}{A}$ — resistivity × length ÷ cross-sectional area. Given in the data booklet (as ρ = RA ÷ L).

The **resistivity** of the material (unit Ω m) — a property of the material itself, independent of the wire's shape.

R **doubles** — resistance is proportional to length (R ∝ L).

R **halves** — resistance is inversely proportional to area (R ∝ 1/A).

R = V ÷ I = 12 ÷ 4.0 = 3.0 Ω.

Components joined in **one single loop**, end to end — only **one path** for the charge.

Components joined **side by side** on separate branches — the charge has a **choice of paths**.

The **current** — one loop means one current everywhere.

The **potential difference (voltage)** — every branch sits across the same two points.

They **add**: $R_s = R_1 + R_2 + \ldots$ — **given** in the data booklet. Total is bigger than any one.

Add the reciprocals then flip: $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \ldots$ — **given**. Total is smaller than any one.

**R ÷ 2** (half of one). N equal resistors in parallel give R ÷ N.

It **splits** between the resistors **in proportion to their resistance**; the separate p.d.s add up to the supply.

It **splits** between the branches; the **smaller** resistance carries the **larger** current. The branch currents add up to the total.

**Combine** the resistors into one equivalent R, then use **I = V/R**.

Forgetting to **flip** 1/R_p back to R_p — or just adding the values as if in series.

**Lowers** it — an extra path makes it easier for charge to flow.

The **rate at which electrical energy is transferred** (turned into heat, light, motion). Unit: the **watt (W)** = 1 joule per second.

The **watt (W)**. 1 W = **1 joule of energy every second** (1 J s⁻¹).

$P = IV = I^{2}R = \dfrac{V^{2}}{R}$ — all **given** in the data booklet.

**P = IV** — current × voltage, the simplest form.

**P = I²R** — avoids having to find V first.

**P = V²/R** — avoids having to find I first.

**P = V²/R**, so P ∝ 1/R — **more** resistance means **less** power (e.g. double R → half the power).

**P = I²R**, so P ∝ R — **more** resistance means **more** power.

$E = Pt$ — energy = power × time. **Given** in the data booklet.

The energy a **1 kW** appliance uses in **1 hour** (= 3.6 × 10⁶ J). Energy bills are charged per kWh.

Energy in **kWh** (power in kW × time in hours), then **× the price per kWh**.

It **doubles** — for a uniform wire R ∝ length (L).

The **energy given to each coulomb** of charge by the cell — its 'pushing voltage'. Unit: **volt (V)**.

The **resistance inside the cell itself** (symbol r). The current flows through it, so some energy is lost inside the cell.

The voltage **actually delivered** across the cell's terminals to the circuit: $V = \varepsilon - Ir$.

$\varepsilon = I(R + r)$ — emf = current × (load + internal resistance). **Given** in the data booklet.

$V = \varepsilon - Ir$ — emf minus the lost volts (Ir).

**Ir** — the volts used up inside the cell by its internal resistance. They grow as the current grows.

Because some of the emf is used to push current through the **internal resistance r**, losing Ir volts inside the cell.

Lost volts = ε − V = Ir, so $r = \dfrac{\varepsilon - V}{I}$.

It **drops** — bigger I means bigger lost volts Ir, so less voltage reaches the circuit.

When the internal resistance **r is negligible** (or the current is very small), so Ir ≈ 0.

The simple **ε = IR** — the emf is just current × external resistance.

The **terminal p.d.** V = ε − Ir (the same as IR, the voltage across the load).