Current Electricity - 13

Coulombs (C)

It is the rate of flow of charge SI Unit is Amperes (A)

Circuit must be closed and there must be a potential between the 2 ends

$I \space (A) = \frac{Q \space (C)}{t \space (s)}$

Use an ammeter connected in series to the circuit

Positive to negative vs negative to positive respectively

The electromotive force (e.m.f) is defined as the work done in driving charges around a complete circuit.

Volt (V) or Joule per coloumb$(JC^{-1})$

$\varepsilon \space (V) = \frac{W \space (J)} {Q \space (C)}$

The potential difference (p.d.) across a component in an electric circuit is the work done to drive a unit charge through the component. SI Unit: Volts (V)

Connect a voltmeter in parallel with the electrical source (e.g. battery) and the positive & negative terminals of the voltmeter should be connected to the same terminals of the battery.

A voltmeter is connected in parallel with the component

Independent of the resistance of the circuit vs proportional to the resistance between the 2 points

The resistance of a component is the ratio of the potential difference across the component to the current flowing through the component. SI unit is Ohms $(\Omega)$

$R \space (\Omega) = \frac{V}{I}$

$Re = R_1 + R_2 + R_3$

$Re = \left (\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right)^{-1}$

It states that the current passing through a metallic conductor is directly proportional to the potential difference across it, provided that physical conditions (e.g. temperature) remain constant.

It is a property of a material and is independent of the dimensions of the material SI Unit ohm metre ($\Omega \space m$)

$R_{wire} = \frac{pL}{A}$ where p is resistivity of wire, L is length of wire in metres, A is cross sectional-area of wire in$m^2$ & R is resistance of wire

Longer length = higher resistance, thicker wire = lower resistance, higher resistivity = higher resistance