Benjamin Franklin began experimenting with electricity during the 18th century - Leaving Cert Physics - Question 11 - 2022

An ammeter, also known as a galvanometer or multimeter, is used to measure electric current.

In the circuit diagram, the battery is connected to the switch, which is then connected to the light bulb in series. The circuit is completed by connecting the terminals of the battery to the light bulb through the switch, as illustrated below:

 +---------+ 
 |         | 
 |   Bulb  | 
 |         | 
 +---+-----+
     |     
     |     
 +---+     
 |Battery   |
 +---+     
     |     
     |     
 +---+-----+
 |  Switch  | 
 +---------+ 

Wires in a circuit are made of metal because metals, such as copper and aluminum, are good conductors of electricity. They allow free movement of electrons, facilitating the flow of electric current with minimal resistance.

The subatomic particle that serves as the charge carrier in a metal is the electron.

The current ($I$) can be calculated using the formula: $I = \frac{Q}{t}$ where $Q$ is the charge in coulombs and $t$ is the time in seconds.

Substituting the values: $I = \frac{30 \, C}{6 \, s} = 5 \, A$

The potential difference ($V$) across a wire can be calculated using Ohm's Law: $V = I \times R$ where $I$ is the current and $R$ is the resistance.

Given $I = 5 \, A$ and $R = 3 \, \Omega$, we find: $V = 5 \, A \times 3 \, \Omega = 15 \, V$

The total resistance ($R_T$) of two resistors in parallel is given by: $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}$ Substituting the values: $\frac{1}{R_T} = \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}$ Therefore, $R_T = \frac{6}{5} = 1.2 \, \Omega$

The resistance of a wire is directly proportional to its length. If a piece of wire 1.5 m has a resistance of 12 Ω, then the resistance ($R$) of a 3 m piece can be calculated as: $R = \frac{12 \, \Omega}{1.5 \, m} \times 3 \, m = 24 \, \Omega$

The resistance of a wire is inversely proportional to its cross-sectional area. This means that as the cross-sectional area increases, the resistance decreases, and vice versa.