An electric current flows in a conductor when there is a potential difference between its ends - Leaving Cert Physics - Question 9 - 2008

1. Heating Effect: When an electric current passes through a conductor, it generates heat due to its resistance. This effect is utilized in devices like electric stoves and heaters.

2. Magnetic Effect: An electric current flowing through a wire generates a magnetic field around it. This principle is applied in electromagnets and electric motors.

A battery is a common source of potential difference, providing the energy needed to drive an electric current in a circuit.

To determine whether a substance is a conductor or an insulator, the following experiment can be set up:

Apparatus: A power source (battery), a bulb, leads, and the test substance.

Procedure: Connect the circuit using the battery, bulb, and the test substance in series. If the bulb lights up, the substance is a conductor; if not, it is an insulator. To further confirm, repeat the experiment with a known conductor (such as copper) and a known insulator (such as rubber) for comparison.

The circuit diagram should illustrate the two headlights connected in parallel to a 24 V power supply, with each headlight connected by leads that illustrate how they branch from the main circuit.

Connecting the headlights in parallel ensures that each headlight receives the full voltage of the power supply. This means that if one headlight fails, the other will continue to operate, enhancing reliability.

A fuse is essential in this circuit for safety reasons. It prevents overheating or excessive current flow which could cause the electrical system to fail or create a fire hazard.

To calculate the total resistance for two headlight resistances in parallel, use the formula:

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ Substituting the values:

$\frac{1}{R} = \frac{1}{20 \Omega} + \frac{1}{20 \Omega}$

This simplifies to:

$\frac{1}{R} = \frac{2}{20 \Omega} \\ R = 10 \Omega$

Using Ohm's Law, we can determine the current flowing in the circuit:

$I = \frac{V}{R}$ Substituting the values:

$I = \frac{24 V}{10 \Omega} = 2.4 A$ Thus, the total current in the circuit is 2.4 A.