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 electric current flows through a conductor, it produces heat due to resistance, which can be utilized in various applications, such as in heating elements.

2. Magnetic Effect: An electric current generates a magnetic field around the conductor, which is the principle behind electromagnets and electric motors.

A common source of potential difference is a battery. Other sources include generators and thermocouples.

1. Apparatus: Use a basic circuit setup that includes a power source (like a battery), an ammeter, leads, and a bulb.

2. Procedure: Connect the circuit using the substance in question. Place the substance in series with the circuit.

3. Observation: If the bulb lights up, the substance is a conductor; if it does not light, it is an insulator.

4. Conclusion: The experiment allows you to determine the conductive properties of the substance by observing the flow of current.

The circuit diagram would show the power supply connected to two headlamps in parallel. Each headlamp would be connected to the supply independently, with lines representing the connections.

The advantage of connecting the headlamps in parallel is that if one lamp fails, the other will still work. This ensures continued operation since each headlamp receives the same voltage directly from the power supply.

A fuse is essential for safety; it prevents overheating or excessive current flow that could damage the electrical components. In case of a short circuit, the fuse will blow and disconnect the circuit, protecting the system.

For headlamps connected in parallel, the total resistance (R_total) can be calculated using the formula:

$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}$

Where each headlamp has a resistance of 20 Ω:

$\frac{1}{R_{total}} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20}$
$R_{total} = \frac{20}{2} = 10 \ \Omega$

Using Ohm's law where V = IR:

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

$I = \frac{24 \ V}{10 \ \Omega} = 2.4 \ A$

Therefore, the total current flowing in the circuit is 2.4 A.