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

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Benjamin Franklin began experimenting with electricity during the 18th century. (i) What is electric current? (ii) Name an instrument used to measure electric curr... show full transcript

Worked Solution & Example Answer:Benjamin Franklin began experimenting with electricity during the 18th century - Leaving Cert Physics - Question 11 - 2022

Step 1

What is electric current?

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Answer

Electric current is defined as the flow of electric charge. It is measured in amperes (A), where one ampere corresponds to one coulomb of charge passing through a point in a circuit per second.

Step 2

Name an instrument used to measure electric current.

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Answer

An ammeter (or galvanometer or multimeter) is used to measure electric current.

Step 3

A torch contains a battery, a light bulb and a switch. Draw a circuit diagram to show how these components are connected in a torch.

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To draw the circuit diagram, you need to represent the battery as a pair of parallel lines (long line for positive, short for negative), the light bulb as a circle with a cross inside, and the switch as a break in the line connecting the battery to the bulb. Connect the components in a series circuit.

Step 4

The wires in a circuit are made of metal. Explain why.

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Wires in a circuit are made of metal because metals are good conductors of electricity. They have free electrons that can move easily through the metal structure, allowing electric current to flow with minimal resistance.

Step 5

Name the subatomic particle that is the charge carrier in a metal.

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

Step 6

A charge of 30 C passes through a wire in a time of 6 s. Calculate the current flowing in the wire.

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Answer

The current (I) can be calculated using the formula:

I=QtI = \frac{Q}{t}

where QQ is the charge and tt is the time. Substituting the values:

I=30C6s=5AI = \frac{30 C}{6 s} = 5 A

Step 7

The wire has a resistance of 3 Ω. Calculate the potential difference (voltage) across the wire.

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The potential difference (V) can be calculated using Ohm's law, which states:

V=I×RV = I \times R

where II is the current and RR is the resistance. Substituting in the known values:

V=5A×3Ω=15VV = 5 A \times 3 Ω = 15 V

Step 8

The 3 Ω wire is connected in parallel with another wire of resistance 2 Ω. Calculate the total resistance of the two wires in parallel.

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The total resistance (RtR_t) of two resistors in parallel can be calculated using:

1Rt=1R1+1R2\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2}

Substituting in the values:

1Rt=13Ω+12Ω\frac{1}{R_t} = \frac{1}{3 Ω} + \frac{1}{2 Ω}

This gives:

1Rt=26+36=56\frac{1}{R_t} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}

Thus, Rt=65=1.2ΩR_t = \frac{6}{5} = 1.2 Ω

Step 9

What is the resistance of a 3 m piece of the same wire?

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To find the resistance of a 3 m piece of wire, we use the formula for resistance, which states that resistance is directly proportional to length:

R=ρLAR = \rho \frac{L}{A}

Given that the resistance of the 1.5 m piece is 12 Ω, the resistance of a 3 m piece (which is double the length) will be:

R=2×12Ω=24ΩR = 2 \times 12 Ω = 24 Ω

Step 10

State the relationship between the resistance of a wire and its cross-sectional area.

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Answer

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. This relationship can be expressed mathematically as:

R1AR \propto \frac{1}{A}

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