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As electrons move through a metal, the metal resists this movement - Leaving Cert Physics - Question 9 - 2021

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As electrons move through a metal, the metal resists this movement. The electrons collide with atoms of the metal and lose kinetic energy. This lost energy is conver... show full transcript

Worked Solution & Example Answer:As electrons move through a metal, the metal resists this movement - Leaving Cert Physics - Question 9 - 2021

Step 1

Define resistance.

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Answer

Resistance is defined as the measure of the opposition to the flow of electric current in a conductor. It quantifies how much a material resists the movement of charges and is calculated using the formula:

R=VIR = \frac{V}{I}

where RR is resistance in ohms (Ω), VV is voltage in volts (V), and II is current in amperes (A).

Step 2

Name the instrument used to measure resistance.

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Answer

The instrument used to measure resistance is called an ohmmeter.

Step 3

State the relationship between resistance and length.

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Answer

The relationship between resistance and length can be stated as follows:

Resistance is directly proportional to the length of the conductor. Thus, if the length doubles, the resistance also doubles. Mathematically, this can be expressed as:

RLR \propto L

where RR is resistance and LL is length.

Step 4

Calculate the circular cross-sectional area of the wire.

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Answer

To calculate the circular cross-sectional area (AA) of the wire, we use the formula:

A=πr2A = \pi r^2

Here, the radius rr is given as 0.2 mm, which converts to 0.2 × 10⁻³ m. Therefore:

A=π(0.2×103)2=1.26×107 m2A = \pi (0.2 \times 10^{-3})^2 = 1.26 \times 10^{-7} \ m^2

Step 5

Calculate the resistance of the wire.

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Answer

The resistance of the wire can be calculated using the formula:

R=ρLAR = \frac{\rho L}{A}

Where:

  • ρ=1.1×106 Ωm\rho = 1.1 \times 10^{-6} \ \Omega m (resistivity of nichrome)
  • L=4.8 mL = 4.8 \ m
  • A=1.26×107 m2A = 1.26 \times 10^{-7} \ m^2 (cross-sectional area)

Substituting the values:

R=1.1×106×4.81.26×1074.20 ΩR = \frac{1.1 \times 10^{-6} \times 4.8}{1.26 \times 10^{-7}} \approx 4.20 \ \Omega

Step 6

Show that the combined resistance of the three resistors in parallel (i.e. resistors R₁, R₂ and R₃) is 1.05 Ω.

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Answer

For resistors in parallel, the total resistance (RtR_t) is given by:

1Rt=1R1+1R2+1R3\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

Given:

  • R1=5 ΩR_1 = 5 \ \Omega
  • R2=3 ΩR_2 = 3 \ \Omega
  • R3=4 ΩR_3 = 4 \ \Omega

Calculating:

1Rt=15+13+14=0.20+0.33+0.25=0.78\frac{1}{R_t} = \frac{1}{5} + \frac{1}{3} + \frac{1}{4} = 0.20 + 0.33 + 0.25 = 0.78

Then:

Rt=10.781.05 ΩR_t = \frac{1}{0.78} \approx 1.05 \ \Omega

Step 7

Calculate the total resistance in the circuit.

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Answer

After calculating the total resistance for resistors in parallel as 1.05 Ω, if any additional resistances are in series, they must be added. Assuming, for example, that a resistor of 3 Ω is in series, the total resistance (RtotalR_{total}) will be:

Rtotal=Rt+Rseries1.05+3.00=4.05ΩR_{total} = R_t + R_{series} \approx 1.05 + 3.00 = 4.05 \Omega

Step 8

Calculate the current flowing through the ammeter, A.

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Answer

To find the current (II) flowing through the ammeter, we can use Ohm's law:

I=VRI = \frac{V}{R}

Where:

  • V=12 VV = 12 \ V (voltage of the circuit)
  • R4.05 ΩR \approx 4.05 \ \Omega (total resistance)

Thus:

I=124.052.96 AI = \frac{12}{4.05} \approx 2.96 \ A

Step 9

What is the function of the wire labelled P?

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Answer

The wire labelled P is yellow and green, which indicates it is the earth wire. Its function is to provide a safe path for excess current to flow into the ground, thereby preventing electrical shock and ensuring safety.

Step 10

Name the wire labelled Q.

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Answer

The wire labelled Q is typically referred to as the neutral wire.

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