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A hair dryer with a plastic casing uses a coiled wire as a heat source - Leaving Cert Physics - Question 8 - 2010

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A hair dryer with a plastic casing uses a coiled wire as a heat source. When an electric current flows through the coiled wire, the air around it heats up and a mot... show full transcript

Worked Solution & Example Answer:A hair dryer with a plastic casing uses a coiled wire as a heat source - Leaving Cert Physics - Question 8 - 2010

Step 1

What is an electric current?

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Answer

An electric current is the flow of electric charge, typically carried by moving electrons in a conductor.

Step 2

Give two other effects of an electric current.

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Answer

  1. Magnetic effect: An electric current generates a magnetic field around the conductor.
  2. Chemical effect: An electric current can cause chemical reactions, such as electrolysis.

Step 3

(i) Describe what happens: (a) when switch A is closed and the rheostat is adjusted

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Answer

When switch A is closed, the circuit is completed, allowing current to flow. Adjusting the rheostat changes the resistance in the circuit, which in turn modifies the speed of the fan by controlling the amount of current flowing to the motor.

Step 4

(i) Describe what happens: (b) when switch A and switch B are closed.

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Answer

When both switches A and B are closed, current flows through the heating coil, causing it to generate heat. At the same time, the fan operates, blowing the heated air out, which aids in drying.

Step 5

(ii) (a) What is the initial resistance of the coil?

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Answer

To find the initial resistance of the coil, we use the formula for power:
P=I2RP = I^2 R
Here, we rearrange it to find R:
R=PI2R = \frac{P}{I^2}
Assuming a maximum power (P) of 2000 W at 230 V:
The current (I) can be derived from Ohm’s law:
I=PV=20002308.7extAI = \frac{P}{V} = \frac{2000}{230} \approx 8.7 ext{ A}
Then, substituting the values:
R=2000(8.7)226.4ΩR = \frac{2000}{(8.7)^2} \approx 26.4 \Omega

Step 6

(ii) (b) Calculate the current that flows through the coil when the dryer is turned on.

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Answer

Using the formula for current:
I=PV=2000extW230extV8.7extAI = \frac{P}{V} = \frac{2000 ext{ W}}{230 ext{ V}} \approx 8.7 ext{ A}

Step 7

(iii) Calculate the length of the coil of wire.

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Answer

To calculate the length of the wire, use the formula for resistance:
R=ρLAR = \rho \frac{L}{A}
where:

  • R=26.4ΩR = 26.4 \Omega
  • ρ=1.1×106Ωm\rho = 1.1 \times 10^{-6} \Omega m (resistivity of nichrome)
  • A=πr2A = \pi r^2, with diameter 0.17 mm so radius r = 0.085 mm = 0.085 \times 10^{-3} m.
    Calculating the cross-sectional area:
    A=π(0.085×103)22.27×109m2A = \pi (0.085 \times 10^{-3})^2 \approx 2.27 \times 10^{-9} m^2
    Thus, substituting into the resistance formula:
    26.4=1.1×106L2.27×10926.4 = 1.1 \times 10^{-6} \frac{L}{2.27 \times 10^{-9}}
    Solving for length L gives:
    L0.5448mL \approx 0.5448 m or 54.48 cm.

Step 8

(iv) Explain why the current through the coil would decrease if the fan developed a fault and stopped working.

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Answer

If the fan stops working, the airflow that helps dissipate the heat generated by the coil diminishes. Consequently, the coil's temperature rises, increasing its resistance due to the temperature coefficient of resistivity.
Higher resistance leads to a reduction in current, according to Ohm’s law, as the voltage remains constant.

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