Figure 7 shows a circuit designed by a student to monitor temperature changes. The supply has negligible internal resistance and the thermistor has a resistance of 750 Ω at room temperature. The student wants the output potential difference (pd) at room temperature to be 5.0 V. The 0.25 kΩ resistor is made of 50 turns of wire that is wound around a non-conducting cylinder of diameter 8.0 mm. Resistivity of the wire = 4.2 × 10⁻⁷ Ω m 1. Determine the area of cross-section of the wire that has been used for the resistor. 2. The student selects a resistor rated at 0.36 W for the 0.25 kΩ resistor in Figure 7. Determine whether this resistor is suitable. 3. Determine the value of R that the student should select. 4. Give your answer to an appropriate number of significant figures. 5. State and explain the effect on the output pd of increasing the temperature of the thermistor.

A-Level AQA Physics Circuits & The Potential Divider: Figure 7 shows a circuit designe

Figure 7 shows a circuit designed by a student to monitor temperature changes - AQA - A-Level Physics - Question 4 - 2018 - Paper 1

Question 4

Figure 7 shows a circuit designed by a student to monitor temperature changes. The supply has negligible internal resistance and the thermistor has a resistance of ... show full transcript

Worked Solution & Example Answer:Figure 7 shows a circuit designed by a student to monitor temperature changes - AQA - A-Level Physics - Question 4 - 2018 - Paper 1

Step 1

Determine the area of cross-section of the wire that has been used for the resistor.

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Answer

To find the area of cross-section (A) of the wire, we can use the resistivity formula:

$A = \frac{\rho L}{R}$

Given:

Resistivity ( $\rho$ ) = 4.2 × 10⁻⁷ Ω m
Resistance (R) = 0.25 kΩ = 250 Ω
Length (L) = 50 turns × circumference of cylinder

To find the circumference: Circumference = π × diameter = π × 0.008 m ≈ 0.0251 m

Therefore, the total length of wire: $L = 50 × 0.0251 m = 1.255 m$

Substituting into the area formula:

$A = \frac{(4.2 × 10^{-7}) (1.255)}{250} ≈ 2.1 × 10^{-9} m²$

Step 2

Determine whether this resistor is suitable.

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Answer

To check if the selected resistor is suitable, we need to calculate the maximum power dissipation at its rated resistance.

Using the power formula: $P = \frac{V^2}{R}$

Substituting known values, where V is the potential difference across the 0.25 kΩ resistor: $V = 9.0 V - 5.0 V = 4.0 V$

Now substituting into the power formula: $P = \frac{(4.0)^2}{250} = 0.064 W$

Since this is less than 0.36 W, the resistor is suitable.

Step 3

Determine the value of R that the student should select.

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Answer

Using the potential divider formula, to find R that gives an output of 5.0 V:

$\frac{R}{R + 250} × 9.0 = 5.0$

Rearranging gives: $R = \frac{5.0}{9.0 - 5.0} × 250 = 250 Ω$

Thus, R should be approximately 250 Ω.

Step 4

Give your answer to an appropriate number of significant figures.

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Answer

The value of R = 250 Ω is already in 3 significant figures as requested.

Step 5

State and explain the effect on the output pd of increasing the temperature of the thermistor.

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Answer

As the temperature of the thermistor increases, its resistance decreases. This leads to a higher current through the circuit. Consequently, applying the potential divider rule, the output potential difference (pd) across the thermistor will decrease. Hence, if temperature increases, the output pd measured will decrease, leading to an inverse relationship between temperature and output voltage.

Figure 7 shows a circuit designed by a student to monitor temperature changes - AQA - A-Level Physics - Question 4 - 2018 - Paper 1

Worked Solution & Example Answer:Figure 7 shows a circuit designed by a student to monitor temperature changes - AQA - A-Level Physics - Question 4 - 2018 - Paper 1

Determine the area of cross-section of the wire that has been used for the resistor.

Determine whether this resistor is suitable.

Determine the value of R that the student should select.

Give your answer to an appropriate number of significant figures.

State and explain the effect on the output pd of increasing the temperature of the thermistor.

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