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.

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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

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 of the wire, we use the formula:

ho \frac{L}{A} $$ Where: - $R$ is the resistance (0.25 kΩ = 250 Ω), - $\rho$ is the resistivity (4.2 × 10⁻⁷ Ω m), - $L$ is the length of the wire, - $A$ is the area of cross-section. First, we need to calculate the length of the wire. The wire is made of 50 turns around a cylinder of diameter 8.0 mm: $$ L = 50 \times \pi \times d = 50 \times \pi \times 0.008 m \approx 1.26 m $$ Now substituting into the resistance formula: $$ 250 = 4.2 \times 10^{-7} \frac{1.26}{A} $$ Rearranging to find A: $$ A = 4.2 \times 10^{-7} \frac{1.26}{250} \approx 2.13 \times 10^{-9} m^2 $$

Step 2

2. Determine whether this resistor is suitable.

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Answer

To determine if the resistor is suitable, we must check the maximum power it can dissipate:

$P_{max} = I^2 R$

To find the current $I$ , we first set up the circuit:

Given that the total supply voltage $V$ is 9.0 V and the desired output voltage across the thermistor must be 5.0 V, the remaining voltage across the 0.25 kΩ resistor is:

$V_{R} = 9.0 V - 5.0 V = 4.0 V$

The current through the resistor is:

$I = \frac{V_{R}}{R} = \frac{4.0 V}{250 \Omega} = 0.016 A$

Substituting $I$ back to find $P_{max}$ :

$P_{max} = (0.016)^2 \times 250 = 0.064 W$

Since the selected resistor is rated at 0.36 W and the power dissipated (0.064 W) is below this, the resistor is suitable.

Step 3

3. Determine the value of R that the student should select.

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Answer

To find the required resistance $R$ , we can use the potential divider formula to determine the necessary values:

$V_{out} = V_{in} \frac{R}{R + R_{thermistor}}$

Rearranging this to find the value of $R$ leads to:

$R = \left( \frac{V_{out}}{V_{in} - V_{out}} \right) R_{thermistor}$

Substituting the known values:

$V_{out} = 5.0 V$
$V_{in} = 9.0 V$
$R_{thermistor} = 750 \Omega$ :

$R = \left( \frac{5.0}{9.0 - 5.0} \right) \times 750 \approx 937.5 \Omega$

Step 4

4. Give your answer to an appropriate number of significant figures.

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Answer

The value of $R$ calculated was about 937.5 Ω. Given the context, we would present this answer in two significant figures, leading to:

$R \approx 940 \Omega$

Step 5

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 change affects the potential difference across the thermistor. According to the potential divider rule, if the resistance of the thermistor decreases, while the resistance $R$ remains constant, the total resistance in the circuit decreases, which leads to an increase in the current flowing through the circuit.

Since $V_{out}$ is dependent on the current, the output potential difference across the thermistor would also decrease as its resistance decreases. Thus, the effect of increasing the temperature of the thermistor results in a reduction of the output pd.

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

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

2. 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.

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