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Figure 11 shows how the resistance of an LDR varies with light intensity - AQA - A-Level Physics - Question 6 - 2021 - Paper 1

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Figure 11 shows how the resistance of an LDR varies with light intensity. Figure 12 shows one proposal for a sensor circuit for this system. The power supply to th... show full transcript

Worked Solution & Example Answer:Figure 11 shows how the resistance of an LDR varies with light intensity - AQA - A-Level Physics - Question 6 - 2021 - Paper 1

Step 1

Show that the current in the sensor circuit when the LDR is not illuminated by the light beam is approximately 16 µA.

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Answer

When the LDR is not illuminated by the light beam, its resistance is approximately 300 kΩ (as per Figure 11).

Using Ohm’s Law, we can calculate the current (I) in the circuit:

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

Substituting the values:

I=5.0 V300 kΩI = \frac{5.0 \, \text{V}}{300 \, k\Omega}

Calculating this gives:

I=5.0300×103=16.67μA,I = \frac{5.0}{300 \times 10^3} = 16.67 \mu A,

which can be approximated to 16 µA.

Step 2

Discuss whether the circuit shown in Figure 12 is suitable. Support your answer with a calculation.

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Answer

To determine the suitability of the circuit, we need to assess the potential difference across the LDR when illuminated and when not illuminated.

The resistance of the LDR when illuminated is about 16.1 kΩ (from Table 3).

Using Ohm’s Law again:

I=5.016.1×103≈310.56 μAI = \frac{5.0}{16.1 \times 10^3} \approx 310.56 \, \mu A

Now we can calculate the voltage across the LDR:

Vs=I⋅R≈310.56μA⋅16.1kΩ≈5.00 VV_s = I \cdot R \approx 310.56 \mu A \cdot 16.1 k\Omega \approx 5.00 \, V

When the LDR is not illuminated, as calculated above, the current is approximately 16 µA, leading to:

Vs=16μA⋅300kΩ≈4.8 VV_s = 16 \mu A \cdot 300 k\Omega \approx 4.8 \, V

Thus, the change in potential is:

5.0 V−4.8 V=0.2 V5.0 \, V - 4.8 \, V = 0.2 \, V

This is 4% of the power supply voltage. Since the alarm triggers for a change of more than 25% (which is 1.25 V), the circuit as shown is not suitable.

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