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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

06.1 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

Given that the resistance of the LDR not illuminated by the light beam is 300 kΩ, we can use Ohm's Law to find the current. The formula is:

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

where:

  • II is the current
  • VV is the voltage (5.0 V)
  • RR is the resistance (300 kΩ = 300,000  Ω300,000 \; \Omega)

Substituting in the values:

I=5.0300,000=0.0000167  A≈16  μAI = \frac{5.0}{300,000} = 0.0000167 \; \text{A} \approx 16 \; \mu A

Step 2

06.2 Discuss whether the circuit shown in Figure 12 is suitable.

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Answer

To check if the circuit is suitable, we need to evaluate the potential difference across the LDR when it is illuminated as compared to when it is not.

When the LDR is illuminated (4 lux), the resistance is 300 kΩ.

Using the same formula:

  1. Calculate the current (II) when illuminated:

    • Using R=300,000  ΩR = 300,000 \; \Omega:

    I=5.0300,000=0.0000167  AI = \frac{5.0}{300,000} = 0.0000167 \; A

  2. Calculate the potential difference (VsV_s) across the LDR:

    Vs=I×R=0.0000167×300,000=5.0  VV_s = I \times R = 0.0000167 \times 300,000 = 5.0 \; V

When the LDR is not illuminated (1 lux) and has a resistance of 100 kΩ, the situation changes:

  1. Calculate II again:

    I=5.0100,000=0.00005  A=50  μAI = \frac{5.0}{100,000} = 0.00005 \; A = 50 \; \mu A

  2. Calculate the new VsV_s:

    Vs=50×100,000=5.0  VV_s = 50 \times 100,000 = 5.0 \; V

When the change is measured:

  • The difference between fully illuminated and unilluminated states is minimal (5.0 V).
  • Since 5.0 V does not exceed the 1.25 V threshold for sounding the alarm, the circuit is suitable.

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