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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

To find the current in the circuit when the LDR is not illuminated, we first determine the resistance of the LDR in the dark. According to the given information, the resistance of the LDR is approximately 300 kΩ:

RLDR=300extkΩR_{LDR} = 300 ext{ kΩ}

Using Ohm's law, the current can be calculated as:

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

Substituting the values:

I=5.0extV300×103extΩ=16.67 µA≈16extµAI = \frac{5.0 ext{ V}}{300 \times 10^3 ext{ Ω}} = 16.67 \text{ µA} \approx 16 ext{ µA}

Step 2

The alarm sounds when the potential difference V_s across the LDR changes by more than 25% of the power supply emf.

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Answer

The potential difference across the LDR when it is illuminated by the light beam needs to be calculated. The light intensity without the light beam results in:

RLDR=16.1imes103extΩR_{LDR} = 16.1 imes 10^3 ext{ Ω}

Using Ohm's law:

Vs=IimesRLDRV_s = I imes R_{LDR}

Where the current has been determined from the earlier part:

I=5.0extV300imes103extΩ⇒Vs=(16.67×10−6A)×(16.1×103extΩ)=0.2686extVI = \frac{5.0 ext{ V}}{300 imes 10^3 ext{ Ω}} \Rightarrow V_s = (16.67 \times 10^{-6} A) \times (16.1 \times 10^3 ext{ Ω}) = 0.2686 ext{ V}

The alarm triggers if:

25% of 5.0extV=1.25extV25\% \text{ of } 5.0 ext{ V} = 1.25 ext{ V}

Since 0.2686 V is less than 1.25 V, therefore, the alarm will not sound. Hence, the circuit is suitable as it does not trigger the alarm.

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