A-level Physics (Advancing Physics)/Sensors/Worked Solutions

Note to the reader: I am not entirely happy with the answers given on this page. --Sjlegg (talk) 15:01, 19 December 2007 (UTC)

An LDR's resistance decreases from a maximum resistance of 2kΩ to a minimum resistance of 0Ω as light intensity increases. It is used in a distance sensing system which consists of a 9V power supply, a 1.6 kΩ resistor, the LDR and a multimeter which displays voltage to 2 decimal places measuring the potential difference across one of the two resistors.

1. Across which resistor should the multimeter be connected in order to ensure that, as the distance from the light source to the sensor increases, the potential difference recorded increases?

As light intensity increases, distance decreases. So, as distance increases, light intensity decreases. As light intensity decreases, the LDR's resistance increases, as does the potential difference across it. So, as distance increases, the potential difference increases. Since we want the potential difference to change in the same direction as the distance, the multimeter must go across the LDR, not the resistor.

2. In complete darkness, what voltage is recorded on the multimeter?

The voltage is split between the 1.6kΩ resistor and the LDR, which currently has a resistance of 2kΩ. Therefore, the potential difference across the LDR is:

$V={\frac {2}{2+1.6}}\times 9=5V$

3. When a light source moves 0.5m away from the sensor, the voltage on the multimeter increases by 2V. What is the sensitivity of the sensing system when using this light source, in V m⁻¹?

$S={\frac {2}{0.5}}=4{\mbox{ Vm}}^{-1}$

4. When the same light source is placed 0m from the sensor, the potential difference is 0V. When the light source is 1m away, what voltage is displayed on the multimeter?

Assuming a linear relationship between distance and potential difference, we know that:

$0.25={\frac {m}{V}}$

Hence:

$V={\frac {m}{0.25}}={\frac {1}{0.25}}=4V$

5. What is the resolution of the sensing system?

The multimeter measures to 2 decimal places, so the smallest measurable voltage is 0.01V.

$0.25={\frac {m}{V}}$

Hence:

$m=0.25\times V=0.25\times 0.01=0.0025{\mbox{ m}}=2.5{\mbox{ mm}}$

6. Draw a circuit diagram showing a similar sensing system to this, using a Wheatstone bridge and amplifier to improve the sensitivity of the system.

7. What is the maximum potential difference that can reach the amplifier using this new system (ignore the amplification)?

The maximum potential difference will occur when the LDR has no resistance. This will result in 9V on the left-hand side of the voltmeter, and 5V on the right-hand side. The difference is 4V.

8. If this signal were to be amplified 3 times, would it exceed the maximum voltage of the system? What would the limits on the signal be?

$4\times 3=12V>9V$

The signal would be limited to the range -9V < V < 9V.