Question

A photoresistor, whose resistance decreases with light intensity, is connected in the circuit of the figure. On a sunny day, the photoresistor has a resistance of 0.52 kΩ . On a cloudy day, the resistance rises to 4.4 kΩ . At night, the resistance is 29 kΩ . (Figure 1)Part A

What does the voltmeter read on a sunny day?

Part B

What does the voltmeter read on a cloudy day?

Part C

What does the voltmeter read at night?

Part D

Does the voltmeter reading increase or decrease as the light intensity increases?

Does the voltmeter reading increase or decrease as the light intensity increases?

The voltmeter reading increases because the current through the resistor decreases.

The voltmeter reading increases because the current through the resistor increases.

The voltmeter reading decreases because the current through the resistor increases.

The voltmeter reading decreases because the current through the resistor decreases.

Answer 1

The resistance the 9v battery sees = Rp+1000 where Rp is the resistance of the photo resistor.

[The diagram is not provided so assuming V = 9V and the circuit resistance to be 1kohm= 1000 ohm]

Therefore the current out of the battery = 9/(Rp+1000) Amps and the voltmeter

reading = 9/(Rp+1000) * 1000Ω = 9000/(Rp+1000) = Vm

a) What does the voltmeter read on a sunny day?

9000/(520+1000) = 5.92 volts

b) What does the voltmeter read on a cloudy day?

9000/(4400+1000) = 1.67 volts

c) What does the voltmeter read at night?

9000/(29,000+1000) = 0.3 volts

d) The voltmeter reading increases because the current through the resistor increases.