A photoresistor, whose resistance decreases with light intensity, is connected i

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 0.60 k? . On a cloudy day, the resistance rises to 3.2 k? . At night, the resistance is 26 k? . (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? (Choose one)

- The voltmeter reading decreases because the current through the resistor increases. - The voltmeter reading decreases because the current through the resistor decreases. - The voltmeter reading increases because the current through the resistor decreases. - The voltmeter reading increases because the current through the resistor increases. Photoresistor 9.0 V 1.0 k

Explanation / Answer

The resistance the 9v battery sees = Rp+1000 where Rp is the resistance of the photo resistor.
Therefore the current out of the battery = 9/(Rp+1000) Amps and the voltmeter
reading = 9/(Rp+1000) * 1000 = 9000/(Rp+1000) = Vm
1) What does the voltmeter read on a sunny day?
9000/(600+1000) = 5.6 volts <----- 1)
2) What does the voltmeter read on a cloudy day?
9000/(3200+1000) = 2.1 volts <----- 2)
3) What does the voltmeter read at night?
9000/(26,000+1000) = 0.33 volts <----- 3)

Does the voltmeter reading increase or decrease as the light intensity increases?
d. The voltmeter reading increases because the current through the resistor increases.
You can see that above: 5.6 > 2.1 > 0.33

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