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

ID: 1562080 • Letter: A

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 k . On a cloudy day, the resistance rises to 3.6 k . At night, the resistance is 29 k

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?

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

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

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

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

Problem 23.30 Enhanced with Feedback 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 k2 On a cloudy day, the resistance rises to 3.6 k2 At night, the resistance is 29 k2 (Figure 1) You may want to review ages 738-740 Figure 1 of 1 Photoresistor 9.0 V 1.0 kn

Explanation / Answer

A photoresistor, whose resistance decreases with light intensity, is connected in the circuit of the figure. On a sunny day

V = IR

I = V/R = 9/1000+0.52 = 0.00899 A

V = I * R
V = 0.0089953 * 1000
V = 9 V

(B)

V = IR
9 = I * (1000 + 3600)
I = 0.001956

V = 0.001956 * 1000

V = 1.95V

(C)

V = IR
9 = I * (1000 + 29000)
I = 0.0003

V = 0.0003 * 1000
V = 0.3 V

(D)

It increases, since the resistance of the photoresistor drops.

option(1) is correct answer

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