A photoresistor, whose resistance decreases with light intensity, is connected i
ID: 1610665 • 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 knExplanation / 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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