Problem 23.30 Enhanced with Feedback A photoresistor whose resistance decreases

Question

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 0.62 k2 On a cloudy day, the resistance rises to 3.0 kn. At night, the resistance is 29 k2. (Figure 1) You may want to review (DA ages 738 740 Figure 1 T of 1 Photoresist or 9.0 V 1.0 kn Part A What does the voltmeter read on a sunny day? Express your answer with the appropriate units. 8.99 V Submit My Answers Give Up Incorrect, Try Again, 4 attempts remaining Part B What does the voltmeter read on a cloudy day? Express your answer with the appropriate units. V Value Units Submit My Answers Give Up Part C What does the voltmeter read at night? Express your answer with the appropriate units. V Value Units Submit My Answers Give Up

Explanation / Answer

1.The voltage drop across a resistor is the current times the resistance of that resistor. The photoresistor and regular resistor are in series, so add their resistances together. This gives a total resistance of 1000.62ohms.

V = IR
9 = I * 1000.62
I = 0.0089944

To find the voltmeter reading, just multiply the current by the resistance of what the voltmeter is connected to:

V = I * R
V = 0.0089944 * 1000 =8.99V

2.

V = IR
9 = I * (1000 + 3000)
I = 0.00225

V = 0.00225 * 1000
V = 2.25V

3.

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

V = 0.00030 * 1000
V = 0.3 v

4.It increases, since the resistance of the photoresistor drops.

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