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 0.58 k? . On a cloudy day, the resistance rises to 4.8 k? . At night, the resistance is 23 k? . (Figure 1)

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

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

c. What does the voltmeter read at night?

d. Does the voltmeter reading increase or decrease as the light intensity increases? (4 choices below)

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

Explanation / Answer

A)

Just remember the equation V = IR. 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 1580 ohms.

V = IR
9 = I * 1580
I = 0.00569

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

V = I * R
V = 0.00569 * 1000
V = 5.69 V (rounded)

B)

V = IR
9 = I * (1000 + 4800)
I = 0.00155

V = 0.00155 * 1000
V = 1.55 V

C)

V = IR
9 = I * (1000 + 23000)
I = 0.000375

V = 0.000375 * 1000
V = 0.375 v

D)

It increases, since the resistance of the photoresistor drops.

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