Question

Photoresistor 9.0 V = 1.0 k 12

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 0.66 kΩkΩ . On a cloudy day, the resistance rises to 4.0 kΩkΩ . At night, the resistance is 26 kΩkΩ

What does the voltmeter read on a sunny day?

What does the voltmeter read on a cloudy day?

What does the voltmeter read at night?

Does the voltmeter reading increase or decrease as the light intensity increases (in terms of current)?

Answer 1

Part A.

On a Sunny day, equivalent resistance of circuit will be

Req = 1.0 k-ohm + resistor of photoresistor

Resistor of photoresistor on sunny day = 0.66 k-ohm = 660 ohm

Req = 1000 + 660 = 1660 Ohm

Now We know from Ohm's law, that

ieq = V/Req = 9/1660 = 0.005422 Amp.

Now we know that in series circuit current in resistor will be equal to total current, So

current in 1 k-ohm = 0.005422 Amp

Voltage in this 1 k-ohm resistor will be

V = i*R = 0.005422*1000

V = 5.422 V

In two significant figures:

V = 5.4 V

Part B.

On a cloudy day, equivalent resistance of circuit will be

Req = 1.0 k-ohm + resistor of photoresistor

Resistor of photoresistor on cloudy day = 4.0 k-ohm = 4000 ohm

Req = 1000 + 4000 = 5000 Ohm

Now We know from Ohm's law, that

ieq = V/Req = 9/5000 = 0.0018 Amp.

Now we know that in series circuit current in resistor will be equal to total current, So

current in 1 k-ohm = 0.0018 Amp

Voltage in this 1 k-ohm resistor will be

V = i*R = 0.0018*1000

V = 1.8 V

Part C.

at night, equivalent resistance of circuit will be

Req = 1.0 k-ohm + resistor of photoresistor

Resistor of photoresistor at night = 26 k-ohm = 26000 ohm

Req = 1000 + 26000 = 27000 Ohm

Now We know from Ohm's law, that

ieq = V/Req = 9/27000 = 0.00033 Amp.

Now we know that in series circuit current in resistor will be equal to total current, So

current in 1 k-ohm = 0.00033 Amp

Voltage in this 1 k-ohm resistor will be

V = i*R = 0.00033*1000

V = 0.33 V

Part D.

From above we can see that as current increase (given light intensity increases) reading of voltmeter increases.

Let me know if you've any query.