Question

A simple circuit can be created using a voltage source of some form, let's say a battery, and a load, something that converts electrical energy into some other form of energy. An example of a load might be the thin wire filament in a light bulb, where the electrical energy is being converted into thermal energy and light energy. The resistance is a property of the load that determines how much current flows through the load.

Ohm's Law relates the voltage across two ends of some object or group of objects to the current through that object or group of objects and the resistance of that object or group of objects.

Suppose we hook up a battery with a voltage of 2.45 V in a simple circuit with a light bulb that has a resistance of 17.5 Ω. Use Ohm’s Law to find the current flowing through the light bulb.

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We can turn the problem around as well. Again, hook up a simple circuit with a battery and a bulb. In this case we measure a current of 0.130 A flowing through the bulb which has a resistance of 12.3 Ω. What is the voltage being applied to the bulb?

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To measure the resistance of a bulb, you can apply a voltage to the bulb and measure the current through the bulb. Suppose you hook the bulb up to a 3.90 V voltage source and measure a current of 0.500 A flowing through the bulb. What is the resistance of the bulb? (enter ohms for the units. Do not try to enter Ω).

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If a bulb of resistance 25.0 Ω is connected to a 3.20-V voltage source, what is the power produced by the bulb?

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This discussion is closed.

Answer #1

here,

A)

voltage of battery , V = 2.45 V

resistance of the bulb , r = 17.5 ohm

let the current flowing through bulb be I

using ohm's law

V = I * r

2.45 = I * 17.5

I = 0.14 A

the current flowing through the bulb is 0.14 A

B)

let the voltage of battery be V

resistance of the bulb , r = 12.3 ohm

the current flowing through bulb , I = 0.13 A

using ohm's law

V = I * r

V = 0.13 * 12.3

V = 1.6 V

the voltage of battery is 1.6 V

C)

resistance of the bulb , r = V/I

r = 3.9/0.5

r = 7.8 ohm

the resistance of the bulb is 7.8 ohm

D)

power , P = V^2 /r

P = 3.2^2/25

P = 0.41 W

the power produced by the bulb is 0.41 W

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