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1 year ago

There is a current of 0.83 A through a lightbulb in a 120 V circuit. What is the resistance of this lightbulb?

Physics

1 answer:

Lemur [1.5K]1 year ago

3 0

Considering the Ohm's law, the resistance of the lightbulb is 144.58 Ω.

<h3>Definition of current</h3>

The flow of electricity through an object, such as a wire, is known as current (I). Its unit of measure is amps (A). So the current is a measure of the speed at which the charge passes a given reference point in a specified direction.

<h3>Definition of voltage</h3>

The driving force (electrical pressure) behind the flow of a current is known as voltage and is measured in volts (V) (voltage can also be referred to as the potential difference or electromotive force). That is, voltage is a measure of the work required to move a charge from one point to another.

<h3>Definition of resistance</h3>

Resistance (R) is the difficulty that a circuit opposes to the flow of a current and it is measured in ohms (Ω).

Ohm's law establishes the relationship between current, voltage, and resistance in an electrical circuit.

This law establishes that the intensity of the current that passes through a circuit is directly proportional to the voltage of the same and inversely proportional to the resistance that it presents.

Mathematically, Ohm's law is expressed as:

$I=\frac{V}{R}$

Where:

I is the current measured in amps (A).
V the voltage measured in volts (V).
R the resistance that is measured in ohms (Ω).

<h3>Resistance of the lightbulb</h3>

In this case, you know that the voltage between two points in a circuit is 120 V and there is a current of 0.83 A.

Replacing in the Ohm's Law:

$0.83A=\frac{120 V}{R}$

Solving:

0.83 A× R= 120 V

$R=\frac{120 V}{0.83A}$

Finally, the resistance of the lightbulb is 144.58 Ω.

Learn more about Ohm's law:

brainly.com/question/13076023

brainly.com/question/17286882

brainly.com/question/2275770

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