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2 years ago

"How do changes in voltage or resistance affect current in an electric circuit?"

1 answer:

5 0

We can answer the question by looking at the Ohm's law, which gives us the relationship between voltage (V), current (I) and resistance (R) of a circuit:
$V=IR$
equivalently, we can rewrite it as
$I= \frac{V}{R}$

by looking at the equation, we can make the following observations:
1) The current is proportional to the voltage: therefore, if the voltage increases, the current increases as well; if the voltage decreases, the current decreases too.
2) The current is inversely proportional to the resistance: if the resistance increases, the current decreases, and if the resistance decreases, the current increases.

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

Lemur [1.5K]

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:

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1 year ago

Which force requires contact

jekas [21]

Gravity?????????? or maybe like any object

5 0

2 years ago

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True of false metals like copper are sometimes used to fill cavities in teeth

vovikov84 [41]

Yes. Copper, mercury, and tin are all used to fill in cavities.

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2 years ago

Which of the following is true about a planet orbiting a star in uniform circular

Mnenie [13.5K]

The velocity vector of the planet points toward the center of the circle is the following is true about a planet orbiting a star in uniform circular motion.

A. The velocity vector of the planet points toward the center of the circle.

<u>Explanation:</u>

Motion of the planet around the star is mentioned to be uniform and around a circular path. Objects in uniform circular motion motion has constant angular speed but the velocity of the object will not remain constant. Since the planet is in circular motion the direction of velocity vector at a particular point is tangential to the circular path at that particular point.

Thus at every point, the direction of velocity vector changes and this means the velocity is never constant. The objects in uniform circular motion has centripetal acceleration which means that velocity vector of the planet points toward the center of the circle.

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3 years ago

An inductor is connected in series to a fully charged capacitor. Which of the following statements are true? Check all that appl

Luba_88 [7]

Answer:

As the capacitor is discharging, the current is increasing

Explanation:

Lets take

C= Capacitance

L=Inductance

V=Voltage

I= Current

The total energy E given as

$E=\dfrac{IL^2}{2}+\dfrac{CV^2}{2}$

We know that total energy E is conserved so when electric energy 1/2 CV² decreases then magnetic energy 1/2 IL² will increases.

It means that when charge on the capacitor decreases then the current will increase.

As the capacitor is discharging, the current is increasing

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