All categories

3 years ago

What would happen if the voltage impressed across a circuit is held constant while the resistance doubles

Physics

1 answer:

RideAnS [48]3 years ago

8 0

Answer:

What would happen if the voltage impressed across a circuit is held constant while the resistance doubles.

<em>The answer is:</em>

A change will occur in the current which is, "The new current will be half the initial value."

Explanation:

Voltage, also called electromotive force, is the measure of specific potential energy between two locations in an electrical field.

V= IR

Where, V= Voltage

I= Current

R= Resistance

The greater the voltage in a circuit, the greater its ability to push more electrons and do work.

Voltage is measured in volts (V). Voltage is the difference in charge between two points, and can also be considered as the pressure that forces the charged electrons to flow in an electrical circuit.

Current is the rate at which charge is flowing. An electric current flows when electrons move through a conductor, such as a metal wire.

Resistance is a material's tendency to resist the flow of charge (current). Resistance is measured in ohms, and can be further explained as a measure of the opposition to current flow in an electrical circuit. Resistance is good because it protects humans from the harmful energy of electricity. Resistance, R in ohms (Ω) is equal to the voltage V in volts (V) divided by the current I in amps (A).

Current is directly proportional to the voltage and inversely proportional to the resistance.

If voltage is increased, the current will also increase. The higher the resistance, the lower the current flow. The lower the resistance, the higher the current flow.

Doubling the voltage will cause the current to be doubled. Also, doubling the resistance will cause the current to be one-half the original value.

You might be interested in

NGSS Physics

vovangra [49]

Answer: 3.5 seconds

EXPLANATION:

Using the formula:
v = u + at
And taking the upwards direction as positive, we have the following information:

u = 35 m/s
a = -10m/s^2 (this is acceleration due to gravity)

At the top of its path, the apple will have a velocity of 0 m/s, therefore:

v = 0m/s

Once you substitute everything into the formula, you get:

0 = 35 + (-10)t

Therefore, t = 35/10 = 3.5 seconds

6 0

3 years ago

Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .

Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

Where;

$G$ is the Gravitational Constant and its value is $6.674(10^{-11})\frac{m^{3}}{kgs^{2}}$

$M=1.9(10^{27})kg$ is the mass of Jupiter

$a=4.22(10^{5})km=4.22(10^{8})m$ is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}$

$T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}$

$T=\sqrt{2.339(10^{10})s^{2}}$

Then:

Which is the same as:

Therefore, the answer is:

The orbital period of Io is 42.482 h

7 0

4 years ago

I drop an egg from a certain distance and it takes the egg 3.74 seconds to reach the ground. How high up was the egg?

Ulleksa [173]

Answer:

Explanation:

<u>Free Fall Motion </u>

When a body is left to move in the air with no friction, the motion is ruled only by the force of gravity. The vertical distance a body travels in the air after a time t is .

$\displaystyle y=\frac{gt^2}{2}$