Describe the basic conditioning necessary to produce a current in a circuit without a voltage source and list the three methods of meeting this condition

Answers

Answer 1

To produce a current in a circuit without a voltage source, it is necessary to induce a voltage in the circuit by changing the magnetic field passing through it.

This is achieved through a process called electromagnetic induction. According to Faraday's law of electromagnetic induction, a voltage is induced in a circuit when there is a change in the magnetic flux through the circuit. To meet this condition, there are three basic methods: Moving a magnet relative to a coil of wire: When a magnet is moved close to a coil of wire or away from it, the magnetic flux through the coil changes and induces a voltage in the coil. This is the basic principle behind generators. Changing the current in a nearby coil: When the current in a coil of wire is changed, the magnetic field surrounding the coil changes, inducing a voltage in a nearby coil of wire. This is the basic principle behind transformers. Moving a coil of wire in a magnetic field: When a coil of wire is moved in a magnetic field, the magnetic flux through the coil changes, inducing a voltage in the coil. This is the basic principle behind electric motors. By using one of these methods, it is possible to produce a current in a circuit without a voltage source.

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Related Questions

a balloon with an internal pressure of 300. torr rises to a height of 30,000 feet, where the pressure is 15. torr. assuming temperature remains constant, by what ratio did the volume change?

Answers

The volume of the balloon increased by a factor of 20 or increased by 20 times.

We can use Boyle's Law to solve this problem, which states that the pressure and volume of a gas are inversely proportional, as long as the temperature remains constant.

Boyle's Law: P1V1 = P2V2

Where P1 and V1 are the initial pressure and volume, and

P2 and V2 are the final pressure and volume.

Also we know that P1 = 300 torr, P2 = 15 torr, and V1 = V

Therefore, after substituting

300 V = 15 (V2)

V2 = 300/15

V2 = 20 V1

Therefore, the volume changed by a factor of 20 or increased by 20 times.

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4x10^-15C charged experiences a 5x10^-16N force. What is the strength of the electric field in that region of space?

Answers

Answer:

Explanation:

The electric force F exerted by the field on the positive charge is F = qE

where q is electric charge and E is electric field.

E=F/q=5 *10^-16/4*10^-15=1.25*10^-1 = 0.125N/C

a 56.2-gram tennis ball is loaded into a 1.27-kg homemade cannon. the cannon is at rest when it is ignited. immediately after the impulse of the explosion, a photogate timer measures the cannon to recoil backwards a distance of 6.1 cm in 0.0218 seconds. determine the post-explosion speed of the cannon and of the tennis ball.

Answers

A photogate timer measures the cannon to recoil backward at a distance of 6.1 cm in 0.0218 seconds. The post-explosion speed of the cannon and of the tennis ball is: 599.9 m/s

The post-explosion speed of the cannon and of the tennis ball can be determined using the impulse-momentum theorem. This theorem states that the impulse (change in momentum) is equal to the product of the average force and the time interval during which it acts.

The impulse is equal to the mass of the cannon times the change in velocity of the cannon. The mass of the cannon and the tennis ball is the sum of the mass of the cannon and the mass of the tennis ball, which is 1.326 kg. The time interval is 0.0218 seconds.

We can calculate the change in velocity of the cannon and the tennis ball using the equation impulse = mass x change in velocity. This gives us:

change in velocity = impulse / mass = (1.326 kg) x (6.1 cm) / (0.0218 s) = 599.9 m/s.

Therefore, the post-explosion speed of the cannon and the tennis ball is 599.9 m/s.

This calculation is based on the assumption that the impulse of the explosion is the same for both the cannon and the tennis ball. Therefore, the post-explosion speed of the cannon and the tennis ball is the same and equal to 599.9 m/s.

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A car wheel of radius 0.3 m has an angular acceleration of 2.45 rad/s2
. Calculate
the linear acceleration of wheel.

Answers

The linear acceleration of the car wheel is 0.735 m/s².

What is the the linear acceleration of car wheel?

The linear acceleration of a point on the rim of a wheel is related to the angular acceleration of the wheel by the equation:

a = rα

Where "a" is the linear acceleration, "r" is the radius of the wheel, and "α" is the angular acceleration.

Given the data in the question;

angular acceleration of α = 2.45 rad/s² radius r = 0.3 m. linear acceleration a = ?

Plug the given values into the above formula and solve for the linear acceleration of wheel.

a = r × α

a = 0.3 m × 2.45

a = 0.735 m/s²

Therefore, the linear acceleration is 0.735 m/s².

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