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Nadya [2.5K]
2 years ago
10

Define;i) Voltageii) Currentiii) Electrical Poweriv) Electrical Energy​

Engineering
1 answer:
nalin [4]2 years ago
7 0

Answer:

I) <u>Voltage</u> - is the pressure from an electrical circuit's power source that pushes charged electrons (current) through a conducting loop, enabling them to do work such as illuminating a light. In brief, voltage = pressure, and it is measured in volts (V).

II) <u>Current</u> - is the movement of electrons through a wire. Electric current is measured in amperes (amps) and refers to the number of charges that move through the wire per second. If we want current to flow directly from one point to another, we should use a wire that has as little resistance as possible.

III) <u>Electrical Power </u>- is the rate, per unit time, at which electrical energy is transferred by an electric circuit. The SI unit of power is the watt, one joule per second. Electric power is usually produced by electric generators, but can also be supplied by sources such as electric batteries.

IV) <u>Electrical Energy </u>- is a form of energy resulting from the flow of electric charge. Energy is the ability to do work or apply force to move an object. In the case of electrical energy, the force is electrical attraction or repulsion between charged particles.

Explanation:

I hope ot helps to you a lot! Correct me if I'm wrong.

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Why is it important to keep hand tools free of oil and grease?
Bezzdna [24]

Answer:

keeps it from rusting

Explanation:

5 0
3 years ago
6. During some actual expansion and compression processes in piston–cylinder devices, the gases have been
Katyanochek1 [597]

During some actual expansion and compression processes in piston-cylinder devices, the gases have been are the P1= P2.

<h3>What is the pressure?</h3>

Pressure is something that has the pressure that is physical and that causes the pressure is piston-cylinder devices.

During a few real enlargements and compression procedures in piston-cylinder devices, the gases were located to meet the connection PV n = C, wherein n and C are constants.

Read more about the pressure :

brainly.com/question/25736513

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5 0
1 year ago
Given asphalt content test data:
VMariaS [17]

Answer:

hello your question is incomplete attached below is the complete question

A) overall mean = 5.535,  standard deviation ≈ 0.3239

B ) upper limit = 5.85, lower limit = 5.0

C) Not all the samples meet the contract specifications

D) fluctuation ( unstable Asphalt content )

Explanation:

B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday

The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85

The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0

attached below is the required plot

C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :

15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20

D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed

5 0
3 years ago
Supercharged engine what it does to the car
geniusboy [140]

Answer:

A supercharger is an air compressor that increases the pressure or density of air supplied to an internal combustion engine. This gives each intake cycle of the engine more oxygen, letting it burn more fuel and do more work, thus increasing power.

Explanation:

3 0
3 years ago
Read 2 more answers
In a 5V system if you were asked to take one input HIGH and another LOW what would you do (i.e. where would you connect them)?
dangina [55]

Answer:

HIGH from the supply voltage

LOW from ground

Explanation:

The answer depends on the kind of system and the purpose of the signal. But for practical reasons, in a DIGITAL system where 5V is HIGH and 0 V is LOW, 5 volts can be taken from the supply voltage (usually the same as high,  BUT must be verified), and the LOW signal from ground.

If the user has a multimeter, it must be set to continuous voltage on 0 to 20 V range.  Then place the probe in the ground of the circuit (must be a big copper area). Finally  leave one probe in the circuit ground and place the other probe in some test points to identify 5 v.

4 0
3 years ago
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