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Sample Problem: AC Power 1

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Linear Circuits 2: AC Analysis

Технологический институт Джорджии

4.6 (оценок: 83) | Зарегистрировано учащихся: 6.6K

This course explains how to analyze circuits that have alternating current (AC) voltage or current sources. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

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Рецензии

4.6 (оценок: 83)

5 stars
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4 stars
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1 star
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AA

Jan 01, 2019

Really beneficial especially with covering Filters and Transformers

MR

Mar 29, 2018

The best course and understandable. Thanks dear teachers

Из урока

Module 4: Power

The description goes here

4.1 Power Module Overview1:18

4.2 Root Mean Square8:29

4.3 Power Factor12:06

4.4 Power Triangles10:10

4.5 AC Power Transfer13:59

Sample Problem: AC Power 15:12

Sample Problem: AC Power 214:30

Sample Problem: AC Power 37:41

Sample Problem: AC Power 49:38

Sample Problem: AC Power 54:42

Преподаватели

Dr. Bonnie H. Ferri
Professor
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center

Текст видео

The topic of this problem is AC steady state power.

The problem is to determine the values for the resistance R and

the inductance L in the circuits shown below, such circuits has a voltage source

is 12cos at 4t volts And has the resistor inductor and in parallel with one another.

We also know that the power delivered by the voltage source, the complex power,

is 18 plus j9 volt amps.

So we're going to use this information to find the appropriate values for R and

L based on the power that's delivered by the source.

We use our well-known equation for complex power,

complex power is equal to the voltage times

the Conjugate of the current divided by two.

So if we substitute first of all to find I of T.

If we substitute end values for the complex power and

the voltage then we can find the value for

the conjugate of the current in the circuit.

And so doing that, we have two times the complex

power 18 plus J9 divided by the voltage which

is 12 with a phase angle of zero degrees.

To solve this we end up with the complex current which is equal to 3 plus j 1.5.

And magnitude angle

format It's 3.35 in an angle of 26.6 degrees for

our complex conjugate of the current.

So we know that If we know the current was flows in to the RL combination,

and we know the voltage drop across the RL parallel combination,

then we can find our values for R and L.

And we do that by first of all recognizing that these are impervious values and

there in parallel with one another, so.

We know that the total impedance for those two elements

is going to be 1 over R.

This is 1 over the complex

impedance plus, 1 over j omega L.

And so if we plug in values for that,

We plug in our omega value, which is four and

we set that equal to V over I, but

it's one over Z, so it's I divided by V.

Then we can find our values for R and L.

So going ahead and doing that, first of all putting in our values for

I and V, we have 3.35 the angle of 26.6 degree for

our current and our voltage value is 12 at an angle of 0 degrees.

Our current I is a complex conjugate of the current value we found here.

So I need to put a negative sign in front of the 26.6 degrees

if we're looking at the current and not the complex conjugate of the current.

And if we continue with that and

break it down into rectangular coordinates that we have.

0.25 minus J 0.125.

And again, that's equal

to one over R plus one over J4L.

And so if we equate the real parts of the two expressions and

also the imaginary parts of the two expressions we can find our value for

R and L.

Our R, which is 1 over 0.25 comes out to be 4 and

our inductance comes out to be 2 Henries.

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