EXERCISE 6 - Phasors and Circuit Analysis

 

·       In this exercise section we explore a more complicated circuit and the use of phasors in its analysis. This in-depth example is from the online site for the EE text Fundamentals of Electrical Circuits by Alexander and Sadiku.

 

Circuit Exercise:

 

Consider the circuit below.

The ultimate task is to find i(t). To begin, we want to transform the circuit into the phasor domain. But there's a problem. One of the sources contains a sine, not a cosine. It will therefore be necessary to convert the sine to a cosine.

120sin(200t - 8°) = 120cos(200t - q)

What is q?    

To convert sine to cosine, subtract 90 degrees.

 The result is 120cos(200t - 8 degrees - 90 degrees)= 120cos(200t-98o).

 

The circuit below now shows both sources as cosines.

Now let's convert the voltage source to its phasor value. It looks like this: M Ðq.

What is M?

The answer is 120.

120cos(200t - 98°) V converts to 120/-98° V.
4cos(200t - 60°) A converts to 4/-60° A.

Now let's convert the inductor to the phasor domain. It looks like this: jX.

What is X?

L converts to j x angular frequency x L.

700mH is 0.7H.

Thus, ZL = jwL = j (200)(.7) = j140 W

The answer is 140.

120cos(200t - 98°) V converts to 120/-98° V.
4cos(200t - 60°) A converts to 4/-60° A.

It is not necessary to convert the resistor value.

Now let's convert the capacitor to the phasor domain. It looks like this: jX.

What is X?

C converts to 1/(j x angular frequency x C).

That's 1/[j(200)(.00004)].

When you move j to the numerator, it changes sign.

The X value will therefore be negative.

Calculate 1/[(200)(.00004)] and change the sign.

Thus, ZC = 1/jwC = -j/wC = -j/(200 x 40 x 10-6) = -j125 W

The answer is -125.

120cos(200t - 98°) V converts to 120/-98° V.
4cos(200t - 60°) A converts to 4/-60° A.
700mH converts to j140 W.
It is not necessary to convert the resistor value.
i(t) converts to the phasor I.

The transformed circuit below reflects these values.

There are several ways to solve for I. One of the best ways (for this particular circuit) is the mesh current method. Notice that mesh currents have been entered on the above circuit. It is only necessary to write one mesh current equation, since the current in the mesh on the right is already known. This equation for the mesh on the left is:

120/-98° = j140I + XI - j125 [I - (-4/-60°)]

What is X?

It's the resistor value.

The answer is 25.

The mesh equation is then:
120/-98° = j140I + 25I - j125 [I - (-4/-60°)]

After grouping terms, the mesh equation becomes:
120/-98° = (j140 + 25 - j125) I - j500/-60°

The j500/-60° term is a bit inconvenient, as it mixes rectangular coordinates with polar coordinates. Nevertheless, the term can easily be converted to straight polar form. This will be:

500/q

What is q?

Multiplying by j is the same as multiplying by one at an angle of 90 degrees.

So just add 90 degrees to the angle.

The answer is 30.

Thus, j500/-60° = 500/-60° + 90° = 500/30°

Our mesh equation is now:
120/-98° = (j140 + 25 - j125) I - 500/30°

After simplifying, the mesh equation becomes:
I = (120/-98° + 500/30°)/(25 + j15)

I can be resolved into polar form, M/q.

What is M? Round your answer to the nearest integer.

The numerator simplifies to 416.3 + j131.2.

In polar form, the numerator is 436.5 at an angle of 17.49 degrees.

In polar form, the denominator is 29.15 at an angle of 30.96 degrees.

Dividing, you get 14.97 at an angle of -13.47 degrees.

Just round off the 14.97.

The answer is 15.

Thus, I = (120/-98° + 500/30°)/(25 + j15) = (416.3 + j131.2)/(25 + j15)
I = 436.5/17.49°/29.15/30.96° = 14.97/-13.47° ~ 15/-13.5° amps

This is the phasor current I.

The phasor I must now be converted back into the time domain.

i(t) = 15cos(wt - 13.5°) A

What is w?

You may have to back up a few screens to find the original angular frequency.

Use the angular frequency of the voltage source and the current source.

The answer is 200.

i(t) = 15cos(200t - 13.5°) A. Therefore the problem is solved.

 

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