Electrical Engineering - AC Circuits Phasors

In summary, the student is struggling to solve part (b) of the homework and does not understand how to draw the voltage phasor diagram and current phasor diagram.
  • #1
YoshiMoshi
228
8
on the homework help forums, please present the problem statement as text or LaTex and not as an image

Homework Statement



Please see attached image 1 and image 2

Homework Equations

The Attempt at a Solution



Alright so I believe I solved part a correctly. It was very easy.

(A) 5 * phasor of 30 degrees

However it's been a while since I took circuits 1, so I'm struggling a bit with part (b). You can see my work in image 3 attached. For the phasor diagram for the voltage I drew IR as the reference along the positive x axis, IX point upwards, and the hypotenuse as ZI=V at an angle of 30 degrees from the reference. I believe this is correct?

I'm not exactly sure how to solve for the current here. I know ohms law V = IR, and I know that generally speaking we can write alternating current in terms of trigonometric functions, so the voltage as a function of time is something along the lines of

V(t) = A*sin(w*t + phase)

The problem specifies that it's at 60 Hz and I know that

w = 2*pi*f = 120*pi

So I know that the voltage is of the form

V(t) = 120*sin(120*pi*t + phase) volts

So

I = V/Z, but I can't seem to remember how to get the voltage into phasor form when I don't know the phase angle, hence I'm not sure how to draw the single phase diagram, or how to get the phasor current in this case. I guess I have a hard time putting the English into mathematical equation.

(C) I'm not exactly sure how to draw these graphs

For the voltage I drew the max as 120 volts, the min as -120 and I found the average as (2*max)/pi. I found this formula online, I believe it's correct?

I guessed that the current was in the form of 120/5 or 24 times some phasor

I(t) = 24 * phasor so I drew the max as 24 the min as -24 and used the formula for average above

For the power S I believe it's of the form sqrt(3)2880*some phasor

I don't see how to draw these graphs without given the initial phase angle of the voltage. Any help in solving either of these parts would be greatly appreciated. Thanks guys!
IMG_20150914_165442478_HDR.jpg

IMG_20150914_165504772_HDR.jpg
IMG_20150914_165616866 (1).jpg
 
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  • #2
YoshiMoshi said:
V(t) = 120*sin(120*pi*t + phase) volts

in phasor form
V(s)= 120 angle phase

YoshiMoshi said:
For the voltage I drew the max as 120 volts, the min as -120 and I found the average as (2*max)/pi. I found this formula online, I believe it's correct?
are you sure about that formula? think about what a sign wave is. sin(x) what is the average. Think about what a sign wave does.

V=I/Z
ohms law should help you find the current.

if no phase is given, typically you can assume zero.

power=voltage*current
 
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Likes YoshiMoshi
  • #3
ok so
V = 120 phase angle
Z = 5 phase angle 30 degrees

V = IZ

so

I = V/Z = 24 phase angle -30 degrees

S = IV = (24 phase angle -30 degrees)(120) = 2880 phase angle -30

IR drop is 24 phase angle -30 degrees ( (5*sqrt(3))/2) ~ 103.923 phase angle -30 degrees

IX drop is 24 phase angle -30 degrees (5/2) ~ 60 phase angle -30 degrees

So I seem to get that with your post, but how exactly do I draw the phase diagram of these things?

I know that I can draw Z = (5*sqrt(3))/2 + j (5/2) with a phasor with the (5*sqrt(3))/2 along the positive x axis, 5/2 along the positive y axis, and the other side of the triangle as Z = (5*sqrt(3))/2 + j (5/2) with a angle of 30 degrees

I can assume that the voltage has a phase angle of 0 degrees? In that case the phase diagram is just 120 along the positive x axis?

For the power phasor diagram, I can express S = 2880 cos(-30 degrees) + j 2880 sin(-30 degrees)
S = 1440 - j 1140*sqrt(3)
so the phasor diagram would just be 1440 in the positive x direction, 1140*sqrt(3) in the negative y direction, the other side of the triangle would be S = 1440 - j 1140*sqrt(3), and the angle would be -30 degrees

But how do I draw a phasor diagram for IR and IX drop is this talking about the real part of the current times the real part of impedance along with the imaginary part of the current times the imaginary part of the impedance? Isn't this just the phasor diagram mentioned earlier of the voltage? Kind of confused about this one.

Thanks for your help!
 

Related to Electrical Engineering - AC Circuits Phasors

1. What is a phasor in electrical engineering?

A phasor is a representation of a sinusoidal waveform that includes both magnitude and phase information. It is often used in the analysis of AC circuits to simplify calculations and visualize the behavior of the circuit.

2. How do you convert between a phasor and a sinusoidal waveform?

To convert from a phasor to a sinusoidal waveform, you can use the trigonometric identities of sine and cosine functions. The magnitude of the phasor represents the amplitude of the waveform, and the phase angle represents the shift in time.

To convert from a sinusoidal waveform to a phasor, you can use the magnitude and phase information to calculate the complex number representation of the phasor.

3. What is the difference between a series and parallel AC circuit?

In a series AC circuit, all components are connected in a single loop, and the same current flows through each component. In a parallel AC circuit, components are connected in multiple branches, and the total current is divided among the branches. Additionally, the voltage is the same across all components in a parallel circuit, while it is divided among components in a series circuit.

4. How do you calculate impedance in an AC circuit?

Impedance is the total opposition to the flow of current in an AC circuit, and it includes both resistance and reactance. To calculate impedance, you can use Ohm's Law (Z = V/I) with complex numbers. The real part of the complex impedance represents the resistance, and the imaginary part represents the reactance.

5. What is the significance of phase angle in an AC circuit?

The phase angle in an AC circuit represents the shift in time between the voltage and current waveforms. It is crucial in determining the power and energy flow in the circuit and can also indicate the type of load (resistive, inductive, or capacitive) in the circuit.

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