# Homework Help: Find phasor voltages, current, phasor diagrams

1. Mar 11, 2013

### Color_of_Cyan

1. The problem statement, all variables and given/known data

http://imageshack.us/a/img842/2954/homeworkprobsg41.jpg [Broken]

a. Find the phasors for the current and the voltages of the circuit

b. Construct a phasor diagram showing Vs, I, VL, Vc, and VR

2. Relevant equations

phasor current i = V/Z

ic(t) = Ipeakcos2πft

Ipeak = Vpeak*2πfc

Z = Re{Z} + j Im{Z} (rectangular form),

or

Z = M (cos a + j sin a) (polar form)

V(t) = Vp cos (w t + q v)

i(t) = ip cos (w t +q i)

3. The attempt at a solution

Not experienced with AC RLC circuits.

All I think I know here is that the phasor voltage is 10V?

But am I going to have to break it up finding the V across each element too? If so, any tips for where I can start with that? Is there an equation I have to differentiate?

Would it be best to put it in polar form or rectangular form?

This has scared me a lot so far.

Last edited by a moderator: May 6, 2017
2. Mar 14, 2013

### Staff: Mentor

Step 1: With the voltage being 10
express the current in polar form (as a magnitude and an angle).

Step 2: Now that you know the current through each element, you can write the polar form of the voltage across each. For example, the voltage across the resistor is in phase with the current; the voltage across the inductor is 90° ahead of that current phasor, etc.

3. Mar 18, 2013

### Color_of_Cyan

Sorry for getting back to this really late

So it that 0° because there was no angle given?

Also it's polar form so

Z = M (cos a + j sin a)

Would the polar form then be

Z = 10 (cos 0 - jsin90)

?

4. Mar 18, 2013

### Staff: Mentor

You have to set something as reference, and I just nominated voltage. You could choose current to be 0° and express the voltage relative to that.

I wouldn't have chosen Z as the symbol for a voltage, especially when there are impedances involved in the analysis.

If V = V0°,

then V = V(cos0° + j.sin0°)

5. Mar 18, 2013

### Color_of_Cyan

I got that as the formula, it said Z instead and didn't mean impedance, but yes.

So V in Polar Form is :

V = 10(cos0° + jsin0°) ?

Is that one of the phasors then?

6. Mar 18, 2013

### Staff: Mentor

Yes.

7. Mar 18, 2013

### Color_of_Cyan

Is that the only expression that accounts for just the current source then? How would I go about then finding it through the inductor, capacitor, etc? Do you need the phasor voltages through them as well?

How would I go about calculating the current from it too? Could I still apply DC laws to get the current phasor? (ie, the phasor voltage over the resistance?)

8. Mar 18, 2013

### Staff: Mentor

To find the current, you divide the voltage by the circuit impedance (at the frequency of the applied AC). First, you may find it easiest to express the total impedance as a complex number.

9. Mar 18, 2013

### Color_of_Cyan

Okay, but there wasn't any given frequency? (Unless you're supposed to get that from the given equation on the voltage source)

So the impedance is ZR + ZL + ZC

ZL = jωL

ZC = 1/jωC

ZR = 100 ohms the same as the resistor

Question, does ω = 104?

If so then

Z = 100 + 1/j(104)(0.2 x 10-6) + j(104)(50 x 10-3)

Z = 100 + j500 + j500

Z = (100 + j500)Ω ?

Then I'm assuming the current phasor would then be

I = 10V(cos0° + jsin0°)/(100 + j500)Ω

?

10. Mar 18, 2013

### Staff: Mentor

Yes.
Correct so far, but the next line can't be right ....

11. Mar 20, 2013

### Color_of_Cyan

Sorry for getting back to this late again.

So it would've been j500 - j500 then, because the formula for Zc was actually

= -j/ωC ?

so then just

Z = 100 + 0jΩ

?

Am I going to have to convert that also?

I heard the way to convert it to express it (ie the impedence) in X∠° form was something like

Zmagnitude = (R + X)1/2[/sub]

then Zangle = tan-1(R/X)

where Z was the total impedance and X was the magnitude of imaginary impedance, and R the real impedance from form:

Z = R + jX

If so, then Z is 100∠ 90°. Then I would be V/Z so

I = (10∠ 0°)/(100 ∠ 90°)

then

I = (0.1∠ -90°)A ?

12. Mar 21, 2013

### Staff: Mentor

Assuming your arithmetic is correct, then this indicates the circuit is resonant at this applied frequency, with the reactances cancelling to leave you with just the resistance of the circuit.

You heard some rumour? But aren't you taking lectures in this topic?

Anyway, you are on the right track but not quite correct. Better check with your textbook to find what it should be.

After that, your method is correct.

13. Mar 22, 2013

### Color_of_Cyan

I am / was but it's kind of difficult.

And that's what the textbook says actually (it even confuses me). Can I ask what is incorrect about the equations there? Or did I just do it for the wrong form?

14. Mar 27, 2013

### Staff: Mentor

You'll find there is a lot of helpful electrical theory on the net. http://www.allaboutcircuits.com/vol_2/chpt_2/5.html