Find phasor voltages, current, phasor diagrams

[Color_of_Cyan]

Homework Statement

http://imageshack.us/a/img842/2954/homeworkprobsg41.jpg


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

b. Construct a phasor diagram showing V_s, I, V_L, V_c, and V_R

Homework Equations

phasor current i = V/Z

ic(t) = I_peakcos2πft

I_peak = V_peak*2πfc



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

or

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


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

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

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.

 

[NascentOxygen]

Step 1: With the voltage being 10_∠0°
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.

 

[Color_of_Cyan]

Sorry for getting back to this really lateSo 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)

?

 

[NascentOxygen]

Sorry for getting back to this really late


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

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.

Also it's polar form so

Z = M (cos a + j sin a)

Would the polar form then be

Z = 10 (cos 0 - jsin90)

?

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

If V = V_∠0°,

then V = V(cos0° + j.sin0°)

 

[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?

 

[NascentOxygen]

So V in Polar Form is :

V = 10(cos0° + jsin0°) ?

Is that one of the phasors then?

Yes.

 

[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?)

 

[NascentOxygen]

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.

 

[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 Z_R + Z_L + Z_C

Z_L = jωL

Z_C = 1/jωC

Z_R = 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

?

 

[NascentOxygen]

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

Yes.

So the impedance is Z_R + Z_L + Z_C

Z_L = jωL

Z_C = 1/jωC

Z_R = 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)

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

Z = 100 + j500 + j500

 

[Color_of_Cyan]

Sorry for getting back to this late again.

So it would've been j500 - j500 then, because the formula for Z_c 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

Z_magnitude = (R + X)<sup>1/2[/sub]

then Z_angle = 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 + jXIf so, then Z is 100∠ 90°. Then I would be V/Z so

I = (10∠ 0°)/(100 ∠ 90°)

then

I = (0.1∠ -90°)A ?</sup>

 

[NascentOxygen]

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

= -j/ωC ?


so then just

Z = 100 + 0jΩ

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.

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

Z_magnitude = (R + X)<sup>1/2[/sub]

then Z_angle = 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</sup>

<sup>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.</sup>

 

[Color_of_Cyan]

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.

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?

 

[NascentOxygen]

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?

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