Find phasor current (impedance, etc.), finding polar form

[Color_of_Cyan]

Homework Statement

A 90Ω resistor, a 32 mH inductor, and a 5μF capacitor are connected in series across the terminals of a sinusoidal voltage source Vs = 750cos(5000t + 30)V.

Calculate the phasor current.

Homework Equations

phasor current i = V/Z

V in polar form = (Magnitude)(cos a + j sin a)

Z (inductor) = wLj

Z (capacitor) = -j/ωC

Z (resistor) = r

total impedance = Z inductor + Z capacitor + Z resistorConversion to other form of Z:

Z(mag) = (R²+X²)^1/2

Z angle = tan^-1(R/X)

from the form Z = R + jx

The Attempt at a Solution

Not sure how I would convert the voltage to polar form and then I could find the current.

But for the impedance it would be:

ω = 5000,

so, Z (inductor) = 5000*(32 x 10^-3) = 160j

Z (capacitor) = -1j/(5000*(5 x 10^-6)) = -40j

Z (r) = 90Ω

So Z = 90 +120jΩ

Then to polar form would be

Z(mag) = (R²+X²)^1/2

Z(angle) = tan^-1(R/X)So Z(mag) = 150

Z(angle) = 36.8 degZ = 150 ∠ 36.8 degrees, for sureNow for the voltage I'm not so sure, but I'm guessing the angle is just 30 so

V = 750∠30 degSo if the voltage is correct then would the phasor current be

I = (750∠30 deg) / (150 ∠ 36.8 deg)

I = (5 ∠ -6.8 deg)A ?Thank you.

 

[gneill]

The polar form of your impedance is not correct. How can the magnitude be less than that of of the individual components that comprise it?

 

[Color_of_Cyan]

I see. I've edited the original post, so the components were supposed to be squared first then?So what about for converting the voltage now too,I got 150 for the impedance magnitude now.

Is the phasor current then I = 5 ∠ -6.8 deg)A now then?

 

[gneill]

I see. I've edited the original post, so the components were supposed to be squared first then?


So what about for converting the voltage now too,


I got 150 for the impedance magnitude now.

Is the phasor current then I = 5 ∠ -6.8 deg)A now then?

The magnitude looks fine now, but your angle does not.

So Z = 90 +120jΩ

Then to polar form would be

Z(mag) = (R2+X2)1/2

Z(angle) = tan-1(R/X)


So Z(mag) = 150

Z(angle) = 36.8 deg $\leftarrow$ atan(120/90) = ?


Z = 150 ∠ 36.8 degrees, for sure


Now for the voltage I'm not so sure, but I'm guessing the angle is just 30 so

V = 750∠30 deg


So if the voltage is correct then would the phasor current be

I = (750∠30 deg) / (150 ∠ 36.8 deg)

I = (5 ∠ -6.8 deg)A ?

Be careful to place the imaginary component over the real component to form the tan of the angle for the complex number.

 

[Color_of_Cyan]

I keep making these silly mistakes ;/ I had the R/X switched around. It's 53.1 deg

So Z = 150 ∠ 53.1 deg then.

So is I = (5 ∠ 16.3 deg)A then? Any other mistakes I made?

 

[gneill]

I keep making these silly mistakes ;/


I had the R/X switched around. It's 53.1 deg

So Z = 150 ∠ 53.1 deg then.

So is I = (5 ∠ 16.3 deg)A then? Any other mistakes I made?

No, your current angle's not correct. The voltage angle is 30°, the impedance angle is 53.1°. You're calculating E/Z, so how should you handle the angles?

 

[Color_of_Cyan]

Would I just subtract them? 53.1 - 16.3 would be 23.1 deg then (don't know how I got 16.3, must have subtracted something wrong). So is I is (5 ∠ 23.1 deg)A then.I didn't know that you could just get the angle from the phase in the given voltage so that's why I asked though, thanks.

 

[gneill]

You do a subtractions, yes, but the order of the operands is important. The sign of your resulting angle is incorrect because you chose to subtract the numerator's angle from the denominator's angle. This is exactly wrong When you divide in polar form, you subtract the denominator's angle from the numerator's angle.

 

[Color_of_Cyan]

Alright, so the phasor current I then is

( 5 ∠ -36.8 )A then, right?

Thanks for the help again.

 

[gneill]

alright, so the phasor current i then is

( 5 ∠ -36.8 )a then, right?

Thanks for the help again.

30 - 53.1 = -36.8 ?

 

[Color_of_Cyan]

This is so bad I keep making all those mistakes :( I = (5 ∠ -23.1)A

 

[gneill]

This is so bad I keep making all those mistakes :(


I = (5 ∠ -23.1)A

That result looks good.