Is this the right way to find impedance using complex power and volta

[pokie_panda]

Homework Statement

The load (Z) in the circuit below absorbs a complex power of 54+j46 VA when the voltage across it is 3cos(400t+-60)V. Use this information to determine the impedance of the load

Homework Equations

P = V^2 / Z

The Attempt at a Solution

first convert V to polar form
3 < -60
convert power to polar form
70.9<40.4
Apply the formula
V^2/P= 9<-120/70.9<-40.4
=.1268<-160.4
convert to rectangular

Z=-0.11947- j0.04248 ohms

I'm not sure if the answer is correct

 

[ShreyasR]

Nope. U can't get such funny angles... take 54 as real power, which is equal to VI cos∅. and take 46 as reactive power which is VI sin∅. You have already found the phase difference (∅) between V and I as 40.4... Now Find RMS value of V from the voltage expression given, and hence the impedance using V and I.

 

[ShreyasR]

And BTW, Z=R+jX... R cannot be negative!

 

[pokie_panda]

Nope. Z cannot be negative... IMPOSSIBLE! Now take 54 as real power, which is equal to VI cos∅. and take 46 as reactive power which is VI sin∅. You have already found the phase difference (∅) between V and I as 40.4... Now Find RMS value of V from the voltage expression given, and hence the impedance using V and I.

Is this what your trying to say ?

So Vrms = 3/√2

Therefore solve for Irms

which is 54 = 3/√2Icos(40.1)
I=1.66

Z=(3/√2) / 1.66

Then find Z but you don't have an imaginary part ?

 

[ShreyasR]

I will be 39.439 A... You missed a bracket there... 54 = (3/√2) I cos(40.4)... You have V=3cos(400t+-60)... Current I = 39.439√2 cos(400t+-60+40.4)... Now u can find the imaginary part? This is a complicated question...

 

[ShreyasR]

About that part... V=3cos(400t+-60)... What do u mean by '+-'? Can you please be clear?

 

[pokie_panda]

I just copied the function off the question, I'm assuming it's negative
Therefore
V(t)=3<-60
I(t)=39.439√2<-20.4

Z= V/I
=.0537<-39.6

X=3/29.432sqrt(2)sin(-40.4)

.0537 -j0.03486

Is that part correct