Complex current along a resistor

[Ry122]

If a current of .587e^j1.12 is passing along a 200ohms resistor, how would I go about calculating the power dissipated in the resistor?

 

[xcvxcvvc]

$$P = (V)(I^*)=(IZ)(I^*)=Z(II^*)$$

Now, I times I conjugate leaves the magnitude of I squared. Z is not complex, only resistive.

 

[Ry122]

the complex conugate of i multiplied by i is .0653 - .5487j isn't it?
Then do I just multiply the real part of this by the resistance?

 

[xcvxcvvc]

the complex conugate of i multiplied by i is .0653 - .5487j isn't it?
Then do I just multiply the real part of this by the resistance?

No. Complex conjugate means you take the vector, keep its magnitude, and add a negative to the angle. You can also think of it as adding a negative to the imaginary component of the vector.

$$I = .587e^{j1.12}$$
$$I^* = .587e^{-j1.12}$$
$$II^*=.587e^{j1.12}.587e^{-j1.12}=(.587)^2e^{j1.12-j1.12}= (.587)^2e^{0}=(.587)^2$$

 

[Ry122]

does Z represent the resistance?
.587^2 x 200 = 68 Watts which is wrong for some reason.

 

[Zryn]

Z = Impedance (complex resistance)

What does your book say is the correct value?

 

[Ry122]

It says the correct answer is 34.5 watts

 

[xcvxcvvc]

It says the correct answer is 34.5 watts

It's because you gave us peak current without saying! Almost always will you be given RMS and when it's not stated, RMS is assumed. Everything I said above is the same for peak values of voltage/current except you need to divide by 2.

$$\frac{I_p}{\sqrt{2}}=I_{rms}$$
So when you have I^2 (in rms) you have 1/sqrt(2) squared = 1/2