# Homework Help: Complex current along a resistor

1. Jul 21, 2010

### 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?

2. Jul 21, 2010

### 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.

3. Jul 21, 2010

### 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?

4. Jul 21, 2010

### xcvxcvvc

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.

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

5. Jul 21, 2010

### Ry122

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

6. Jul 21, 2010

### Zryn

Z = Impedance (complex resistance)

What does your book say is the correct value?

7. Jul 21, 2010

### Ry122

It says the correct answer is 34.5 watts

8. Jul 21, 2010

### xcvxcvvc

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

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