Solving A.C Circuits with complex numbers

[mattakun]

Hi, here is the problem..

The potential difference across a circuit is represented by 40 + j25 volts, and the circuit consists of a coil with an inductance of 0.06H in series with a resistance of 20 Ohms. If the frequency is 80Hz find the complex number in rectangular form that represents the current in amperes.

I think that I am able to solve this problem, but I'm just not sure how to retrieve the voltage, is it as simple as converting the 40 + j25 into polar form? Which would make the voltage 47 Volts at an angle of 32? Is this correct?

I understand that the potential difference is the voltage, but I'm just not sure what to do with the complex number to begin with.

Thanks.

 

[Curious3141]

Why go to polar form at all? The question specifically asks you to give your answer in rectangular form, and the input voltage is in rectangular form.

Just work out the complex impedance of the circuit (in rectangular form) and divide the voltage by that value.

 

[mattakun]

Okay, this is what I've done:

I got the inductive reactance using the formula 2piFL so XL = 30.16 Ohms.
I believe that the resistor being 20 Ohms would make the complex impedance Z = 20 - j30.16 because XL is negative on the phaser diagram.

Now, I find that converting both the complex impedance and voltage into polar form makes for a less error prone calculation when dividing them. Then I convert the answer back into rectangular form to meet the criteria of the question.

Please correct me if I've gone wrong.

Thanks

 

[mattakun]

Ahh, my bad, Z should equal 20 + j30.16 instead of minus, because XL should be in the positive j axes not the negative.

 

[Curious3141]

Ahh, my bad, Z should equal 20 + j30.16 instead of minus, because XL should be in the positive j axes not the negative.

Right. It's simple when you realize that the complex reactance of a pure inductance is $$j\omega L$$ and that of a pure capacitance is $$\frac{1}{j\omega C} = -\frac{j}{\omega C}$$. You don't have to worry about the phasors when you work things out this way.