Solving A.C Circuits with complex numbers

In summary: Just remember to use the negative j axis for inductances and the positive j axis for capacitors. In summary, the problem is that 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.
  • #1
mattakun
7
0
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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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
 
  • #4
Ahh, my bad, Z should equal 20 + j30.16 instead of minus, because XL should be in the positive j axes not the negative.
 
  • #5
mattakun said:
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 [tex]j\omega L[/tex] and that of a pure capacitance is [tex]\frac{1}{j\omega C} = -\frac{j}{\omega C}[/tex]. You don't have to worry about the phasors when you work things out this way.
 

1. How do complex numbers help in solving A.C circuits?

Complex numbers are used in A.C circuits because they allow us to represent both the magnitude and phase of the voltage and current. This is important in A.C circuits because the current and voltage are constantly changing and have both real and imaginary components.

2. What is the difference between a real and imaginary component in A.C circuits?

In A.C circuits, the real component represents the resistance and the imaginary component represents the reactance. The reactance is dependent on the frequency of the A.C signal and can be inductive (positive) or capacitive (negative).

3. How do you represent A.C circuits using complex numbers?

A.C circuits can be represented using phasors, which are complex numbers that represent the magnitude and phase of the voltage or current. These phasors can be added, subtracted, and multiplied using the rules of complex numbers to solve A.C circuits.

4. What are the advantages of using complex numbers in A.C circuits?

Using complex numbers in A.C circuits allows for a simplified and efficient way to solve complex problems. It also allows for a better understanding of the behavior of A.C circuits, especially when dealing with reactive components like inductors and capacitors.

5. Are there any limitations to using complex numbers in A.C circuits?

One limitation of using complex numbers in A.C circuits is that it assumes ideal conditions, such as no resistance in the wires or perfect components. In reality, there will always be some resistance and imperfections in the circuit, which may affect the accuracy of the results obtained using complex numbers.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
9
Views
2K
  • Precalculus Mathematics Homework Help
Replies
21
Views
707
  • Precalculus Mathematics Homework Help
Replies
31
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
26
Views
2K
Replies
12
Views
1K
Replies
3
Views
789
  • Precalculus Mathematics Homework Help
Replies
3
Views
8K
Replies
1
Views
695
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
0
Views
415
Back
Top