Basic AC Circuit Question on Equivalent Impedance

[captjackAV]

Homework Statement

If R1 = 2 Ω, X1 = 1 Ω, R2 = 2 Ω, and X2 = -2 Ω , the equivalent impedance Zeq (in ohms) at terminals a and b is:

(the circuit diagram is attached to this post)

a. 3 + j 0
b. 3 + j 2
c. 3 - j 2
d. 0 - j 2
e. 0 + j 2

The answer is a. I have no idea how to get the answer. I have tried to simplify the circuit but I do not think I am doing it right.

Homework Equations

Series: Zeq = Z1 + Z2
Parallel: Zeq = (Z1*Z2)/(Z1 + Z2)
Total Impedance is Z= r +jX
Zc = -jXc
ZL = j(XL)

The Attempt at a Solution

I tried to:
1. combine R2 and and the capacitor in parallel (i think my errors occur here) i got (2+j) + (-4j/(2-2j))
2. combine R1 and the inductor in series (i got 2 +j1)
3. combine the rest in series.

 

[vela]

Homework Statement

If R1 = 2 Ω, X1 = 1 Ω, R2 = 2 Ω, and X2 = -2 Ω , the equivalent impedance Zeq (in ohms) at terminals a and b is:

(the circuit diagram is attached to this post)

View attachment 61215

a. 3 + j 0
b. 3 + j 2
c. 3 - j 2
d. 0 - j 2
e. 0 + j 2

The answer is a. I have no idea how to get the answer. I have tried to simplify the circuit but I do not think I am doing it right.

Homework Equations

Series: Zeq = Z1 + Z2
Parallel: Zeq = (Z1*Z2)/(Z1 + Z2)
Total Impedance is Z= r +jX
Zc = -jXc
ZL = j(XL)

The Attempt at a Solution

I tried to:
1. combine R2 and and the capacitor in parallel (i think my errors occur here) i got (2+j) + (-4j/(2-2j))
2. combine R1 and the inductor in series (i got 2 +j1)
3. combine the rest in series.

It looks like the result you have to step 1 already includes R1 and the inductor. It is, in fact, correct and simplifies to the correct answer for the problem. For just R2 and the capacitor, the impedance is given by the second term.

 

[captjackAV]

Thanks for pointing out my mistake. So when I try to simplify (2+j) + (-4j/(2-2j)), I am getting a fraction. How do I simplify this correctly? and what happens if there is a j^2.

 

[vela]

First, $j^2 = -1$. You want to get j out of the denominator of the second term. The standard technique would be to multiply the top and bottom by the conjugate of the denominator, i.e. by $\frac{2+2j}{2+2j}$.

 

[rezeew]

(i got 4 + j1)

Your attempt at solving the problem is partially correct. Here is a step-by-step solution to help you understand how to get the answer:

1. Simplify the components in series: R1 and X1.
Z1 = 2 + j1

2. Simplify the components in parallel: R2 and X2.
Z2 = (2 * (-2j)) / (2 + (-2j))
Z2 = (-4j) / (2 - 2j)
Z2 = (-4j) * (2 + 2j) / (2 - 2j) * (2 + 2j)
Z2 = (-8 - 8j) / (4 + 4j - 4j - 4j^2)
Z2 = (-8 - 8j) / (4 + 4)
Z2 = (-8 - 8j) / 8
Z2 = -1 - j

3. Combine the simplified components in series: Z1 and Z2.
Zeq = Z1 + Z2
Zeq = (2 + j1) + (-1 - j)
Zeq = 1 + j

Therefore, the equivalent impedance at terminals a and b is 1 + j ohms. This can also be written as 1 + j0 ohms, which is the same as option a.

It is important to note that when combining components in parallel, you need to use the formula Zeq = (Z1*Z2)/(Z1 + Z2). You also need to be careful with the signs of the imaginary components (j) when simplifying in parallel. In this case, the negative sign on X2 becomes positive when combined with R2 in parallel.

I hope this helps you understand how to solve this type of problem. Remember to always double check your calculations and pay attention to the signs of the components. Good luck with your studies!