Engineering 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 $\frac{2+2j}{2+2j}$.