# complex impedance of parallel RLC circuit

#### b_andries

Joined Nov 19, 2022
4
Hi,

I have a question which i can't solve. "Suppose that we connect a resistor of 25 Ohms, capacitor of 0.002µf, and a coil of 7,7 µH in parallel. We operate the circuit at 2 MHZ. What's the complex impedance?"
a) 8.1 + j22
b) 8.1 - j22
c) 22 + j8.1
d) 22- j8.1

So my reasoning is that the answer would be:
Z = R + (XL + Xc)
This gives : 25 + (j96.76 - j39.79) = 25 + j56.97
but this is not in one of the answers. What am I missing?

Thank you!

#### WBahn

Joined Mar 31, 2012
29,878
How do reactances combine when they are in parallel?

#### b_andries

Joined Nov 19, 2022
4
So the theory in my book says that Resistance and inductance combine in parallel (as in a parallel DC circuit), capacitances add up. But i don't think i can combine R with XL as in (1/ ((1/25)+(1/96.76)) or even if i take the net reactance (XL +XC) and then do do following (1/ ((1/25)+(1/56.97). I think i'm mixing up everything.

#### WBahn

Joined Mar 31, 2012
29,878
One of the beauties of complex impedance is that you can treat impedances just like resistances as far circuit equations are concerned. The math is just a bit more complicated because the values are complex numbers.

How would you combine three resistors, R1, R2, and R3, that are in parallel?

You do the exact same thing for three impedances, Z1, Z2, and Z3.

Don't make it harder than it is.

#### b_andries

Joined Nov 19, 2022
4
Ok thanks guys i Think i found it after your help. I came up with answer D

#### ericgibbs

Joined Jan 29, 2010
18,662
hi b_,
Good news, please post your results, they will help others with similar problems.
E

#### Irving

Joined Jan 30, 2016
3,810
I just 'wasted' an hour refreshing my ability to do this calculation from 1st principles, getting all the sin/cos/tan in the right places. lol. Suffice to say I got the right answer - well one that's in the list! Its quite instructive to do the LTSpice simulation as well... Its definitely easier to do the KCL/Ohms Law approach (as per the AAC article @ericgibbs linked to) than trying to manipulate raw re/im tuples1.

#### b_andries

Joined Nov 19, 2022
4
So as WBahn said, we combine Z1,Z2 and Z3 as resistances in paralallel:
1/Z = 1/ZR + 1/ZC + 1/ZL
1/Z = 1/R + j/Xc - j /XL
1/Z = 1/R + j(1/Xc - 1/XL)

Then we search the modulus of the complex number:
p = squareroot[((1/R)² + (1/Xc - 1/XL)²)] = 23.45
then we find the angle given by Arctan(R/Xc - R/XL) = -0.3543 (in radians)

if we then want to find the complex impedance:
Zr= 23.45 * cos(-0.3543) = 22 (rounded)
Zx = 23.45 sin(-0.3543) = -8.1 rounded

This gives Z = 22 - j 8.1

#### WBahn

Joined Mar 31, 2012
29,878
So as WBahn said, we combine Z1,Z2 and Z3 as resistances in paralallel:
1/Z = 1/ZR + 1/ZC + 1/ZL
1/Z = 1/R + j/Xc - j /XL
1/Z = 1/R + j(1/Xc - 1/XL)

Then we search the modulus of the complex number:
p = squareroot[((1/R)² + (1/Xc - 1/XL)²)] = 23.45
then we find the angle given by Arctan(R/Xc - R/XL) = -0.3543 (in radians)

if we then want to find the complex impedance:
Zr= 23.45 * cos(-0.3543) = 22 (rounded)
Zx = 23.45 sin(-0.3543) = -8.1 rounded

This gives Z = 22 - j 8.1
You really should get in the habit of tracking units throughout your work.

#### ericgibbs

Joined Jan 29, 2010
18,662
@WBahn
Would appreciate a double check.
The Article in post#5, appears to have an error in the Inductance current, could be just a transposition error of the 489.71mA, I make it 847.53mA.
E

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#### WBahn

Joined Mar 31, 2012
29,878
@WBahn
Would appreciate a double check.
The Article in post#5, appears to have an error in the Inductance current, could be just a transposition error of the 489.71mA, I make it 847.53mA.
E
Since the magnitude of the inductive impedance (245 Ω) is very close to the magnitude of the resistance (250 Ω), the magnitude of both currents should be very similar, with the inductor current being about 2% higher. So having a resistor current of 480 mA and an inductor current of 490 mA seems quite reasonable, while having it be nearly 850 mA would not make sense unless the resistor current were comparably higher, too.

How did you arrive at the 847.53 mA?

#### ericgibbs

Joined Jan 29, 2010
18,662
hi,
Resolved.
Comparison LTS/Spice result
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