How to computer the impedance of parallel resistor/cap combo

So in trying to learn how to quickly compute the impedance of parallel resistor combo:


using the impedance for a capacitor Xc=1/(2*pi*f*C), with
C = 0.47uF
R1 = 1k
R2 = 1K
f = 60 Hz

Seems like I should have a total impedance value of 459 ohms. But running the simulation gives me a much different value. Can anyone tell me what I should get with this circuit?

Your source voltage is RMS so peak would be 1.41 time higher. Did you consider that?

ronv said:

Your source voltage is RMS so peak would be 1.41 time higher. Did you consider that?

Click to expand...

im computing impedance here, so the amount of voltage shouldn't really be taken into consideration right?, only the frequency is used from the voltage source. I mean if I did just resistors with a DC source, voltage doesn't really come into play when measuring total resistance.

Also Im getting values of 14-17kOhms, much higher than 1.41x the computed ~500 ohms.

and if anyone cares, here's how I computed the impedance.
First I found the impedance of the cap @ 60Hz, so

Xc = 1/(2*pi*60*0.47*10^-6) = 5.64 kOhms. As all three passive components are parallel to each other, I computed this as

1/[1/5644 + 1/1000 + 1/1000] = 459 ohms.

And whats up with the latex editor? When I enter any text, i get a broken image link below my script.

Hello guys,


Well, the peak voltage, rms voltage, or peak to peak voltage will not affect the impedance unless the circuit is non linear.

The impedance of the cap zC is:
zC=1/(j*w*C1)
and then put that in parallel with the two resistors also in parallel and find the magnitude and you should come up with close to 498 ohms at 60 Hz.
The impedance is really complex though, so it would be:
Z=496.1062091109588-43.95149413880021*j
as the 498 is the magnitude of the impedance.

Note that if you dont take the complex form into consideration you get 5643.8 in parallel with 500 ohms which works out to 459.3 ohms which is not correct. That's because the impedance magnitude is not the parallel combination of the capacitive reactance and the resistive part, it is the norm of the complex impedance.

MrAl said:

Hello guys,


Well, the peak voltage, rms voltage, or peak to peak voltage will not affect the impedance unless the circuit is non linear.

The impedance of the cap zC is:
zC=1/(j*w*C1)
and then put that in parallel with the two resistors also in parallel and you should come up with close to 498 ohms at 60 Hz.
The impedance is really complex though, so it would be:
Z=496.1062091109588-43.95149413880021*j
as the 498 is the magnitude of the impedance.

Click to expand...

I realize its complex, but trying to avoid using Laplace transforms and the ilk, as contantly computing those made me lose sight of what I was doing. So I was trying to simplify the subject and get back to the roots.
I computed 458 ohms compared to your 496 ohms, but close enough relative to what I actually got from the simulation.

No idea how multisim is getting values in the 14-17 kohms range, which is way too high.

LTSpice likes Mr. Al's number. Maybe it is just multisim.

ronv said:

LTSpice likes Mr. Al's number. Maybe it is just multisim.

Click to expand...

Here is how to get LTSpice to show you the complex impedance numbers directly. Same circuit, except I used AC analysis. I forced it to compute the AC values at just three points, 59.99Hz, 60Hz and 60.01Hz (.ac lin 3 59.99 60.01)

Now we want the impedance of the network which is the voltage at node A divided by minus of the current through the source. So when you see the expression (V(a)/-I(V1)), think impedance, Z.

I used two LTSpice plot panes set to Cartesian mode (as opposed to Bode mode). The upper one shows a Plot of the Magitude of the complex impedance, which in LTSpice's Plot arithmetic uses the Mag() function, so I plot Mag(V(a)/-I(V1)). Ignore the red dashed line which would normally show the imaginary term which is zero by definition. The cursor shows that Mag(Z) is 498.049Ω at 60Hz.

The lower plot pane shows the Cartesian version (Real and Imaginary parts separately) where the solid green line is the real part, and the dashed green line is the imaginary part. The cursor shows that Re(Z)=496.106Ω and that Im(Z)=-43.9515Ω

Not bad for a free program, huh?

shosh said:

I realize its complex, but trying to avoid using Laplace transforms and the ilk, as contantly computing those made me lose sight of what I was doing. So I was trying to simplify the subject and get back to the roots.
I computed 458 ohms compared to your 496 ohms, but close enough relative to what I actually got from the simulation.

No idea how multisim is getting values in the 14-17 kohms range, which is way too high.

Click to expand...

Hi again,

You dont have to use Laplace Transforms to get the required result. You only need to use complex numbers. Using the complex numbers is just a little bit harder, but not too bad really. If you want an example i can provide one for you a little later today. You can also check using the simulators as other posters here have shown.

MrAl said:

You dont have to use Laplace Transforms to get the required result. You only need to use complex numbers. Using the complex numbers is just a little bit harder, but not too bad really. If you want an example i can provide one for you a little later today. You can also check using the simulators as other posters here have shown.

Click to expand...

Using the complex impedance is the same as using the Laplace transform with s --> jw

So I figured out my problem, I [mood=shameful] used a ohm meter in multisim to measure impedance [/mood].
So this leads me to another question, whats the point of equations such as

Xc = 1/(2*pi*f*C) and
X_L = 2*pi*f*L

If you can't get accurate results from them without Laplace.

Why not? Like dougy said, you don´t need laplace to solve them, just normal calculation and remembering that 1/i=-i and i*i=-1.
500Ω || Zc
Zc=1/jwC=-i/wc=-i/(60*0.47u)=-5646iΩ
500Ω|| -5646iΩ= (500*-5646i)/(500+ -5646i)=496 -44i