Find the Impedance and phase in polar form

I know that [tex]\omega=2\pi*f=4555.3[/tex] the [tex]Z_tot=5*10^3+\frac{1}{4555.3*47*10^{-9}}[/tex]

so overall R=2.33*10^7 and \theta=\frac{\pi}{2}?

 

Please check your math
http://www.wolframalpha.com/input/?i=-i+*4.67+++5

 

You lost j.

 

Find the Impedance and phase in polar form

I know that [tex]\omega=2\pi*f=4555.3[/tex] the [tex]Z_tot=5*10^3+\frac{1}{4555.3*47*10^{-9}}[/tex]

so overall R=2.33*10^7 and \theta=\frac{\pi}{2}?

Why are you forcing your tex code to display as code instead of rendering it? Unless you are asking something about LaTeX code, which doesn't appear to be the case.

Which is easier for your audience to read:

[tex]\omega=2\pi*f=4555.3[/tex] the [tex]Z_tot=5*10^3+\frac{1}{4555.3*47*10^{-9}}[/tex]

or

\(\omega=2\pi*f=4555.3\) the \(Z_tot=5*10^3+\frac{1}{4555.3*47*10^{-9}}\)

Notice how this makes it much easier to spot the missing j in the denominator.

But even with out that, I don't see how you are getting a magnitude of 23 MΩ and an angle of 90°. This is were some quick estimation can really play off. For the numerical part, call 4555.3 ~5000 and call 47 ~50. That means that the denominator has 25 x 1000 x 10 x 10^-9 which is 25 x 10^-5 Call that 25 100/4 and now your fraction is 4/(100x10^-5) = 4/10^-3 = 4 k. Add that to 5 k and you have 9 k (for the numerical part) (since you dropped the j). This is a long, long way from 20+ MΩ !

As for the angle, you know that the capacitor has an angle of -90° and that the resistance can only reduce it towards the real axis, hence the total impedance angle must be between 0° and -90°. With the estimate above that the magnitude of the capacitive reactance is about 4 kΩ, you know that you can expect the impedance angle to be close to, but probably less (in magnitude), than -45°.

Here would be my recommendation for how to present something like this:

\(
Z_{tot} \; = \; R \; + \; \frac{1}{j \omega C}
\)

With

\(
\omega \; = \; \( 2 \pi \, \text{\frac{rad}{cycle}} \) \( 725 \, \text{Hz} \) \; = \; 4555 \, \text{s^{-1}}
\)

we have

\(
Z_{tot} \; = \; 5 \, \text{k\Omega} \; + \; \frac{1}{j \( 4556 \, s^{-1} \) \. \( 47 \. \text{nF} \) }
\;
Z_{tot} \; = \; \( 5 \; - \; j \, 4.67 \) \, \text{k\Omega }
\)

 

Why are you forcing your tex code to display as code instead of rendering it? Unless you are asking something about LaTeX code, which doesn't appear to be the case.

Which is easier for your audience to read:

[tex]\omega=2\pi*f=4555.3[/tex] the [tex]Z_tot=5*10^3+\frac{1}{4555.3*47*10^{-9}}[/tex]

or

\(\omega=2\pi*f=4555.3\) the \(Z_tot=5*10^3+\frac{1}{4555.3*47*10^{-9}}\)

Notice how this makes it much easier to spot the missing j in the denominator.

But even with out that, I don't see how you are getting a magnitude of 23 MΩ and an angle of 90°. This is were some quick estimation can really play off. For the numerical part, call 4555.3 ~5000 and call 47 ~50. That means that the denominator has 25 x 1000 x 10 x 10^-9 which is 25 x 10^-5 Call that 25 100/4 and now your fraction is 4/(100x10^-5) = 4/10^-3 = 4 k. Add that to 5 k and you have 9 k (for the numerical part) (since you dropped the j). This is a long, long way from 20+ MΩ !

As for the angle, you know that the capacitor has an angle of -90° and that the resistance can only reduce it towards the real axis, hence the total impedance angle must be between 0° and -90°. With the estimate above that the magnitude of the capacitive reactance is about 4 kΩ, you know that you can expect the impedance angle to be close to, but probably less (in magnitude), than -45°.

Here would be my recommendation for how to present something like this:

\(
Z_{tot} \; = \; R \; + \; \frac{1}{j \omega C}
\)

With

\(
\omega \; = \; \( 2 \pi \, \text{\frac{rad}{cycle}} \) \( 725 \, \text{Hz} \) \; = \; 4555 \, \text{s^{-1}}
\)

we have

\(
Z_{tot} \; = \; 5 \, \text{k\Omega} \; + \; \frac{1}{j \( 4556 \, s^{-1} \) \. \( 47 \. \text{nF} \) }
\;
Z_{tot} \; = \; \( 5 \; - \; j \, 4.67 \) \, \text{k\Omega }
\)

Hi,

Not to be nit picky here but isn't:

\(
\omega \; = \; \( 2 \pi \, \text{\frac{rad}{cycle}} \) \( 725 \, \text{Hz} \) \; = \; 4555 \, \text{s^{-1}}
\)

supposed to be:
w=a*(rads/cycle)*b*(cycles/sec)=x*a*b*(rads/sec) ?

 

Radians are dimensionless because they are the ratio of arc length to radius.

Example: The circumference of the unit circle is \(2 \pi r\)
The angle of one revolution is \( \frac {2 \pi r} {r}\)

 

Radians are dimensionless because they are the ratio of arc length to radius.

Example: The circumference of the unit circle is \(2 \pi r\)
The angle of one revolution is \( \frac {2 \pi r} {r}\)

Hi,

Sorry, cant agree with that, but i an still open to hear your argument for that reasoning.

For one example, go around the circle one time, that's 2pi radians. Go around the circle two times that is 2*2pi radians. Three times, 3*2pi radians, etc.
Now go around the circle one time in one second, that's 2pi radians per second. Go round the circle two times in one second, that's 2*2pi radians per second. Three times, 3*2pi radians per second.
If we dont know the rate we dont know the distance traveled in one second if this was a physical circle.

For another example, w is always written as in units of radians per second, and sometimes in Hertz which is cycles per second.

For another example, shaft rotation. w is in radians per second.

For another example, 100/sec. 100 what per second, feet, inches, rotations?

It's something per unit time. We can even have seconds per second, or minutes per second, etc. But without specifying the upper units we dont know what we have.

However, feel free to show a counter argument, i'd like to hear your reasoning maybe with an example or two. In any case though i think it is more concise in electronics to label it as radians per second.

 

The fact is that angles measured in radians have no dimensions since they are the ratio of an arc length, measured in units of length, to the radius, also measured in units of length. If you go around the circle twice then two times the circumference is \(4 \pi r\) and the radius is still \(r\) and the angle is still \(4 \pi\). I'm sorry but I don't see where your major malfunction comes from.

 

Hi,

Not to be nit picky here but isn't:

\(
\omega \; = \; \( 2 \pi \, \text{\frac{rad}{cycle}} \) \( 725 \, \text{Hz} \) \; = \; 4555 \, \text{s^{-1}}
\)

supposed to be:
w=a*(rads/cycle)*b*(cycles/sec)=x*a*b*(rads/sec) ?

1 Hz = 1 cycle/sec

So you can use either form. Just as it is not necessary to explicitly convert farads to seconds/ohm in order to combine it with 1/seconds to get 1/ohm in the denominator.

The unit of radian is dimensionless, so it can be included or not as makes sense. If is very reasonable to express the units of radian frequency as rad/sec. I chose not to so as to keep it out of the next step which is the impedance calculation.

 

Hi,

Sorry, cant agree with that, but i an still open to hear your argument for that reasoning.

For one example, go around the circle one time, that's 2pi radians. Go around the circle two times that is 2*2pi radians. Three times, 3*2pi radians, etc.
Now go around the circle one time in one second, that's 2pi radians per second. Go round the circle two times in one second, that's 2*2pi radians per second. Three times, 3*2pi radians per second.
If we dont know the rate we dont know the distance traveled in one second if this was a physical circle.

For another example, w is always written as in units of radians per second, and sometimes in Hertz which is cycles per second.

For another example, shaft rotation. w is in radians per second.

For another example, 100/sec. 100 what per second, feet, inches, rotations?

It's something per unit time. We can even have seconds per second, or minutes per second, etc. But without specifying the upper units we dont know what we have.

However, feel free to show a counter argument, i'd like to hear your reasoning maybe with an example or two. In any case though i think it is more concise in electronics to label it as radians per second.

If you have 100/sec, then you know that you DON'T have feet, inches, or rotations because those are dimensioned quantities and you know that you have 100 of a dimensionless quantity per second. By very widely accepted convention (among those that actually make any effort to be dimensionally consistent), unless the context clearly establishes something different, the dimensionless quantity used is the radian.

As stated before, the unit 'radian' floats in an out. For instance, consider something like

\(
\omega \; = \; \frac{1}{sqrt{LC}}
\)

The units on the right work out to 1/seconds. If you want the radians there you have to artificially introduce them. But that's fine, because whether they are there or not, nothing changes.

 

If you have 100/sec, then you know that you DON'T have feet, inches, or rotations because those are dimensioned quantities and you know that you have 100 of a dimensionless quantity per second. By very widely accepted convention (among those that actually make any effort to be dimensionally consistent), unless the context clearly establishes something different, the dimensionless quantity used is the radian.

As stated before, the unit 'radian' floats in an out. For instance, consider something like

\(
\omega \; = \; \frac{1}{sqrt{LC}}
\)

The units on the right work out to 1/seconds. If you want the radians there you have to artificially introduce them. But that's fine, because whether they are there or not, nothing changes.

Hi,

As expected, that is very reasonable

I am used to making it clear with rads/sec i guess because that's what most textbooks i have ever come across use. In other words you might consider it a little bit redundant, but that happens to be my preference so we can tell right away what is being talked about. It seems harder to assume that there could never be another dimensionless quantity, but yes the context should indicate that somehow.

Sometimes it matters, such as when we are talking about response vs frequency. Are we talking Hertz or rads/second in some graph. I suppose the text should make it clear, but sometimes they dont, but i guess we can blame them for that not really the convention of "to use or not to use" rads per second.

 

I tend to put radians into the units whenever it makes sense to do so, as well. I chose not to do so in this case only because of the following calculation. It was a judgment call and very arguably I chose wrong.

There should be no confusing 1/sec with Hz because Hz is cycles/sec -- a dimensioned quantity in the numerator.

One of my pet-peeves, but also one that I am guilty of because it is so common, is saying that

\(
\omega \; = \; 2\pi f
\)

This implies that ω and f have the same units (since 2π is dimensionless), but that makes no sense because for two quantities that have the same dimensions to be equal, the numerical portions must be equal -- but the very form of this equation guarantees that they won't be!

Since we aren't trying to change the frequency we must multiply it by something that has a value of one. Just like when we convert a distance in feet to inches we do not want to change the value of the distance. So we do NOT multiply the distance by 12, when instead multiply it by (12 inches)/(1 foot) because this ratio is equal to one.

The resolution to this lies quandary for the frequency equation above is to recognize that the 2π is NOT dimensionless -- it is a conversion factor and, like all conversion factors, is a ratio of one thing in the numerator expressed in one set of units to the same thing in the denominator expressed in a different set of units.

There are two completely equivalent ways of expressing this conversion factor.

\(
\omega \; = \; \( \frac{2\pi \frac{rad}{s}}{1 \, Hz}\) f
\)

or

\(
\omega \; = \; 2\pi \frac{rad}{cycle} f
\)

Notice that ω and f are about the only instances (that I can think of) in which the choice of the symbol for a quantity locks in the units to be used. Personally, I believe that this is precisely because we throw around this seemingly dimensionless factor of 2π to relate them instead of tracking the units properly. If we tracked the units properly, then there would be no problem.

For instance, it is 100% perfectly correct to say v(t) = (10 V)·sin(30Hz·t) and then ask what the value of v(2 minutes) is. We multiply 30 Hz by 2 minutes and we get 60 Hz·minutes. We then convert that to radians because we must have a dimensionless quantity as the argument to a transcendental function. No problem -- just properly track the units.