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.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 }
\)
Hi,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 \( \frac {2 \pi r} {r}\)
1 Hz = 1 cycle/secHi,
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) ?
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.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.
by Jake Hertz