# Camara Sample Exam



## jdd18vm (Oct 13, 2007)

Question 158 States What is the Characteristic impedance of a high freq trans line with Zl =.85 ohms/mile and Yl =7.00 z10^-6 S/m. The answer 348 Ohms

Solution Zo= Sqrt of Zl/Yl so They show Sqrt of .85 ohms/m divided by (7.000 X10^-6S/m)(1 x1/ohm/S)

I took the reciprocal os Siemens to get Ohms (wrong).

Hard to visualize, sorry for that in effect the solution is just using the .85 divided by 7 micro S

What am not getting?

John


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## mudpuppy (Oct 13, 2007)

John, what you're missing here is the denominator of the equation is already an admittance, Yl (in Siemens), not Z. Since the Yl is given directly in the problem statement, you don't have to convert it.

You could convert the Yl to ohms if you wanted, but then the characteristic impedance equation would be Z0=sqrt(Zl-series*Zl-shunt), which is equivalent to sqrt(Zl-series*1/Yl)=sqrt(Zl/Yl).

It's hard to explain this in words on a message board so let me know if that explanation didn't make sense.


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## jdd18vm (Oct 13, 2007)

mudpuppy said:


> John, what you're missing here is the denominator of the equation is already an admittance, Yl (in Siemens), not Z. Since the Yl is given directly in the problem statement, you don't have to convert it.
> You could convert the Yl to ohms if you wanted, but then the characteristic impedance equation would be Z0=sqrt(Zl-series*Zl-shunt), which is equivalent to sqrt(Zl-series*1/Yl)=sqrt(Zl/Yl).
> 
> It's hard to explain this in words on a message board so let me know if that explanation didn't make sense.


That helps Mud. I get that its already an admittance, got a little lost on the conversion, but that appears to be just what they did only canceled out the units though? I hadn't done Transmission lines in detail, need to. Not sure what you mean by shunt...but will.

Are we allowed to scan stuff?

John


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## benbo (Oct 13, 2007)

jdd18vm said:


> That helps Mud. I get that its already an admittance, got a little lost on the conversion, but that appears to be just what they did only canceled out the units though? I hadn't done Transmission lines in detail, need to. Not sure what you mean by shunt...but will.
> Are we allowed to scan stuff?
> 
> John


If I recall, a transmission line can be modeled as a series impedance (which is basically along the transmission line in the direction of transmission) and a shunt, which is the impedance between the transmission line and the ground.


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## mudpuppy (Oct 14, 2007)

Yes, benbo is exactly right. In power terms, we typically think of a power transmission line as only a series impedance, but a line is also a capacitor between the conductor and ground, with air being the dielectric. This capacitance is referred to as the shunt impedance or shunt admittance, which is the Yl in the problem. In power systems, this shunt admittance (at 60 Hz) is usually so insignificant that we ignore it but it does come into play at higher frequencies like lightning strikes. The same theory applies to all conductors/cables and is much more significant in communications where high frequencies are transmitted on relatively small cables with high capacitance.

If the equations I wrote out in my last post don't make sense, just ignore them--better to go to your references directly. As for the units, they should work out to Ohms. Keeping in mind that Siemens = 1/Ohms, from the original equation of sqrt(Zl/Yl) you get sqrt(Ohms/Siemens) = sqrt (Ohms/(1/Ohms)) = sqrt(Ohms*Ohms) = Ohms.

As for scanning, do you mean to take into the test? You'd have to check with your state board to be sure, but as far as I know as long as it's in a three-ring binder it is ok.


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## jdd18vm (Oct 14, 2007)

mudpuppy said:


> Yes, benbo is exactly right. In power terms, we typically think of a power transmission line as only a series impedance, but a line is also a capacitor between the conductor and ground, with air being the dielectric. This capacitance is referred to as the shunt impedance or shunt admittance, which is the Yl in the problem. In power systems, this shunt admittance (at 60 Hz) is usually so insignificant that we ignore it but it does come into play at higher frequencies like lightning strikes. The same theory applies to all conductors/cables and is much more significant in communications where high frequencies are transmitted on relatively small cables with high capacitance.
> If the equations I wrote out in my last post don't make sense, just ignore them--better to go to your references directly. As for the units, they should work out to Ohms. Keeping in mind that Siemens = 1/Ohms, from the original equation of sqrt(Zl/Yl) you get sqrt(Ohms/Siemens) = sqrt (Ohms/(1/Ohms)) = sqrt(Ohms*Ohms) = Ohms.
> 
> As for scanning, do you mean to take into the test? You'd have to check with your state board to be sure, but as far as I know as long as it's in a three-ring binder it is ok.


Thanks for the help again.

I was referring to scanning a particular problem from any one of the books for posting here.

I took the morning Camara test...WAY not good, just as I feared...crap! Literally didn't know where to begin with a lot of them

John


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## jdd18vm (Oct 16, 2007)

mudpuppy said:


> Yes, benbo is exactly right. In power terms, we typically think of a power transmission line as only a series impedance, but a line is also a capacitor between the conductor and ground, with air being the dielectric. This capacitance is referred to as the shunt impedance or shunt admittance, which is the Yl in the problem. In power systems, this shunt admittance (at 60 Hz) is usually so insignificant that we ignore it but it does come into play at higher frequencies like lightning strikes. The same theory applies to all conductors/cables and is much more significant in communications where high frequencies are transmitted on relatively small cables with high capacitance.
> If the equations I wrote out in my last post don't make sense, just ignore them--better to go to your references directly. As for the units, they should work out to Ohms. Keeping in mind that Siemens = 1/Ohms, from the original equation of sqrt(Zl/Yl) you get sqrt(Ohms/Siemens) = sqrt (Ohms/(1/Ohms)) = sqrt(Ohms*Ohms) = Ohms.
> 
> As for scanning, do you mean to take into the test? You'd have to check with your state board to be sure, but as far as I know as long as it's in a three-ring binder it is ok.


okay now that i've done a few more problems that makes sense, no need to invert the value associated with the

Admittance just the units. Which is why in Camara Practice Problems 37-4 and the Ohms got Squared?

I just keep thinking from earlier circuit problems (ill have to go look) if it was Siemens you inverted to get ohms sorry if i am :deadhorse:


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## mudpuppy (Oct 17, 2007)

Yes, there are times when you need to invert Siemens to get Ohms. It comes down to what the equation you are using is asking for. If the equation has an impedance quantity (Z, R or X) then you need Ohms. If it is calling for an admittance (Y, G or B ), then you need Siemens. For a characteristic impedance Z0, the equation is mixed with a Z in the numerator and a Y in the denominator.

Problem 37-4 in the Camara practice problems book is asking for an A parameter, and the A paramater is actually unitless. They are multiplying Ohms^-1 times Ohms and end up with no units. Note that they are not inverting anything--Ohms^-1 is the same thing as Siemens. In other words, Siemens * Ohms = 1.


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