Power P=IE

Where P is power in watts, I is current in amperes, and E is voltage in volts.

Ohm's Law E=IR

Where E is voltage in volts, I is current in amperes, and R is resistance in ohms.

[Because of type limitations, ^2 means to square)

Peak Envelope Power = (Peak Envelope Voltage x 0.707)^2

**/**R

Where R is resistance in ohms.

Peak Envelope Voltage = Peak to Peak Voltage / 2

Peak voltage = RMS voltage x 1.414

A 1 dB loss is approximately equal to 20.5%.

Peak Envelope Power = (RMS voltage)^2 / R

Where R is resistance in ohms.

RMS voltage = Peak voltage x 0.707

dB = 10Log(P1/P0)

Where P0 is the reference power and P1 is the power compared to the reference power.

Good luck. If you have any comments, suggestions, or questions, please feel free to leave them in the comments box.

73,

Andy

KE4GKP

I'm having a problem with questions 6 & 13.

ReplyDelete6: The answer is in watts so that's Power (P),

200 Volts (E) and 50ohm (R) Wouldn't that be E^2 over R? or 200 x 200 = 40k /50 = 800?

The correct answer is B. 100 watts

13: The answer is in watts so that's Power (P), 500 Volts (E) and 50 ohm (R) So wouldn't that be E^2 over R? or E x E = Answer /R? or 500 x 500 = 250k divided by 50 = 5,000 watts?

The correct answer is B. 625 watts

What am I doing wrong?

Thank you for your help,

Sally

Sally,

ReplyDeleteI racked my brain a bit and I think I can point you in the right direction. First, you helped me catch a typo on question #6. The correct formula for PEP with the values in the slide is (100 volts * 0.707)^2/50 ohms. I'll have to fix that. Unfortunately, that's not going to help you. I also believe the other question you have deals with question 14 vice 13. It looks like you are trying to figure these questions using Ohm's Law and the P=IE equations. You would be correct if we were dealing with "simple" power and voltage. Unfortunately, the question is looking for Peak Envelope Power (PEP) which is the average power a transmitter sends to the transmission line. This deals with a signal's sine wave and goes a bit beyond Ohm's law and the power equation. Because the sine wave is not sending a constant stream of power throughout the entire wavelength, to make the PEP calculation you need to determine the average power over the course of the wavelength. To determine the average power, you need to determine the peak envelope voltage, multiply it by the 0.707 constant, square the product, and then divide by the resistance. If you were looking for the power at a singular specific point on the sine wave, then you would be on the right track with the Ohm's law and P=IE equations, but because the question is looking for the average power over the course of the wavelength you will need the PEP equation.

For Question 14, you can use the same formula as in question 6. I just added the RMS equation for what I thought was going to be necessary for later explanations. RMS voltage is simply (Peak Envelope Voltage x 0.707)^2. Divide the RMS voltage by resistance and you will get PEP. The formulas for #6 and #14 are essentially the same.

Hope that helps. 73 -Andy

Thank you, Andy. That's exactly what I was doing. I still haven't been able to make the formula work. I'm not giving up. The good news is, I've definitely memorized the answer trying to figure out how to get to it.

ReplyDeleteThank you for a quick answer.

Hey Andy, it's 2:10 am and I'm cramming for my General Class which is in less than 8 hours. Anyway I noticed that in the material for Q6 you have the formula for PEP written and spoken incorrectly the second time you mention it, more towards the end of the slide although it is correct earlier. You say (PEV/0.707)sq/R when it is (PEVx0.707)sq/R. Not sure if maybe this is why Sally had trouble or not, but it through me for a loop for a minute.

ReplyDeleteBrent

KF7SMT

(don't have any gear so you won't hear me on the waves just yet)

Andy when will you be putting together a presentation for the extra class. I have finished both the technician and general with your help. Thank you so much for these presentations.

ReplyDeleteDavid

Hello,

ReplyDeleteI am working through both exams to help get ready to participate in emergency communications. I sincerely appreciate all the work you put into this site.

This lesson made me put my math head on, and I got all the math questions right when I remembered to do the fundamental rules:

1) Without doing any calculations write all the equations down at the top and define what the symbols mean: I = current in amps etc. [I always keep the units because it can help me see where to go]

2) Then for each specific question do two things:

a) write down what you want to get to - what are you solving for? PEP? I?

b) write down what the question has given you.

I (current ) =

R (resistance ) =

3) Get all the mili, kilo, etc into their basic units of volts, amps, etc. so you are not talking about thousandths in the same equation you are also using thousands.

Once I know where I want to go, see what I have, and have all the units at their basic value (instead of milli etc.) then I can select the formulas that will get me there. Having them written out at the header really helps me "shop" for the right one(s) and see the connections between some of them.

Alina

I know this is years later, but I was having similar trouble to Sally. I think we were both plugging in 200 without realizing that peak-to-peak voltage must be divided by 2 to become the peak voltage to be used in the equation. Now it makes sense!

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ReplyDeleteThe quantity of immunity, (R) a substance needs into blood circulation of electrical current is known as resistivity and that can be quantified in Ohms a metre, (Ω/p ). The reverse of resistivity is known as conductance. The immunity of the content is directly proportional with its own length, and inversely proportional to the crosssectional place. Thus in case the substances span has been doubled its immunity fades of course in the event the substance diameter ambigu its own immunity loops.Try this ohms law calculator app through which you can easily came to know about the calculations of voltage,current and resistance.

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