tag:blogger.com,1999:blog-5676523852041991979.post1334135965969393561..comments2019-05-18T10:55:34.756-07:00Comments on The Ham Whisperer: Lesson 21: General Class Exam Course G5BAndynoreply@blogger.comBlogger11125tag:blogger.com,1999:blog-5676523852041991979.post-86974669207212592722018-12-10T23:46:49.122-08:002018-12-10T23:46:49.122-08:00If you are interested in a game that has a sequel ...If you are interested in a game that has a sequel (or two), go for the newest version of the game. In general, the newer it is, the more interesting the title will be. Therefore, don't start at the beginning and work your way through. Simply go for the best game at the beginning, especially if you have a limited budget. <a href="https://hackeroffice.com/" rel="nofollow">https://hackeroffice.com/</a><br /><br />Jessica Wilhttps://www.blogger.com/profile/15556151086192143138noreply@blogger.comtag:blogger.com,1999:blog-5676523852041991979.post-66260887399523473132018-09-10T05:56:48.937-07:002018-09-10T05:56:48.937-07:00The absolute most crucial statistical connection a...The absolute most crucial statistical connection among voltage, current and immunity at power is traditionally clarified as"<a href="https://play.google.com/store/apps/details?id=com.ohms.law.calculator&hl=en_US" rel="nofollow">Ohm's Law</a>". Ohm's law can be utilised to learn the proper resistor price within a circuit as soon as the voltage is understood and you'd really like to confine the existing to some certain price.Patrice Britthttps://www.blogger.com/profile/04342306232028095136noreply@blogger.comtag:blogger.com,1999:blog-5676523852041991979.post-73642832844471121632018-08-16T03:30:44.075-07:002018-08-16T03:30:44.075-07:00The quantity of immunity, (R) a substance needs in...The 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 <a href="https://play.google.com/store/apps/details?id=com.ohms.law.calculator" rel="nofollow">ohms law calculator</a> app through which you can easily came to know about the calculations of voltage,current and resistance.Jeff Hardizhttps://www.blogger.com/profile/06510938812488601309noreply@blogger.comtag:blogger.com,1999:blog-5676523852041991979.post-12597141079814442422017-10-19T05:56:48.053-07:002017-10-19T05:56:48.053-07:00In different countries of the world people are eng...In different countries of the world people are engaged in different societies to train themselves in a better way. If you want to see the ninja girls in the world then you can see in <a href="https://customessaysreviews.com/essaytigers-com-review/" rel="nofollow">essaytigers writing service</a> that Iran is the country giving the special trainning in martial arts to their girls.<br />tysonhttps://www.blogger.com/profile/04844980553124679779noreply@blogger.comtag:blogger.com,1999:blog-5676523852041991979.post-11307584738088958462017-07-31T09:19:03.196-07:002017-07-31T09:19:03.196-07:00I know this is years later, but I was having simil...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!me4sunshinehttps://www.blogger.com/profile/12158964741780084917noreply@blogger.comtag:blogger.com,1999:blog-5676523852041991979.post-57287704964671088892014-05-21T17:16:24.689-07:002014-05-21T17:16:24.689-07:00Hello,
I am working through both exams to help ge...Hello,<br /><br />I 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.<br /><br />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:<br />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]<br />2) Then for each specific question do two things:<br /> a) write down what you want to get to - what are you solving for? PEP? I?<br /> b) write down what the question has given you.<br /> I (current ) = <br /> R (resistance ) = <br />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.<br /> <br /> 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.<br /><br />AlinaAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-5676523852041991979.post-15609188298059563142012-09-14T02:31:19.087-07:002012-09-14T02:31:19.087-07:00Andy when will you be putting together a presentat...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.<br /><br />DavidAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-5676523852041991979.post-64994430305545779872012-02-25T03:12:34.887-08:002012-02-25T03:12:34.887-08:00Hey Andy, it's 2:10 am and I'm cramming fo...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.<br /><br />Brent<br />KF7SMT<br />(don't have any gear so you won't hear me on the waves just yet)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5676523852041991979.post-10581773289296593652011-03-23T15:18:46.400-07:002011-03-23T15:18:46.400-07:00Thank you, Andy. That's exactly what I was doi...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.<br /><br />Thank you for a quick answer.SallyWhttps://www.blogger.com/profile/06549126636957379198noreply@blogger.comtag:blogger.com,1999:blog-5676523852041991979.post-79397446164493229912011-03-23T13:55:35.241-07:002011-03-23T13:55:35.241-07:00Sally,
I racked my brain a bit and I think I can p...Sally,<br />I 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.<br /><br />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.<br /><br />Hope that helps. 73 -AndyAndyhttps://www.blogger.com/profile/07972687504772876019noreply@blogger.comtag:blogger.com,1999:blog-5676523852041991979.post-32283727547470293072011-03-23T12:14:28.204-07:002011-03-23T12:14:28.204-07:00I'm having a problem with questions 6 & 13...I'm having a problem with questions 6 & 13.<br /><br />6: The answer is in watts so that's Power (P),<br />200 Volts (E) and 50ohm (R) Wouldn't that be E^2 over R? or 200 x 200 = 40k /50 = 800?<br />The correct answer is B. 100 watts<br /><br />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?<br />The correct answer is B. 625 watts<br /><br />What am I doing wrong?<br />Thank you for your help,<br />SallySallyWhttps://www.blogger.com/profile/06549126636957379198noreply@blogger.com