Amateur Extra Class Lesson 4.2 Part 1: Electrical Principles

Here’s your video introduction to the ARRL Extra Class License Manual for Ham Radio, Section 4.2, Electrical Principles. I’ve divided this rather lengthy section into three videos, of which this is the first; this video covers fields, energy, and time constants.

After you have watched the video, studied the material in the text (up to but not including the section on Phase Angle), are sure that you understand the answers to the relevant questions from the question pool, click here to return to the list of Amateur Extra videos.

Please leave a comment if you have questions, comments, concerns, or suggestions.

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7 Responses to Amateur Extra Class Lesson 4.2 Part 1: Electrical Principles

  1. Dave says:

    No problem! Glad you found the issue. 73, Dave, KEØOG

  2. Art Womer says:


    I found my mistake. I was doing ll the steps at once, instead of stopping, clearing and reentering. After re-reading both your notes and the book, I caught it. Guess old age is really catching up,and not the “new math”…

    Thanks & 73

  3. Dave says:

    Hi Art, First, apologies for a misprint in my reply. I put a tilde in front of 0.2 to indicate it was approximate, however WordPress rendered it as a negative sign. I have updated the comment to be correct. I think where there is divergence is that exp(-1.6) (e raised to the negative 1.6th power) is 0.2, not 0.02. Also, I wrote a short piece of software in basic and ran it. Here’s the software and the results:

    REM * Program to calculate equation 4-2
    REM * on page 4-8 of the ARRL Extra Manual
    rem * clear the screen
    rem * tau is time constant in seconds
    tau = 5
    print "tau: " ;tau; " seconds"
    rem * time is time in seconds after beginning of charge
    time = 8
    print "time: ";time; " seconds"
    rem * appvolt is the applied voltage (charging voltage)
    appvolt = 20
    print "applied voltage: ";appvolt; " volts"
    rem * exp(x) is the function that raises e to the x-th power
    rem * calculate voltage at time t
    volts = appvolt * ( 1 - exp(-(time/tau)))
    print "voltage across capacitor at time is: ";volts;" volts"
    rem * hold open until dismissed
    input "Press Enter to end", x

    tau: 5 seconds
    time: 8 seconds
    applied voltage: 20 volts
    voltage across capacitor at time is: 15.96207 volts
    Press Enter to end

    This is what I came up with before: 16v.

  4. Art Womer says:

    Hello again,
    First, thanks for getting back to me and explaining. I’m following along with both the video and print and getting everything up to -0.021896518. When I “subtract” this from 1 I’m getting -7.92485….. If I change the sign to positive I still get 7.9. Close enough for the test “IF” I can remember that its 16v. Used 2 calcs and get the same answer. Maybe new math was built into these 2. LOL
    Anyway, Thanks again and I’m plugging along for Extra.

    Art Womer

  5. Dave says:

    Hi Art, I went over the example a couple times again, plus compared it with Figure 4.8A, and I’m pretty sure what’s shown in the video is correct. The equation in this case is Eq 4-2. The values are tau = 5 seconds, t = 8 seconds, and E = 20 volts. Eq 4-2 says the voltage at time 8 seconds is equal to the supply voltage (E) multiplied by one minus e raised to the -t/tau power. So, here are the steps:

    • t over tau is 8 over 5, which is 1.6
    • Make this negative, so we have -1.6
    • Use this as the exponent of e. e raised to the -1.6 power is approximately 0.2
    • [note change; WordPress renders a tilde as a hyphen]

    • Subtract this from 1, giving us 1 – 0.2 = 0.8
    • Multiply this by the supply voltage, gives us 20 x 0.8 = 16 volts
    • So the charging capacitor will equal 16 volts approximately 8 seconds after charging begins.
    • Compare this with Figure 4-8A. The charging voltage is 100 volts, so we will have to scale. At a time of 1.6 tau, the charge curve hits about 80 volts. We scale this back to one fifth of this amount, since our charging voltage of 20 volts is one fifth of 100 volts. One-fifth of 80 vols is 16 volts, the same answer as that obtained by running the numbers through the equation.

    Let me know if that still doesn’t work for you. 73, Dave, KEØOG

  6. Art Womer says:

    Hi Dave, Doing the calculations using the TI-30Xa model, almost the same as yours, but not solar. I’m getting “almost” the same answers, but when I do the math getting from -0.02.. it comes up as -7.9…. I,ve watched the video a few times to make sure I’m following correctly and used another calc. with the same results. Any suggestions?

  7. Andy M says:

    Thank you Dave for your continued production of these inspirational videos, technician , general and Extra. I passed the Section 4 test over the weekend, due in no small part to your ability to break down the ARRL chapters into meaningful bytes and communicate the general over view and context. The ARRL hand book is, at best, a dry read, you through your videos served as my Virtual Elmer , thank you.



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