Amateur Extra Lesson 4.2 Part 3: Resonance

Here’s the third of three video introductions to Section 4.2, Electrical Principles, in the ARRL Extra Class License Manual for Ham Radio. This video focuses on resonance, Q, and magnetic cores, particularly on resonance. Resonance is the fundamental building block that Mother Nature provides that makes radio possible.

NOTE: Lester Wetherell points out an error: “In your video amateur extra lesson 4.2 part 3 @ ~ 24 minute, made a mistake saying 10 pF = 1 X 10^-12. Just thought you should know. WD4IFU Lester.” In fact, of course 10 pF is 10 x 10^-12 farads, or in engineering notation, 10E-12 farads.

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8 Responses to Amateur Extra Lesson 4.2 Part 3: Resonance

  1. Jon says:

    I think your two examples (5 μH & 8 MHz) of moving the decimal point could be confusing. Why? Because you don’t start at the decimal point but rather at one end or the other. Move to the right yields 10-x and move to the left yields 10^x. You say “move to the right” but your drawing shows moving to the left.

    For me I would start with the decimal point and move it one space to the right.
    5 μH = 0.000005 H
    move once -> 0.00005 x 10^-1 H
    move twice -> 0.0005 x 10^-2 H

    I know this is “picky”. However for those with limited math background it could be confusing.

  2. Dave, I truly enjoy all of your videos. When I was studying for my General Class ticket I used a combination of the ARRL Book for the General License and would use your videos for extra support and understanding of the topics. I have been studying for the Extra Class license for the last few months using the ARRL Book, Gordon West’s book and CD’s and also YOUR videos! I am scoring 80% on the practice tests and hope to soon feel confident to take the test. I just wanted you to know that I am sure there are MANY others out there, like myself, who really appreciate all the research, preparation, set-up, filming, editing and uploading to YouTube that I am sure requires many HOURS of your time. Thank you very much, Dave! 73, Spence

  3. Richard A Purdy ke0cew says:

    Dave, 2014-12-13

    re: Extra Lesson 4.2, part 3, at min. 31.31

    When you were talking about how skin-effect-resistance can be reduced by using fat conductors — you briefly showed a picture of a high-voltage power transmission line, and said:

    “The very high tension wires that take electricity across hundreds of miles are actually hollow.”

    – not exactly ‘hollow’ –
    Yes, this is an important example of using fat conductors to overcome skin-effect.
    And from an electrical perspective, the center of the conductor is irrelevant (hollow).

    However, as an electric power engineer, I feel a need to point out that — those aluminum conductors are not hollow, to the contrary — they contain a core of high-strength steel cable.

    This type of conductor is called ‘ACSR’ (Aluminum Conductor, Steel Reinforced). The steel provides most of the strength and the aluminum provides most of the electrical conductivity.

    obtw: There are additional reasons for using fat conductors in power transmission —

    * reduced electric field gradient, which decreases corona loss,

    * reduced self-inductance, which reduces the impedance of the transmission line.

    I’m enjoying your video series on The Amateur License Manual.
    Thank you for sharing your knowledge, experience, and talent.
    Richard ke0cew.

  4. Frank Dambach says:

    Thanks sooo much! This really helped, I’d forgotten how to treat the square and multiplication of exponents. I figured out how to get the answer on a scientific calculator but this refresher is greatly appreciated. Thanks again!

  5. Dave says:

    Hi Frank. See a video of the worked example here.

  6. Frank Dambcah says:

    Thanks Dave, much appreciated and looking forward to it! (Note new vanity call in email address).

  7. Dave says:

    Frank, yes, I’ll create a video that shows the step-by-step process, hopefully within the next few days.

  8. Frank Dambcah says:

    Hi Mark,

    My name is Frank and I’m studying for my Amateur Extra license exam. I haven’t had to use this kind of math for some years now and I’ve been struggling with the following equation. I’d greatly appreciate it if you could show me step by step how to solve it and I know it will come in handy throughout your Amateur Extra course.

    fr=1/6.28 √(50 X 10^-6)(10 X 10^-12)

    I’m not sure this is clear but it should read: 1 divided by 6.28 times the square root of (50 X 10^-6)(10 X 10^-12)

    Thanks in advance for any help you can offer.

    Frank Dambach

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