Only a semi-tricky quiz

Discussion in 'Radio Circuits, Repair & Performance' started by KL7AJ, Sep 10, 2017.

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  1. AI3V

    AI3V Ham Member QRZ Page

    I see what you did there! :)

  2. N2EY

    N2EY Ham Member QRZ Page

    But you need a physical voltage source and a meter to do the measurement!


    Bad question, Eric. Moving goalposts fallacy.

    Read some "Quist Quiz" items from QST to get the hang of it.
  3. K8ERV

    K8ERV Ham Member QRZ Page

    I'm going back to sleep. I don't parallel resistors, I just demand that one is the exact value I need.
    Works for me.

    TOM K8ERV Montrose Colo
  4. WA7PRC

    WA7PRC Ham Member QRZ Page

    "Using only Ohm's Law to find the total resistance" implies that, in order to solve for R, you MUST know at least two of the three others, E, I, and P:
  5. KA9JLM

    KA9JLM Ham Member QRZ Page

    Sometimes you need to.

    When I do, I just measure the value and let my meter do the math.

    It is not rocket science.
  6. K8ERV

    K8ERV Ham Member QRZ Page

    Here is a fixture I made for V Fixture.jpg easily connecting parts to a meter. You can put several resistors across it to get the
    right trim group.

    TOM K8ERV Montrose Colo
    N0TZU and KA9JLM like this.
  7. G4COE

    G4COE Ham Member QRZ Page

    Daves law..... er sorree-:

    Kirchhoff's law

  8. WA7ARK

    WA7ARK Ham Member QRZ Page

    To solve it, you can use Ohm's Law, superposition, and some simple algebra.

    If you have a constant-voltage source of E Volts, and you connect a resistor R1 across it, the current from the supply is I1 = E/R1

    If you connect a resistor R2 across E, the current from the supply is I2 = E/R2.

    Now if you connect both resistors at the same time, E is still E Volts because it is ideal, so the supply's total current is I1 + I2. This is obvious, but it is also implied by the principle of superposition. This is also stated by Kirchoff's Current Law, but we got here without using it.

    By extension, if you have five resistors, the total current
    Itot = I1 + I2 + I3 + I4 + I5

    By substitution,
    Itot = E/R1 + E/R2 + E/R3 + E/R4 + E/R5

    Itot = E/(1/R1 + 1/R2 +1/R3 +1/R4 +1/R5)

    We are trying to find the effective Resistance Rtot which would draw the same current from E as the five parallel ones.

    By Ohms law,
    Rtot = E/Itot

    By substitution,
    Rtot = E/(E/(1/R1 + 1/R2 +1/R3 +1/R4 +1/R5))

    Rtot =1/(1/R1 + 1/R2 +1/R3 +1/R4 +1/R5)

    Which is the "reciprocal of the sum of the reciprocals" we all remember...

    Notice that I never had to assume E was anything in particular, like 1V
  9. WR2E

    WR2E Ham Member QRZ Page

    But you don't, and you don't need it.

    Yes, and it's still using strictly only Ohm's Law, so Tom was correct.

    The 'tin star' was awarded to a guy who used a 1V test source and a means to measure current.
    Last edited: Sep 14, 2017
  10. K8ERV

    K8ERV Ham Member QRZ Page

    You expected less?

    TOM K8ERV Montrose Colo
    WR2E likes this.

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