# Only a semi-tricky quiz

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

1. ### AI3VHam MemberQRZ Page

I see what you did there!

Rege

2. ### N2EYHam MemberQRZ Page

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

FAIL

Bad question, Eric. Moving goalposts fallacy.

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

3. ### K8ERVHam MemberQRZ 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. ### WA7PRCHam MemberQRZ 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. ### KA9JLMHam MemberQRZ 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. ### K8ERVHam MemberQRZ Page

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

TOM K8ERV Montrose Colo

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7. ### G4COEHam MemberQRZ Page

Daves law..... er sorree-:

Kirchhoff's law

8. ### WA7ARKHam MemberQRZ 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

Factoring,
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))

Simplifying,
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. ### WR2EHam MemberQRZ 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. ### K8ERVHam MemberQRZ Page

You expected less?

TOM K8ERV Montrose Colo

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