Old Time Radio Engineers

[Nowhere-Man]

I figured it might be a good reality trip to try and get my head around how the old radio engineers of the Thirties and Forties did their diagnostics. I chanced upon a book that was written in the hungry Thirties and then updated to 1943. It's less technical than my sixties Ham book but more practical from the actual engineering perspective.

Did you know, a lot of repairmen at that time relied on a Milliameter with a full-scale deflection of 0.005 Amp? Much of their work was done reading Milliamp deflections. According to the book, it was typical for the said Milliameter to have a resistance of 100 Ohms and deflection of the full 0.005 Amp at 5 volts. The book then launches into a method for testing voltages. Shunt resistors are wired in series with the circuit where the voltage might be expected to be, say, 50 volts. Then, it seems you would divide 50 by 5 to get 10, multiply the 100 Ohms of the Milliameter and then insert a shunt resistor of 900 Ohms. Presumably then if the device shows a deflection of the full 0.005 Amp. you can assume the circuit voltage will be 50 volts.

as Terry says

Series resistors would be used with the ammeter to measure volts.
Shunt resistors would be used to enable the meter to be able to read larger current values.

Ohm's law states V = I x R
Where I is the amps ( I forget why he used I)
so
100 ohms x 0.005 amps = 0.5 volts for full scale deflection, not 5 volts as you said.

Another useful little formula to always remember is in a DC circuit,  Amps x Volts = Watts.
(an AC circuit is similar but other factors influence the power.)

Substituting for amps in the 1st formulae above gives Volts squared, divided by Resistance = watts.

These 3 little formulae are very useful in all forms of electrical work, understanding them is in my view essential.


Mike

Correction. I'll have to check the book again but I'm sure it stated 5 volts. I agree it doesn't add up if you multiply the ohms and current. Not sure why it says 5 volts but I'm pretty sure it does as I memorised the calculation and it all adds up the same as the book by dividing 5 into 50 which gives the 1000 ohms. Funny that! If I used 0.5 volts it wouldn't have been the same as the 1000 Ohms the book gives. I have no explanation for now till I recheck the figures given. Strange.