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

I think you misunderstood what I said about the 5 volts. Having said that, it's probably the way I explained myself. According to the paragraph, the meter used as a milliameter had a full scale deflection of 0.005 Amp at 5 volts. The resistance of the meter without any added is 100 Ohms. Therefore, if the meter reads the full five milliamps you know you have 5 volts across the meter. However, the voltage in the circuit may need to to be more than 5 volts. You may wish to test a terminal that's designated as 50 volts. In order to do that, a series resistor needs to be used but it has to be calculated. Divide 5 into 50 to get 10 and you can multiply the 100 ohms you already have. Therefore, 1000 Ohms but you need to allow for the 100 Ohms we already have so 900 Ohms in series ought to allow for a 50 volt reading. The meter will still read 0.005 Amp but that then means at full scale deflection, the 900 Ohm resistor is taking the remaining current.

This is the same system that is used with ammeter shunts on boats. The ammeter itself doesn't take more than, say, 100 milliamps but the current in an engine circuit will be at least 100 amps max. Therefore, a large, high watt, low resistance shunt is used in parallel with whatever the meter resistance may be. This is like a long bar with terminals that is placed in the alternator + battery negative line.

I assume this method is explained because back then voltmeters were expensive.