# New member with maths question

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#### Lee7534

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Welcome to ElectriciansForums.net - The American Electrical Advice Forum

New to the forum so hi everyone.
I'm due to start an electrician course in July so I thought I'd revise some maths as it's been some years since I've done any. I bought the book 'Electrical installation calculations'. I've flown through resistors in series and parallel so far but have come up against a brick wall with the following question. Any help with solving it and showing step by step workings would be massively appreciated. As follows:

To vary the speed of a d.c. series motor it is usual to connect a diverter resistor in parallel with the field winding. The field of a series motor has a resistance of 0.6Ω and the diverter resistor has three steps, of 5Ω, 4Ω and 2Ω. Assuming that the total current is fixed at 28A, find out how much current flows through the field winding at each step of the diverter.

Many thanks
Lee

#### telectrix

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calculate the currents for resistors in parallel, e.g 0.6 parallel 5.0.

#### marconi

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You have probably covered the 'potential divider' circuit and the rule for how the total potential across resistors in series is divided over the individual resistors.

Look up the 'current divider circuit' and a similar rule but this time for the way the total current is divided between the individual resistors/conductors.

#### rolyberkin

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Lee, join the trainee section of this forum, it will be invaluable if you are just starting out your studies.

#### Lee7534

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Thanks for the replies guys. I'm fairly happy with calculating total resistances in parallel and then currents for each. I think where I'm getting stuck is which resistor values or combination of, I need to use to initially calculate the voltage.

Thanks rolyberkin, I did look at that section but it says I need to have started the course to join.

#### rolyberkin

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Thanks rolyberkin, I did look at that section but it says I need to have started the course to join.
You have started by the looks of it! #### marconi

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You need to consider the field resistance in parallel with:

for Step 1 - 5Ohms

for step 2 - 4 Ohms

for step 3 - 2 Ohms

#### Lee7534

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OK, thanks Marconi, so this is what I have so far:
3 settings on the diverter giving 3 different values when in parallel with the field resistance;
0.6 parallel 5 = 0.54
0.6 parallel 4 = 0.52
0.6 parallel 2 = 0.46

So in order to calculate the current for each, I require the voltage. As I understand it (maybe this is where I'm getting confused) the voltage in a parallel circuit is constant so I don't get which of the above resistance values I use to calculate the voltage as they would each give a different value. Once I'm past this stage I'm fairly happy working out the individual currents.
Cheers.

#### marconi

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Not quite the correct reasoning. Now you know what the effective resistance of the field winding is with say 5 Ohms in parallel, you now need to use the information about the current being constant through them combined in parallel to arrive at the common (not constant) voltage across them - which does change with step position.

Then, for each step, knowing the common voltage across both resistors in parallel you can work out the current which flows through the field coil and the diverter resistance separately. As a check they should add to 28Amps.

(As an extra to tuck away in your mind for later, the motor is powered by a constant current source not a constant voltage source which is unusual but not uncommon. It will all make sense in a few months.)

#### marconi

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If Lucien Nunes reads this perhaps he might show an image of a stepped speed/torque controller for a series wound dc motor as used in an electric traction locomotive/tube train of yesteryear.

#### Richard Burns

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You can work it out the long way starting with
3 resistances on the diverter giving 3 different total resistances when in parallel with the field resistance;
A 0.6Ω in parallel with 5Ω = 0.54Ω
B 0.6Ω in parallel with 4Ω = 0.52Ω
C 0.6Ω in parallel with 2Ω = 0.46Ω
with a current of 28A the voltages in each case will be V=IR
A V= 28*0.54 = 15.12V
B V= 28*0.52 = 14.56V
C V= 28*0.46 = 12.88V
Then for each option current through the 0.6Ω winding is I=V/R
A I= 15.12/0.6 = 25.2A
B I= 14.56/0.6 = 24.27V
C I= 12.88/0.6 = 21.47A

However a quicker method if you have two resistances A and B in parallel.
You know the total resistance is product over sum i.e. AB/(A+B)
so your voltage would be I*(AB/(A+B)) or IAB/(A+B)
you will then take that voltage and divide by the resistance you want (say A) so you have (IAB/(A+B)) /A, the As cancel out to give IB/A+B.
Therefore you can say that the fraction of current taken by A is the resistance B over the sum of the two resistances A+B.
so for your calculations you can say that in each case
A 0.6Ω+5Ω =5.6Ω, current through winding is 28 *(5/5.6) =25A
B 0.6Ω+4Ω =4.6Ω, current through winding is 28* (4/4.6) = 24.35A
C 0.6Ω+2Ω =2.6Ω, current through winding is 28* (2/2.6) = 21.54A
even though these values vary from the ones above this is due to multiple rounding at each separate stage and calculation of the first set without rounding gives the same values.

#### Lee7534

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Step 1: 25A
Step 2: 24.4A
Step 3: 21.6A

Marconi, thank you for your help.

#### Lee7534

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Thanks Richard, just beat me to it with the post. Managed to work it out with the long method you demonstrated. I'll probably try the quick method once I get my head back into this maths malarkey #### marconi

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For Lee7534: Will the series motor increase or decrease in speed as the divertor is moved from step 1 to 2 and then 3?

#### Lee7534

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Hi Marconi, I know nothing about motors yet, this was simply the theme of the question in the 'resistors in parallel' section of the maths book I'm using.
At first glance I thought greater current would = greater speed; on researching it the case is the opposite. I think understanding how at this stage would be getting ahead of myself as I'm still trying to re-learn the maths that I haven't touched for about 25 years and haven't yet progressed past various uses for Ohm's Law Reply to New member with maths question in the The Welcome Forum area at ElectriciansForums.net