Beginners Question: Calculate tot. Amps after it goes through total resistance in DC series circuit

Thread Starter

G_Raas

Joined Feb 12, 2017
20
My first post. Hello all, glad to have found this site!

I have just started first year at college for Electronics Engineering. I get Ohms law, eg; if I have voltage and Ohms I can figure out Amps, etc.

If I am asked to calculate total current in a circuit being supplied by a source with 10V /.20A going across a circuit with 3.8kΩ of resistance and am then asked to calculate total current, would current just be the same as supplied from the source (.20A)? I am confused...

If I try to calculate current by dividing source voltage 10V by 3.8kΩ to get current, I dont arrive back at the source supplied current of .20A, instead I get 2.632mA .

I am missing something here and have puzzled over it for awhile now and cant figure it out. I checked the forums but wasnt sure what search terms to use to find the answer to this specific question.
 

GopherT

Joined Nov 23, 2012
8,012
My first post. Hello all, glad to have found this site!

I have just started first year at college for Electronics Engineering. I get Ohms law, eg; if I have voltage and Ohms I can figure out Amps, etc.

If I am asked to calculate total current in a circuit being supplied by a source with 10V /.20A going across a circuit with 3.8kΩ of resistance and am then asked to calculate total current, would current just be the same as supplied from the source (.20A)? I am confused...

If I try to calculate current by dividing source voltage 10V by 3.8kΩ to get current, I dont arrive back at the source supplied current of .20A, instead I get 2.632mA .

I am missing something here and have puzzled over it for awhile now and cant figure it out. I checked the forums but wasnt sure what search terms to use to find the answer to this specific question.

The label on a battery or power supply is the maximum that the supply should be used for. The actual current will be volts divided by total load (ohms).
 

Thread Starter

G_Raas

Joined Feb 12, 2017
20
Thank you sooo much! I had hoped that was the answer but I have to do about 20 questions with ever increasing resistance and do the calculations for power for each and didnt want to get through them all only to find out I had to do them all over again because I didnt calculate the first one correctly...

Thank you, thank you, thank you!
 

bertus

Joined Apr 5, 2008
20,092
Hello,

It looks like there is a current limited voltage source used.
When the load is high enough, it can deliver the 10 Volts.
When the load gets to low, the current will be limited to the 0.20 Amps and the voltage will drop.

Bertus
 

bertus

Joined Apr 5, 2008
20,092
Hello,

You can calculate the point where the source goes from voltage to current.
The load would draw 0.20 Amps and have 10 Volts accross it.

Bertus
 

Tonyr1084

Joined Sep 24, 2015
3,635
Use Excel or some other math package to automate all the calculation.
I must admit I often go this route too. However, I think that when you're learning something new it's going to be more beneficial if you do the math every time instead of just plugging in numbers. I can do Trig, but when I was doing it on a regular basis I used Excel. I know as much about trig as any student just starting out in it. Tangent = opposite over adjacent (or something like that) and so on. But if I did it every time using a calculator I'd probably be able to answer with greater authority.

My advice, use the spreadsheet as a backup plan for checking your work. But do the work using your calculator, pencil and paper. Or whatever medium you choose. I sometimes go to my white board.
 

Thread Starter

G_Raas

Joined Feb 12, 2017
20
I must admit I often go this route too. However, I think that when you're learning something new it's going to be more beneficial if you do the math every time instead of just plugging in numbers. I can do Trig, but when I was doing it on a regular basis I used Excel. I know as much about trig as any student just starting out in it. Tangent = opposite over adjacent (or something like that) and so on. But if I did it every time using a calculator I'd probably be able to answer with greater authority.

My advice, use the spreadsheet as a backup plan for checking your work. But do the work using your calculator, pencil and paper. Or whatever medium you choose. I sometimes go to my white board.
That is quite coincidental... I had just started plugging the Ohms law wheel formulas into excel to create an automated calculator... I am finding the most important thing right now is to observe, observe, observe... what happens when this value goes up/down to that value, etc...

Thanks for all the help and advise all, sincerely. It is really encouraging. I hope to become a regular contributor/participant to this site/forum as my experience and expertise develops in this field.
 

GopherT

Joined Nov 23, 2012
8,012
That is quite coincidental... I had just started plugging the Ohms law wheel formulas into excel to create an automated calculator... I am finding the most important thing right now is to observe, observe, observe... what happens when this value goes up/down to that value, etc...

Thanks for all the help and advise all, sincerely. It is really encouraging. I hope to become a regular contributor/participant to this site/forum as my experience and expertise develops in this field.
We would be happy to double check one or two of you problems if you post your answer with them.
 

Thread Starter

G_Raas

Joined Feb 12, 2017
20
We would be happy to double check one or two of you problems if you post your answer with them.
Thats a fantastic offer!

I will always endeavour to answer the problem first and capture my work to the point of failure and post this work along with any questions I might post in the future... I get that this site and its forum members are not here to do my homework for me :) Besides, i wont learn anything If I dont try and fail first and learn from where I failed right?

Thanks again everyone, it is good knowing such a helpful community exists!
 
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