current calculations

Thread Starter

Neil Groves 1

Joined May 3, 2016
21
I am doing some circuit revision and ohms law stuff and I have forgotten my units, I calculate the current through a resistor with voltage divided by resistance but is my result in amps, milliamps or microamps?

Neil.
 

WBahn

Joined Mar 31, 2012
25,923
I am doing some circuit revision and ohms law stuff and I have forgotten my units, I calculate the current through a resistor with voltage divided by resistance but is my result in amps, milliamps or microamps?

Neil.
Just track the units as you go.

For example: 800 mV is applied across a 12 kΩ resistor.

You have

\(
I \; = \; \frac{V}{R} \; = \; \frac{800 \, mV}{12 \, k\Omega}
\)

For the purpose of figuring out what the result works out to, we can break this into three pieces -- the numerical values, the scaling prefixes, and the base units.

\(
I \; = \; \( \frac{800}{12} \) \( \frac{m}{k} \) \( \frac{V}{\Omega} \)
\)

We can tackle each separately. The numerical value is easy -- it works out to 66.7.

\(
I \; = \; \( 66.7 \) \( \frac{m}{k} \) \( \frac{V}{\Omega} \)
\)


You have three issues -- the numerical value, the scaling prefixes, and the base nits and the scaling prefixes. The units are easy to take care of since one ohm is, by definition, a resistance that produces one volt per ampere, or

\(
1 \, \Omega \; = \; 1 \; \frac{V}{A}
\)

So we can replace the Ω in the last factor with V/A since they are equivalent units.

\(
I \; = \; \( 66.7 \) \( \frac{m}{k} \) \( \frac{V}{\Omega} \)
I \; = \; \( 66.7 \) \( \frac{m}{k} \) \( \frac{ \( V \) }{ \( \frac{V}{A} \) } \)
I \; = \; \( 66.7 \) \( \frac{m}{k} \) \( \frac{VA}{V} \)
I \; = \; \( 66.7 \) \( \frac{m}{k} \) \( \frac{\strike{V}A}{\strike{V}} \)
I \; = \; \( 66.7 \) \( \frac{m}{k} \) \( A \)
\)

Now we can deal with the scaling prefixes. In this case we know that 1 k = 1000 and 1 m = 1/1000. So that gives us

\(
I \; = \; \( 66.7 \) \( \frac{m}{k} \) \( A \)
I \; = \; \( 66.7 \) \( \frac{ \( \frac{1}{1000} \)}{ \( 1000 \) } \) \( A \)
I \; = \; \( 66.7 \) \( \frac{1}{1,000,000} \) \( A \)
\)

We also know that 1 μ = 1/1,000,000, so we know have

\(
I \; = \; \( 66.7 \) \( \frac{1}{1,000,000} \) \( A \)
I \; = \; \( 66.7 \) \( \mu \) \( A \)
\)

Finally we combine the single surviving scaling prefix with the surviving base units and get

\(
I \; = \; \( 66.7 \) \( \mu \) \( A \)
I \; = \; 66.7 \, \mu A
\)

That was all very explicit, but I wanted you to see each step and that there is nothing very difficult involved. With just a small amount of practice you will be able to reduce the scaling prefixes and base units by inspection.
 

#12

Joined Nov 30, 2010
18,210
I wanted you to see each step and that there is nothing very difficult involved.
Merely 16 equations to do something the rest of us do in our heads?:eek: I don't think that's simplifying it.
Look on the bright side, Neil. You will be doing this in your head in about 3 days.:)
 

WBahn

Joined Mar 31, 2012
25,923
The point wasn't to simplify it -- the point was to examine the process in detail so that it can be readily seen for what it is and internalized. Note my last sentence -- "with just a small amount of practice you will be able to reduce the scaling prefixes and base units by inspection."
 

JoeJester

Joined Apr 26, 2005
4,264
Merely 16 equations to do something the rest of us do in our heads?:eek: I don't think that's simplifying it.
Look on the bright side, Neil. You will be doing this in your head in about 3 days.
I agree he will be doing it shortly by inspection. It also reminded me of the prefix tables and knowing the values of going from one to another, Sometimes I'd use scientific notation vice prefixes.

Six of one, half-a-dozen of the other.

I thought it was a good illustration of tracking the units .... WBahn style.
 
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