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Some basic Ohms law question

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V= 25V
I= 50mA
R= ?

25V / 50 x 10^-3 = 500 Ohms

i took 25/50 = .5
then said
.5 x10^-3 or .0005 (what I did)

.5 x 10^3 = 500 (now this is the correct answer.)


I am just confused... why would it turn into a positive power of 10.

btw I got a text book from the library. It seems pretty good. Electronic Fundamentals by FLOYD
 
I do 25/50 = 0.5.
It's not ohms because it was mA so the 0.5 becomes 0.5 x 1000 = 500.

If it were 50 µA, then it's still 0.5. Then it's 0.5 x 1,000,000 = 500 kΩ.

Practice, and confirm with a calculator.

Memorize the reciprocal of 1 through 20.

Memorize the product of 1 through 10 times 1 through 10; go to 1 through 20 if you're up to it, then you'll be within 5% of any answer.

For most work two significant figure is enough, so PI is 3.1.
 
I never use "1 divided by 2pi fRC". Instead I use 0.16/fRC.
 
1 divided by 1x10^-3 = 1000

Your equation is 25V/(50 x 10^-3). The 10^-3 in the denominatior becomes 10^3 in the numerator
 
When you have a powered number at the denominator of a fraction, you can multiply that power into a negative and put that powered number at the numerator.
An example:

500/(8 x10^-9) = 500 x 10^9/8

so:
25V / 50 x 10^-3 = 25V x 10^3/50

You could do this too:
25V/0.05= 500
or
25000/50= 500 (I guess you are able to see what I have done here according to the first paragraph).

So you see that the calculation of those kinds of numbers is so easy. So please do several exercise to be clever regarding to them
 
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Use a common preferred value like 510R or 470R rathing that 500R which isn't easy to get hold of.
 
Try it like this: (25)/(50 * 10^-3)

Remember, 1/(x^-n) = x^n

and

1 divided by 1x10^-3 = 1000

Your equation is 25V/(50 x 10^-3). The 10^-3 in the denominatior becomes 10^3 in the numerator


Thanks this is exactly what i was looking for. I was missing a simple algebra rule that i just didnt see.

thanks
A lot!
 
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