Units while dividing

Thread Starter

squiggly_line

Joined Feb 6, 2019
3
Hey guys, super simple question. It would help so much if you could explain what I am doing wrong/ not doing.

So in this series circuits
E = 12 V
R1: 2kΩ
R2: 3kΩ
R3: 4kΩ

Rt: 9 kΩ (which i found by simply adding them all)
Now it ask for current. Which I would find by,
I = E / Rt
I = 12 V/ 9kΩ
I = ?
(a)0.10mA
(b)1.0mA
(c)1.0μA
(d)10μA
(e) none of these
------ I'm confused as to whether i just carry that kilo over to the answer or do i apply it (E^3) to the answer i just got? Its a hassle to have to convert everything to a base Amp, Ohm, Watt for every single problem so how would you do it without the converting everything first. How do you treat this problem in a calculator?
I = E / Rt
I = 12 V/ 9kΩ
I = 1.33333 (kA?)

(calculator) I tried converting after that number after to compare it so that it could be in the form of either mA or μA, as those are the only options in the answers but I really don't know if i'm doing this right.
1.3333E^3 = 1333.3333 mA
or
1.3333E^-6 = 1.3333E^-6 μA
Any help is appreciated!


 

wayneh

Joined Sep 9, 2010
17,496
Well obviously 12/9 is never going to equal 1.0 no matter the units, so the answer doesn't require any more analysis.

When in doubt, it's always best to stick to the base units and figure out the power of ten you need. When you are used to dealing with milliamps, you can do the conversions in your head; I = 12V/9K = 1.33 mA just as 12mV/9Ω = 1.33mA. But that's dangerous and it's easy to get confused. If you stick rigorously to the base units and the exponents you're given, you'll always get to the right answer and it won't really take much longer. You don't need as much time spent double-checking and self-questioning.
 

WBahn

Joined Mar 31, 2012
29,976
You just need to build familiarity with unit prefixes and also learn to evaluate whether an answer makes sense.

In this case, you need to just learn and internalize things like k = 1000; 1/k = 1/1000; m = 0.001 = 1/1000. Therefore 1/k = m.

Then, you need to treat the prefixes just like the units or any other variable. You tried to go from V/(kΩ) to (kV)/Ω. Can't just magically move something from the denominator to the numerator -- in this case it throws your answer off by six order of magnitude.

Next you need to consider what makes sense. If 12 V / 9 Ω would yield a bit over an amp of current. Does it make sense that increasing the resistance by a factor of 1000 could somehow increase the current? Let alone increase it by a factor of 1000.

But, to answer this question, you don't even need to look at the units at all. All of the options (except 'none of these') is some power of 1. But 12 / 9 is NOT going to yield 1. So you can just pick (e) and move on.
 

Jony130

Joined Feb 17, 2009
5,487
If you have a current in milliamperes and the resistance in kiloohms then the result will be in Volts.

V = 1.33mA * 2kΩ = (1.33 * 2) = 2.66V

The same is true for current:

I = 10V/2kΩ = (10/2 = 5) = 5mA
I = 12V/9kΩ = (12/9 = 1.33) = 1.33mA

I = 3V/300Ω = 3V/0.3kΩ = (3/0.3 = 10) = 10mA

I = 300mV/1kΩ = 0.3V/1kΩ = (0.3/1) = 0.3mA = 300μA
 

Thread Starter

squiggly_line

Joined Feb 6, 2019
3
If you have a current in milliamperes and the resistance in kiloohms then the result will be in Volts.

V = 1.33mA * 2kΩ = (1.33 * 2) = 2.66V

The same is true for current:

I = 10V/2kΩ = (10/2 = 5) = 5mA
I = 12V/9kΩ = (12/9 = 1.33) = 1.33mA

I = 3V/300Ω = 3V/0.3kΩ = (3/0.3 = 10) = 10mA

I = 300mV/1kΩ = 0.3V/1kΩ = (0.3/1) = 0.3mA = 300μA

So now im trying to find resistance across a resistor but the answer are in mW. If i were to do 1.33mA(5.32V). Why does it just jump to Watts?
 

djsfantasi

Joined Apr 11, 2010
9,156
So now im trying to find resistance across a resistor but the answer are in mW. If i were to do 1.33mA(5.32V). Why does it just jump to Watts?
Because the definition of watts is VxI, amperes times volts.

As a refresher, this is Ohm’s law

V = I x R

And watts is one of the following

W = V x I
W = I^2 x R

The second form can be derived from the first and Ohm’s law

W = V x I
W = (I x R) x I
W = I^2 x R
 

MrChips

Joined Oct 2, 2009
30,708
There are different ways to approach this and still arrive at the correct results.

Firstly, know the applicable formula and the associated base units.

Base units are:
V = volt (V)
I = amp (A)
R = ohm (Ω)
C = farad (F)
L = henry (H)
t = second (s)
f = cycles per second (Hz)
P = watt (W)

I = V / R (amp)
V = I x R (volt)
R = V / I (ohm)

P = I x V (watt)
P = I x I x R (watt)
P = V x V / R (watt)

t = R x C (second)
f = 1 / t (Hz)

1) Write out all the values in their base units.
For example:
I = V / R
I = 12mV / 6kΩ
= 0.012 / 6000 A
= 0.000012 / 6 A
= 0.000002 A
= 0.002 mA
= 2 μA

2) Use scientific notation:
For example:
t = R x C
t = 10kΩ x 0.47μF s
= 10 x 10^3 x 0.47 x 10^-6 s
= 10 x 0.47 x 10^-3 s
= 4.7 x 10^-3 s
= 4.7 ms

3) Learn to cancel out exponents:
For example:
R = V / I
R = 500mV / 25mA Ω
= 500V / 25A Ω
= 20 Ω

V = I x R
= 2 mA x 5kΩ V
= 2 x 5 V
= 10 V
 
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