I am looking at your formula in post#9 but I am trying to get the answer for R2. I need a formula for using Rt and R1 to come up with R2.
Ohm's law question
- Thread starter Kid347
- Start date
Hello Kid347
One of the things you should know to delve into the world of electronics is Mathematics.
If shteii01 tells you: RT = R1 x R2 / R1 + R2. (For two resistances in parallel only) from this formula You must know immediately that: R2 = R1 x RT / R1 – RT.
It can be solved in another way:
The data we know are:
V = 12
Itotal = 300 mA = IT
R1 = 60 Ohms.
We also know that the two resistors are connected in parallel, so both resistors are connected to 12 volts.
Surely you've seen this formula: I = V / R.
12 / R1 = I(R1) = 12/50 = 0.240A = 240 mA.
So by R2 circulate the difference: IT-I(R1) = 300-I(R1) = 300-240 = 60 mA. = I(R2).
Again: R=V/I = R2=V/I(R2) = 12/0.060 = 200 Ohms = R2.
Could you please calculate the power dissipated by R1 and R2?
and also the total power ??
One of the things you should know to delve into the world of electronics is Mathematics.
If shteii01 tells you: RT = R1 x R2 / R1 + R2. (For two resistances in parallel only) from this formula You must know immediately that: R2 = R1 x RT / R1 – RT.
It can be solved in another way:
The data we know are:
V = 12
Itotal = 300 mA = IT
R1 = 60 Ohms.
We also know that the two resistors are connected in parallel, so both resistors are connected to 12 volts.
Surely you've seen this formula: I = V / R.
12 / R1 = I(R1) = 12/50 = 0.240A = 240 mA.
So by R2 circulate the difference: IT-I(R1) = 300-I(R1) = 300-240 = 60 mA. = I(R2).
Again: R=V/I = R2=V/I(R2) = 12/0.060 = 200 Ohms = R2.
Could you please calculate the power dissipated by R1 and R2?
and also the total power ??
You have two resistors. One of them is 50 Ω and the other is 200 Ω. What is the total resistance of these two resistors if placed in parallel?
I would use the formula 50x200/50+200=40 ohms that formula I have not problem with. I still don't know how to come up with the missing value for R2. I don't know what formula to use.
If you have the formula
\(
x \; = \; \frac{y \times z}{ y \; + \; z}
\)
And you know that x = 4 and y = 10, can you solve for z?
If not, then your problem has nothing to do with electronics, it has to do with math -- namely 7th grade algebra. Electronics is very math intensive, so if you lack the basic math skills, particularly algebra, then you will always struggle (and often fail) in your electronics pursuits (unless you are in the very small handful of people that can develop a sufficient intuitive feel for the material). So you need to put together a plan to learn the math, either via self-study or through any of a number of brick-and-mortar and online options available.
Thank you very much for the help. This is the first time seeing subtracting in this formula. where you say R1-Rt gives you R2. it works great, it is just that in my 10 months or so of trying to learn electronics on my own, I have not seen in any of my books or online where you subtract on value from another. I will now work on getting the rest of the answers.
Well, first off, 50x200/50+200 is not 40, it is 400. Multiplication takes precedence over addition (and multiplication/division are done left to right), so your equation is the same as
( (50x200) / 50 ) + 200 = (10000 / 50) + 200 = 200 + 200 = 400
You MEANT to use the formula
(50x200)/(50+200) = 10000 / 250 = 40
You can't be sloppy with the math and you can't shrug it off by saying, "You know what I meant." Don't force people -- including yourself -- to guess at what you meant.
You seem to want a book of formulas that tell you exactly what formula to use in each and every situation you come across. Such a book does not exist, though many books aimed at the "technician" level have tried to go down that road. Instead, learn the math.
The bedrock of algebra is that if you have two expressions that are equal to each other, then you can do whatever you want (as long as it is a valid mathematical operation) to one side and as long as you do the same to the other side then both resulting expressions will still be equal.
So if we have
\(
R_t \; = \; \frac{R_1 \; \times R_2 }{R_1 \; + \; R_2}
\)
And let's say that we know Rt and R1 but want to find R2.
We can proceed along the following path:
\(
R_t \times \( R_1 \; + \; R_2 \) \; = \; \frac{R_1 \times R_2 }{R_1 \; + \; R_2} \times \( R_1 \; + \; R_2 \)
\;
R_t \times \( R_1 \; + \; R_2 \) \; = \; R_1 \times R_2
\;
\( R_t \times R_1 \) \; + \; \( R_t \times R_2 \) \; = \; R_1 \times R_2
\;
\frac{\( \( R_t \times R_1 \) \; + \; \( R_t \times R_2 \) \)}{R_1} \; = \; \frac{ \(R_1 \times R_2 \)}{R_1}
\;
\frac{\( \( R_t \times R_1 \) \; + \; \( R_t \times R_2 \) \)}{R_1} \; = \; R_2
\;
\frac{\( R_t \times R_1 \)}{R_1} \; + \; \frac{\( R_t \times R_2 \) }{R_1} \; = \; R_2
\;
R_t \; + \; \( \frac{R_t}{R_1} \) \times R_2 \; = \; R_2
\;
R_t \; + \; \( \frac{R_t}{R_1} \) \times R_2 \; - \; \( \frac{R_t}{R_1} \) \times R_2 \; = \; R_2 \; - \; \( \frac{R_t}{R_1} \) \times R_2
\;
R_t \; = \; R_2 \; - \; \( \frac{R_t}{R_1} \) \times R_2
\;
R_t \; = \; R_2 \times \( 1 \; - \; \frac{R_t}{R_1} \)
\;
R_t \; = \; R_2 \times \( \frac{R_1}{R_1} \; - \; \frac{R_t}{R_1} \)
\;
R_t \; = \; R_2 \times \( \frac{\( R_1 \; - \; R_t \)}{R_1} \)
\;
R_t \; \times \; \( \frac{R_1}{\( R_1 \; - \; R_t \)} \) \; = \; R_2 \times \( \frac{\( R_1 \; - \; R_t \)}{R_1} \) \; \times \; \( \frac{R_1}{\( R_1 \; - \; R_t \)} \)
\;
R_t \; \times \; \( \frac{R_1}{\( R_1 \; - \; R_t \)} \) \; = \; R_2
\;
\frac { \( R_1 \times R_t \) }{ \( R_1 \; - \; R_t \) } \; = \; R_2
\;
R_2 \; = \; \frac { \( R_1 \; \times \; R_t \) }{ \( R_1 \; - \; R_t \) }
\)
You should be able to follow step by step and see how I got from each line to the next (though I did combine a couple simple steps in a few places).
With practice, you can combine a lot of these steps and do everything in about five lines or so.
Power dissipated by R1= 12volts/50ohms =0.240mA
Power dissipated by R2 =12volts/200ohms =0.060mA
When you add 0.240mA and 0.060mA you get 0.300mA which is the current going into the resistors.
This is where I get confused, on the first line where the formula says R1xR2 how can I do that if I don't know the value for R2 ? And then go down the line. I don't know what value to plug in for R2. I don't want to seem inpatient, but I am trying my best. I am not wanting to become an engineer, but running a small business we have a lot of electronic equipment that I would like to get a understanding of how it all works. And I just like the stuff. When I do figure out a problem I feel satiffaction and want to celebrate. my problem is this book I purchased complete electronics self-teaching guide is really showing my weakness in electronics and math, I have been trying to get through the DC pre test for three weeks now, and I am only on problem 8 of 12. Can you recommend a self paced online course where I can learn without bothering everyone on this site? Thank you again for all the help.
Power is measured in watts, not milliamps (which is a measure of current).
Power in an electric circuit is the product of a voltage difference and the current flowing through that difference.
P = V·I
Since, for a resistor, you have (via Ohm's Law) V = I·R, you can write this two other was:
P = (V)·I = (I·R)·R = I²·R
P = V·(I) = V·(V/R) = V²/R
Algebra is about applying the rules of arithmetic to symbolic expressions.Notice that I didn't plug in ANY values anywhere in the work that I showed at all. I worked with the symbolic values only and, as a result, came up with a result that works for ANY value of Rt and R2.
Consider this simple example:
The formula for calculating the fuel economy of your car, M, is to take the total distance driven, D, starting from a full tank of gas and divide it by the amount of fuel, G, it took to refill the tank. This is captured by the formula:
M = D/G
Now you are asked how much gas it will take to drive 100 miles if the car gets 20 miles per gallon.
So in the above formula, we know M and D and we want to solve for G.
First, we multiply both sides by G to get
M·G = D
And then we divide both sides by M to get
G = D/M
Now we have the formula we need for what we want and only now do we plug in numbers:
G = (100 miles) / (20 miles/gallon) = 5 gallons
