How to calculate the value of a resistor in a circuit?

[PedalChase]

I am trying to answer a question given in an electronics self-teaching guide I am going through.
I am not having any luck working it out myself with the information given. I must not be applying the correct equations. I am admittedly algebra challenged.
The question asks for the value of Resistor 2.

The information given is a diagram of a simple battery powered current divider circuit with two resistors (R1 and R2) connected in parallel.
Voltage source is 12 V DC battery.
Current Total is 300mA.
Resistor 1 is 50ohms.

    /--------------------------/---------------------/       Total Current: 300 mA
   /                                 /                           /
12V DC                        R1 50 Ohm          R2 ?
 /                                 /                          /
/-------------------------/--------------------/

What equation would you use to determine the value of R2 with the information given?

Thanks

[Simmed]

i = V/ R
0.3 = 12/ R
R = 40

R// = 1/(1/R1+1/R2)
R2 = 1/(1/40 - 1/50)
R2 = 200

[PedalChase]

I had worked out i=V/R. Ohms law.
The second part I will examine further.
If it was in the material, I had not realized how to apply it in this situation.

Thanks for the help.
Much appreciated.

[Konkedout]

Another way to think about it:

R1 will conduct 12VDC/50 ohms = 240 mA.

You want another (300 mA - 240 mA) = 60 mA

 R2 = 12VDC/60 mA = 200 ohms.

We can all go ohm now.   

[golden_labels]

This is a current divider. The current counterpart of a voltage divider.

You may approach the problem by thinking not about resistances, but about conductances. Conductance (G) is inverse of resistance (R):

G = 1 / R

Speaking simply: where resistance indicates opposition to current, conductance tells how easy it is for current to pass.

Mathematically using both works the same way in equations. But thinking in terms of conductances sometimes makes things easier to wrap your head around and to calculate values.

Where with resistances you get “ugly” stacked fractions from parallel resistors, with conductances you have a simple sum:

R1 = 50 Ω, so: G1 = 1/50 S = 0.02 S.
I = 0.3 A
V = 12 V

Total conductance must be: G_TOTAL = 0.3 A / 12 V = 0.025 S

G2 = G_TOTAL - G1 = 0.025 S - 0.02 S = 0.005 S

G1 = 0.005 S, so R1 = 1/0.005 Ω = 200 Ω


“S” above is the unit of Siemens, equal to 1/Ohm.

Note that deep inside the equations are identical to parallel resistances. They just put the problem in a different perspective and let you focus on the important part.

[Solder_Junkie]

If you want to work out Voltages and currents in "awkward" chains of resistors, as opposed to the relatively simple question posed earlier, Spice modelling software can provide the answer.

My personal favourite is the student version of Tina, which unlike LTSpice, it is not free. The student version is currently $41 for USA buyers:
https://www.tina.com/

SJ

[Zero999]

If you want to work out Voltages and currents in "awkward" chains of resistors, as opposed to the relatively simple question posed earlier, Spice modelling software can provide the answer.

My personal favourite is the student version of Tina, which unlike LTSpice, it is not free. The student version is currently $41 for USA buyers:
https://www.tina.com/

SJ

There's also TINA-TI, which is free.
https://www.ti.com/tool/TINA-TI

I prefer LTSpice myself. I find it much easier to use.

[tooki]

I must not be applying the correct equations. I am admittedly algebra challenged.

I highly suggest reviewing and practicing algebra until you really get the hang of it — algebra is inescapable in electronics. It’s sadly much too common for datasheets to include formulas in forms that aren’t immediately useful. (For example, tons of adjustable voltage regulator datasheets give the formula for calculate the voltage-programming resistors as “Vout = …” (for example, Vout = Vref * (1+R2/R1) ),  which is fine if you already have both resistor values and want to figure out what the output voltage is, but more commonly you know what voltage you need, and need to calculate the resistor values to get that voltage. So you need, for example, R2 = R1 * (Vout/Vref - 1). Being able to rearrange formulas to solve for the value you need is an essential skill. Sure, you can use a modern higher-end scientific calculator, or a math program for your computer or phone, that has a “solver” to solve for a particular value. But I think knowing how to do it is a very important skill for electronics.

(Note to self: review calculus.)

[tooki]

If you want to work out Voltages and currents in "awkward" chains of resistors, as opposed to the relatively simple question posed earlier, Spice modelling software can provide the answer.

My personal favourite is the student version of Tina, which unlike LTSpice, it is not free. The student version is currently $41 for USA buyers:
https://www.tina.com/

SJ

There's also TINA-TI, which is free.
https://www.ti.com/tool/TINA-TI

I prefer LTSpice myself. I find it much easier to use.

Having used both quite a bit — during my apprenticeship, one instructor required us to use TINA, the other LTspice — I think they both have their places. LTspice’s user interface is… unique. Fast to use once you’re used to it, but very weird. And editing a schematic after the fact is way more fussy than it needs to be.
TINA’s UI is more intuitive, but has its quirks too. Editing existing schematics is way easier. Where it totally beats LTspice is in interactive digital (logic) simulation, which LTspice simply does not do.

At the end of the day, I ended up preferring LTspice for analog simulations, TINA for digital.


But for a beginner — and frankly, for anyone, when facing an analog circuit you don’t understand — I also highly recommend Falstad, the online simulator, because it is interactive (even for analog circuits!) and above all because it actually animates the current flows in a circuit. It’s tremendously instructive to actually see both the magnitude and direction of current flows and voltages visualized.

http://falstad.com/circuit/

[golden_labels]

Simulators are great and indispensable tools. But they are tools. One has to know, how to use a tool. And be the one driving the tool, not being driven by the tool.

For electronic simulators the limitation is that they’re nothing more than very fancy calculators. They don’t represent reality. They just package mathematical equations into something that is easy to manipulate and interpret, and automate calculations. Forget about that and you get burned. Same for not learning the models in the first place.

Falstad is indeed an amazing simulator. It has many shortcomings, so don’t use it as a proper design or testing tool. But as a tool for education and quick sketching: unbeatable.

[shapirus]

You may approach the problem by thinking not about resistances, but about conductances. Conductance (G) is inverse of resistance (R):

This is an advice that should be given way more often than it is. Understanding and using the concept of conductance makes many things much easier to grasp and they start feeling more natural.

[PedalChase]

Yes, algebra keeps being my hang up as I try to progress. I have not used it in 40 years and need to start from scratch again. I can plug numbers into a formula and do ok but when I need to try and say solve for an unknown, I get locked up.
My issue is that I am interested in getting to where I can do some hands-on work and understand what I am seeing rather than just soldering part a to part b. Algebra was never easy for me so I keep trying to slide by with just the minimum interaction with it. Kind of a case of putting my time into algebra which is of little interest or into electronics which is of great interest.
On the other hand...I now have an "end use" for algebra which I never had in school. I might absorb it much better now that it means something to me...

[PedalChase]

I have heard about conductance in a lecture I listened to once but in the reading I do and my areas of interest (guitar pedals/ham radio) I have never once seen it explained or discussed.

Thanks for the insights.

[Analog Kid]

Yes, algebra: hard to do those kind of calculations until you once again become comfortable with it.

Here's a little taste of it that might help. Start with the one formula you might want to memorize:

I = E / R

(I meaning current, E meaning voltage)

Now derive the other two forms of this equation:

Multiply both sides of the equation by R:

R * I = (E / R ) * R

Since R / R cancels out (= 1), remove it:

R * I = E

reverse it to get it into its usual form:

E = IR

I'll leave the remaining form as an "exercise for the student".

It's really not so bad ...

[PedalChase]

That example is interesting.
I memorized the formula as a triangle years ago and have forgotten the actual steps involved.
I can work a problem out using the triangle for any one of the three

     E
---------
I   x   R

But I can't remember how to do the canceling or reversing.
Same with a few other formulas

   P
-------
I  x  E

I can fill in a number and get the correct answer (enough to get me by up to now) but can't re-arrange the formula to solve for an unknown.

Thanks.

[Analog Kid]

That example is interesting.
I memorized the formula as a triangle years ago and have forgotten the actual steps involved.
I can work a problem out using the triangle for any one of the three

     E
---------
I   x   R

But I can't remember how to do the canceling or reversing.
Same with a few other formulas

   P
-------
I  x  E

I can fill in a number and get the correct answer (enough to get me by up to now) but can't re-arrange the formula to solve for an unknown.

Thanks.

OK, let's see if I can help.
If I end up just annoying you and not helping you, let me know.

Take that last one: P = I E

Forget the triangle for now.
Let's say you know P and E (watts and voltage) and need to know I (current): how do you do that?
You want to get I by itself on the left side, right?
So we need to hack off E from the right side. How do we do that?
Well, remember the rule that anything divided by itself equals 1. Very important.
So if you divide both sides by E

P / E = (I E) / E

the two Es on the right (which are E / E) cancel out and disappear, so we have

P / E = I

which you can just flip around so you get

I = P / E

and as they say, Bob's your uncle.

Does that help?

[PedalChase]

Yes that helps.
I remember the "rule" you mentioned now.

I need to do some research and find a good source for a basic algebra course I can review as a refresher.
Then again...I am good at procrastinating.

[Solder_Junkie]

You can create a spreadsheet with the various formulas, once done it will be there as your own calculating tool.

Daft as it may seem, I have a spreadsheet to calculate the X, Y and Z figures to input into EZNEC, the antenna calculator. The calculations are for an inverted V dipole where you need the length along the ground from the (sloping) wire length. Basic school kid maths, but a spreadsheet with a diagram, etc. is so easy to use and not difficult to setup in the first place.

SJ

[SteveThackery]

Yes that helps.
I remember the "rule" you mentioned now.

I need to do some research and find a good source for a basic algebra course I can review as a refresher.
Then again...I am good at procrastinating.

Check out Khan Academy - some great courses in beginner maths, including algebra.

[PGPG]

I have a spreadsheet to calculate the X, Y and Z figures to input into EZNEC, the antenna calculator.

I have a spreadsheet to calculate EMC field limits from 10m to other distance.
The relationship is not very simple. If you for example calculate limits from 10m to 3m you have two frequencies at which emission changes from near field to far field.

Times changes.
I was able to do simple calculations (like P=I²R) when I was 10, but to use Internet to find answers to my questions when I was 35. Now the order looks being reversed

[Analog Kid]

You can create a spreadsheet with the various formulas, once done it will be there as your own calculating tool.

You could, but for really simple stuff like this (Ohm's Law, etc.), I would recommend not doing that.
Problem is you'll be leaning on that crutch forever. Better to learn it once and have it at your fingertips forever afterwards.

Spreadsheets might be good for more complex stuff, like frequency/reactance calculations. Or you can find lots of calculators online that do all kinds of things, including designing various filters (lowpass, highpass, etc.) for you.