How do i calculate missing resistance readings in a parallel circuit using R1 and Rt

[MikeGawan]

Hi

I am a current student studying to become an electrician and a question has thrown me. I'm wondering if someone could help, there is a question on one of my home work books. "complete the following table assuming the resistors are connected in parallel?" the example I will give is R1 is 120 ohms R2 ? R3? and Rt is 48 ohms. How would I calculate the missing R2 and R3 readings with no voltage or current to complete the table?

I have spoken to my tutor but I believe there was some miscommunication and got an answer which was relating to another topic. I don't want the answer I would just like to know the formula to help proceed with this if someone could help that would be great.

I understand 1/Rt = (1/R1) + (1/R2) + (1/R3).

 

[SparkyChick]

The problem is that if you have 2 unknowns (R2 and R3), there can be a very large number of possibilities.

What answer did your tutor give you?
[automerge]1585942820[/automerge]
I should add... perhaps message @Dan about getting access to the trainee only area... only trainee's and mentors (and staff) can post in there and there is a lot of useful info

 

[MikeGawan]

I did not receive an answer from my tutor. if I give another example there are 3 resistors R1 R2 and R3 and Rt . R1 40 ohms R2 ? R3 20ohms and Rt is 10 ohms? what formula would I use to calculate the missing resistance reading?

The problem is that if you have 2 unknowns (R2 and R3), there can be a very large number of possibilities.

What answer did your tutor give you?

 

[Andy78]

I'm no maths whizz by any means, but surely the only way that question can be answered accurately is if R2 and R3 are known to be equal, or some other predetermined relationship between them ?

 

[SparkyChick]

@MikeGawan ,

You have to transpose the 1/Rt = 1/R1 + 1/R2.... formula. Which can be done if you know all but one of the values. Unless you made a typo in your original post, if you have more than one unknown value (and assuming there are no assumptions you can make as @Andy78 suggests - which should be stated in the question if there are) you can't work out both missing values because they depend on each other.

 

[MikeGawan]

I have just blown the print screen of the question up to full size on a word document to post in the thread and seen there is a tiny hyphen in R3 which I am guessing means there was no R3 needed to be calculated. I did think 2 unknown values must have some relationship to one another but nothing was indicated. Transposing the equation does make sense! Thank you very much!

@MikeGawan ,

You have to transpose the 1/Rt = 1/R1 + 1/R2.... formula. Which can be done if you know all but one of the values. Unless you made a typo in your original post, if you have more than one unknown value (and assuming there are no assumptions you can make as @Andy78 suggests - which should be stated in the question if there are) you can't work out both missing values because they depend on each other.

 

[Wilko]

Hi - for your first problem with R2 and R3 being unknown - you can calculate what their value in parallel, which is 80 Ohms (from 1/48 - 1/120). Then a reasonable pick might be 160 Ohms each, say.

 

[SparkyChick]

Hi - for your first problem with R2 and R3 being unknown - you can calculate what their value in parallel, which is 80 Ohms (from 1/48 - 1/120). Then a reasonable pick might be 160 Ohms each, say.

I didn't think about it like that. Nice, thanks @Wilko

 

[Lucien Nunes]

But a numerical question would not normally require you to guess, therefore Mike's observation that the dash was intended to mean that there was no R3 present in that example, seems more likely.

 

[Lister1987]

The missing value should be lower than your lowest value resistor irrc.

Is this a series (R1) and parallel (R2 & R3) circuit, sounds like it, in which case you'll be wanting product over sum to help with missing value.

When ive had sleep, I'll dig out my notes and revisit this.
See here; Series and Parallel Resistors - https://learnabout-electronics.org/Resistors/resistors_20.php

 

[marconi]

MikeGawan: You might want to learn a new wrinkle to deal with series and parallel problems of resistors which only requires addition(+) and subtraction(-).

In series circuits the total resistance is easily calculated by adding together the individual resistance values viz: 3 + 2 +10 = 15Ohms.

In parallel circuits one can also add together the individual conductances; the conductance(C) of a resistor(R) is the reciprocal of its resistance ie C = 1/R - a resistor with a high resistance has then a low conductance and vice versa.

So, by way of example, if the 3, 2 and 10 resistors were all in parallel one adds together the reciprocals of these values to find the total conductance ie: 1/3 + 1/2 + 1/10 = whatever it comes to.

In the problem you are set, if you work with conductances from the start you just have to remember to take the reciprocal at the end to turn the unknown conductance into a resistance.

Example: 1, 2 and R are all in parallel and the total resistance is 0.1 Ohms or total conductance 10Mhos/Siemens - what is R?

1/1 + 1/2 + 1/R = 1/0.1

=> 1 + 0.5 + C = 10

==> C = 10 - 1- 0.5 = 8.5Mhos or Siemens

R =1/C = 1/8.5 =.... whatever Ohms

The algebra is no different to be honest but if one thinks in terms of conductance for items in parallel and resistance for items in series then it may play out more easily in your mind when tackling these problems. Always put the units in so you know whether a number is resistance value or a conductance value.

It helps too if you have a 'mental model' of the way electrons move through a conductor and what restricts their easy flow - what we call resistance. So, there are more restrictions to easy flow when resistors are in series because all the electrons leaving the first resistor then have to pass through the next and then the next...

When the resistors are in parallel not all the electrons go through all the resistors - the electrons have as many options as there are parallel paths to take - so overall there is less impediment because their flow from the input to the output side of the parallel resistor network is via as many routes as there are parallel resistors but those routes with lower resistance (higher conductance) take a proportionately higher number of electrons. Or something like this - you get the gist...