Dumb question about resistors in series/parallel

[TheN00b]

Hey folks. Just joined EEVBlog's forum, but I've been a big time viewer for 2 years. I'm somewhat new to the world of electronics (just picked up my first oscilloscope 2 months ago!) and I am barely a hobbyist; just a guy who likes sticking probes in places they probably shouldn't be to find out how things work!

With that out of the way, we're learning about circuits in high-school physics right now (so no doctoral dissertations please!). One thing I cannot understand, is resistors in series/parallel. When we find the amperage through a resistor, whether it be the 2nd or 3rd resistor in the series, we always use the source voltage divided by the resistance of the resistor. Maybe I don't understand how resistors work, but shouldn't we subtract the calculated voltage drop of the resistors before them from the source, and then use that for our voltage?

Let me know if I can provide any more information. Thanks in advance folks!

[spitta]

The current through a series circuit will be the same through every resistor.

If you are taking the source voltage divided by a single resistor to find current in a SERIES circuit, you're doing it wrong. The current should be the source voltage divided by the sum of all the resistances.
This technique, however, would work to find currents for each branch of a PARALLEL circuit.

I recommend watching some youtube videos on series parallel circuits to get the ball rolling...

[IanB]

With that out of the way, we're learning about circuits in high-school physics right now (so no doctoral dissertations please!). One thing I cannot understand, is resistors in series/parallel. When we find the amperage through a resistor, whether it be the 2nd or 3rd resistor in the series, we always use the source voltage divided by the resistance of the resistor. Maybe I don't understand how resistors work, but shouldn't we subtract the calculated voltage drop of the resistors before them from the source, and then use that for our voltage?

Voltage is the potential difference between two points. So if you are trying to find the current through a resistor in a circuit, you need to take the voltage difference between each end of the resistor and divide that by the resistance. In this case the "source" voltage is the voltage across the resistor.

On the other hand, if you want the current through a network of resistors, you need to take the voltage across each side of the network and divide that by the equivalent resistance of the network taken as a whole. Now for a simple case of three resistors in series the equivalent resistance is just the sum of the individual resistances, and the current through each resistance is the same. So you can find the current through a single resistor by finding the current through the network (in this one special case).

[TheN00b]

The current through a series circuit will be the same through every resistor.

If you are taking the source voltage divided by a single resistor to find current in a SERIES circuit, you're doing it wrong. The current should be the source voltage divided by the sum of all the resistances.
This technique, however, would work to find currents for each branch of a PARALLEL circuit.

I recommend watching some youtube videos on series parallel circuits to get the ball rolling...

Oh, so what I was explaining would work for parallel, because the voltage from the source stays the same when something is wired in parallel! But let's say I am asked to find current through one specific resistor in parallel(and that can be done using Ohm's law), but before that set of the parallel resistor, I had a resistor in series that dropped the voltage. For finding the current of the single resistor in parallel, using Ohm's law again, do I use the source voltage or the source voltage minus the Vdrop of the resistor that was in series before the resistor in parallel?

With that out of the way, we're learning about circuits in high-school physics right now (so no doctoral dissertations please!). One thing I cannot understand, is resistors in series/parallel. When we find the amperage through a resistor, whether it be the 2nd or 3rd resistor in the series, we always use the source voltage divided by the resistance of the resistor. Maybe I don't understand how resistors work, but shouldn't we subtract the calculated voltage drop of the resistors before them from the source, and then use that for our voltage?

Voltage is the potential difference between two points. So if you are trying to find the current through a resistor in a circuit, you need to take the voltage difference between each end of the resistor and divide that by the resistance. In this case the "source" voltage is the voltage across the resistor.

On the other hand, if you want the current through a network of resistors, you need to take the voltage across each side of the network and divide that by the equivalent resistance of the network taken as a whole. Now for a simple case of three resistors in series the equivalent resistance is just the sum of the individual resistances, and the current through each resistance is the same. So you can find the current through a single resistor by finding the current through the network (in this one special case).

Is that why, when simplifying circuits, we simplify the parallel elements into a single resistance and then add all of the resistors together like it were a series to find a single equivalent resistance?

[IanB]

Is that why, when simplifying circuits, we simplify the parallel elements into a single resistance and then add all of the resistors together like it were a series to find a single equivalent resistance?

Yes, exactly.

[Brumby]

Is that why, when simplifying circuits, we simplify the parallel elements into a single resistance and then add all of the resistors together like it were a series to find a single equivalent resistance?

Yes, exactly.

Snap.

Here is an example involving multiple application of the series and parallel calculations.

Say you wanted to work out the voltage across R6...

You start out by working from the right to left to get the resistances resolved so you can determine the current drawn from the source.  You then use this, together with Ohms Law and the resistance calculations to work from left to right, calculating currents and voltages.

[Kirr]

Is that why, when simplifying circuits, we simplify the parallel elements into a single resistance and then add all of the resistors together like it were a series to find a single equivalent resistance?

We only add them together if they actually are in series.

Some circuits can't be simplified by just parallel and serial replacements. The simplest example is a bridge circuit.

[Brumby]

RA = R6 and R7 in series
RB = RA and R5 in parallel
RC = RB and R4 in series
RD = R2 and R3 in series
RE = RC and RD in parallel
RF = R1 and RE in series

Use Ohm's Law with RF as the resistance and a source voltage of 5V to get the current draw.  Then work backwards through the resistor network...

First step after calculating the total current draw is that you see that ALL of this current will flow through R1.  You then calculate the voltage drop across this resistor, subtract this from the source voltage to get the voltage across the rest of the network.

... and so you continue, addressing each step as a separate calculation.

[BravoV]

If you saw above circuit, as sometimes you can move around to make it "mentally" easier to digest and do the calculation using series/parallel formula like this below.

Both are identical but different looks ..

[TheN00b]

Is that why, when simplifying circuits, we simplify the parallel elements into a single resistance and then add all of the resistors together like it were a series to find a single equivalent resistance?

We only add them together if they actually are in series.

Some circuits can't be simplified by just parallel and serial replacements. The simplest example is a bridge circuit.

We're not that far along yet in my class, but I just looked it up. Looks like an interesting idea for sure though, would explain how sensored brushless motors sense where they are if it uses Hall effect sensors in that sort of circuit, right?

RA = R6 and R7 in series
RB = RA and R5 in parallel
RC = RB and R4 in series
RD = R2 and R3 in series
RE = RC and RD in parallel
RF = R1 and RE in series

Use Ohm's Law with RF as the resistance and a source voltage of 5V to get the current draw.  Then work backwards through the resistor network...

First step after calculating the total current draw is that you see that ALL of this current will flow through R1.  You then calculate the voltage drop across this resistor, subtract this from the source voltage to get the voltage across the rest of the network.

... and so you continue, addressing each step as a separate calculation.

Oh man, I wish I could do more for you than just say "thanks!" This answered my question, for sure. But what about R4, that looks to be in series and would affect R5 through R7 (or in this case RA and RB) in terms of voltage? Or maybe I'm just not seeing something... I swear I'm trying to not be stubborn. Lol


EDIT; That new picture made sense and looked a little more like what we did in class. Not saying that I couldn't do it the other way, but yeah, that makes it a lot easier to see the types of circuits. Thanks for re-drawing that too!

[BravoV]

EDIT; That new picture made sense and looked a little more like what we did in class. Not saying that I couldn't do it the other way, but yeah, that makes it a lot easier to see the types of circuits. Thanks for re-drawing that too!

Similar case with you also happened -> Circuit Analysis Question Has Me Stumped

It started as the OP was confused with this ...



Moved the layout here and there ..


Moral story, especially for beginner, don't be scared by weird layout. 

[TheN00b]

I'll do my best to not be scared. Can't guarantee much though. Lol

[BravoV]

I'll do my best to not be scared. Can't guarantee much though. Lol

LOL .. yeah, it will take a while, practice frequently will make it feels natural, trust me.

One thing for sure, now you've learned or at least realized on how NOT to draw a schematic that sucks. 

[Brumby]

EDIT; That new picture made sense and looked a little more like what we did in class. Not saying that I couldn't do it the other way, but yeah, that makes it a lot easier to see the types of circuits. Thanks for re-drawing that too!

  to BravoV on both diagrams.  That sort of presentation does make it easier to 'get it' ... and once you do understand, your ability to see these things will improve as your experience grows.

This has also been a lesson in shuffling the physical layout around without changing the circuit.  You may hear the word 'topology' mentioned when doing this sort of thing.

[IanMacdonald]

Interesting point is that some of these resistor networks are solvable, some are not, or are incredibly difficult to solve. For example an extra resistor between the junctions of R2/R3 and R6/R7 turns this into an entirely different proposition.

[BravoV]

[That sort of presentation does make it easier to 'get it' ... and once you do understand, your ability to see these things will improve as your experience grows.

Yes, lots of practices and once you get enough experience, handling circuit like that can be easily solved by virtually moving the nodes around in the brain. Of course, when I started this method, I used to sketch a lot on paper, but it will become natural once practiced enough.

That circuit of yours took me for about maybe just 10 seconds to come out that the whole resistors network equal to 10 Ohm, as there are multiple parallel resistor pairs are using identical values and integer, hence the instantaneous result as there is no need to go into fractional calculation that may take a while.

[Brumby]

Yes - the values do work out very neatly ... and it is quickly solvable.

In the real world, however, it is almost never that easy.

[Zero999]

Is that why, when simplifying circuits, we simplify the parallel elements into a single resistance and then add all of the resistors together like it were a series to find a single equivalent resistance?

We only add them together if they actually are in series.

Some circuits can't be simplified by just parallel and serial replacements. The simplest example is a bridge circuit.

That can still be solved using simple resistors in series/parallel calculations, along with Thevenin's theorem and Kirchhoff's law.

It can be treated as two potential dividers, which are equivalent to two voltage sources, each with a series resistance equal to both resistor values in series: Thevenin's theorem.



Remove R5 from the circuit.
Calculate:
V(pd1_open) and V(pd2_open)

V(pd1_open) = V1×R2/(R1+R2) = 5×100/(100+100) = 2.5V
V(pd2_open) = V1×R2/(R3+R4) = 5×100/(100+200) = 1.667V

The output impedance of pd1 and pd2, which is simply equivalent to the resistances in parallel.

R_pd1 = R1×R2/(R1+R2) = 100×100/(100+100) = 50R
R_pd2 = R3×R4/(R3+R4) = 200×100/(200+100) = 66.67R

Reinstate R5. The current can now be calculated using Ohm's law.
The voltage is equal to the open circuit voltages of the potential dividers.
It's the voltage difference, which is important, so they're subtracted.
V = V(pd1_open) - V(pd2_open) = 2.5-1.667 = 0.833V

To get the current, through R5, the impedance of the potential dividers needs to be added to R5:
R(total) = R(pd1) + R(pd2) + R5 = 50 + 66.67 + 50 = 166.67R
I = 0.833/166.67 = 0.005 = 5mA

We know the current through R5 and therefore pd1 and pd2 is 5mA, so the voltages with R5 in place can be calculated using Ohm's law. V(pd1) is higher than V(pd2) so we know that the voltage on V(pd1) will fall (hence the - sign) and the voltage on V(pd2) will rise (hence the + sign) when R5 is connected:  Kirchhoff's voltage law.

V(pd1) = V(pd1_open) - R_pd1×I(R5) = 2.5-50×0.005 = 2.25V
V(pd2) = V(pd2_open) + R_pd2×I(R5) = 0.833+166.67×0.005 = 2V

The voltages on R2 and R4 equal v(pd1) and V(pd2) respectively, so the currents can be calculated:
V(R2) = V(pd1)
I(R2) = V(R2)/R2 = 2.25/100 = 0.0225 = 22.5mA
V(R4) = V(pd2)
I(R4) = V(R4)/R4 = 2/100 = 0.02 = 20mA

The total supply current is simply the sum of the currents in R2 and R4: Kirchhoff's current law.
I(V1) = I(R2)+I(R4) = 22.5mA + 20mA = 42.5mA

The total resistance can be calculated using Ohm's law:
R = V1/I = 5/0.0425 = 117.65R

If V1 is unknown, just replace it with 1. The end resistance value will be the same.

There are other ways to solve this, but this is my favourite one.

[Brumby]

To TheN00b - Don't let that ^ ^ ^ scare you.

You will be able to cope with that a bit further down the road.  When you get there, it will be another challenge to learn, but it will provide you with the ability to solve seemingly impossible problems.

[Zero999]

To TheN00b - Don't let that ^ ^ ^ scare you.

You will be able to cope with that a bit further down the road.  When you get there, it will be another challenge to learn, but it will provide you with the ability to solve seemingly impossible problems.

Apologies if I got carried away. I wanted to demonstrate that this can be used to solve more complex problems.

The other methods such as nodal and mesh analysis work too and arguably are easier to apply, once one is familiar with them, but I prefer my method which I find more intuitive. Indeed, I'd have to consult a textbook if I had to perform mesh and nodal analysis, as I've not done them since college.

[IanB]

The other methods such as nodal and mesh analysis work too and arguably are easier to apply, once one is familiar with them, but I prefer my method which I find more intuitive. Indeed, I'd have to consult a textbook if I had to perform mesh and nodal analysis, as I've not done them since college.

Another method I'd like to try myself when I have a free moment is to use the Δ-Y formula to convert the Δ arrangement of R1, R3 and R5 into the equivalent Y network, which then makes the whole network much easier to reduce.

[IanB]

To TheN00b - Don't let that ^ ^ ^ scare you.

You will be able to cope with that a bit further down the road.  When you get there, it will be another challenge to learn, but it will provide you with the ability to solve seemingly impossible problems.

This. You are seeing things here that are way beyond what you are expected to learn in high-school physics 

[Kirr]

The other methods such as nodal and mesh analysis work too and arguably are easier to apply, once one is familiar with them, but I prefer my method which I find more intuitive. Indeed, I'd have to consult a textbook if I had to perform mesh and nodal analysis, as I've not done them since college.

Another method I'd like to try myself when I have a free moment is to use the Δ-Y formula to convert the Δ arrangement of R1, R3 and R5 into the equivalent Y network, which then makes the whole network much easier to reduce.

Yes, Δ-Y as well Y-Δ has to be mentioned. So let me just expand a bit on this method, for completeness.

Suppose all available serial and parallel arrangements of resistors are already simplified, and the network is still not a single resistor. What to do next? One simple approach is to apply Δ-Y and Y-Δ transforms. Y-Δ is actually a 3-resistor case of a more general star-mesh transform. While Δ-Y is sometimes more convenient, star-mesh transform has some nice properties: 1. It is always applicable. 2. It always removes one node, leaving you with a smaller network. This means the repeated application of star-mesh transform (followed by parallel and serial simplifications when possible) will slowly but surely solve any finite network.

This method is quite mechanical - it does not require thinking about equations or Kirchhoff's laws or much of anything else. So this approach is used in my solver (linked in my sig) - this tool can walk you through this method step by step.

[Kirr]

Interesting point is that some of these resistor networks are solvable, some are not, or are incredibly difficult to solve. For example an extra resistor between the junctions of R2/R3 and R6/R7 turns this into an entirely different proposition.

Decided to try it. Spoiler link - click to check answer or give up. (Since the new resistor is unspecified, I took the liberty to choose a value of 147 Ω).

[TheN00b]

I'll do my best to not be scared. Can't guarantee much though. Lol

LOL .. yeah, it will take a while, practice frequently will make it feels natural, trust me.

One thing for sure, now you've learned or at least realized on how NOT to draw a schematic that sucks. 

And if I do encounter one (probably won't) at least I know how.. who knows, it could be extra credit?

To TheN00b - Don't let that ^ ^ ^ scare you.

You will be able to cope with that a bit further down the road.  When you get there, it will be another challenge to learn, but it will provide you with the ability to solve seemingly impossible problems.

Goodness, I hope so! I want to pursue an EE major someday, so it might be best to learn now though.