# Homework Help: Current through a resistor

1. Apr 21, 2010

1. The problem statement, all variables and given/known data
Given that the souce voltage = Vo, find the Current at resistor 4.

2. Relevant equations
For resistors in series, I = constant
For resistors in parallel V = constant

3. The attempt at a solution

So I gather that since the block containing resistors 2-5 is in series with R1, then whatever current leaves R1 must enter that block as well.

So at R1 we have: V1 = Vo = I1*R1
so that the current leaving R1 and entering the block is I1 = Vo/R1

Now I need to find out how much of that current diverges into R4. Here is what I am thinking:

The amount of current entering R2 and R4 is weighted by one of two things, either the values of R2 and R4 OR the values of (R2 + R3) and (R4 +R5), that is the total resistances in each 'leg' of the block.

I think it is the latter, but I am hoping someone can confirm.

Any thoughts?

2. Apr 21, 2010

### Oddbio

For resistors in series the resistances simply add.
$$R_{tot}=R1+R2+R3+...$$

For resistors in parallel:
$$R_{tot}=\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}+...$$

Also the voltage across R1 will not be Vo because there will be a voltage drop accross the block of other resistors.

3. Apr 21, 2010

I don't see why the voltage across R1 is not Vo ? Maybe i just don't understand circuits that well (it's been a long time ) but I thought that since R1 is the first resistor that the current encounters, then the voltage has not changed yet.

4. Apr 21, 2010

### jdc15

When you use I1 = V1/R1, V1 is equivalent to the voltage drop across R1. Try looking at the circuit as a whole by combining the resistors together.

5. Apr 21, 2010

### Integral

Staff Emeritus
You would measure V0 with respect to the negative battery terminal at the end of R1 closest to the battery, but if you measure the voltage at that same point with respect to the other side of the resistor you find something different.

What you need to find is the current through R4. Start by combining resistors to find the equivalent resistance. Then apply Ohms law.

6. Apr 21, 2010

### clombard1973

Because the source "sees" all the resistance attached to it, so it distributes all of its potential across every element within the circuit, some more than others if they are higher in resistance. If all the input voltage was dropped across R1 that would leave none for everything else in the circuit, plus it would imply the input current is just Vin/R1 which must go into the split series resistors that are in parallel with each other; but no current can flow through them as that would cause voltage drops and the sum of the voltage drops within the entire circuit would be greater than the battery attached providing the input power.

The input voltage is divided up and distributed by the value of the resistors across those resistors. By value I mean a greater valued resistor will take more of the input voltage than a smaller one attached to the circuit, but they all take some portion of the input voltage, not one has all of the input voltage dropped across it.

Many Smiles,
Craig

7. Apr 24, 2010

OK. Let's talk about this some more. I am sorry I am really bad at this, but I am trying here. I am trying to just go through this qualitatively. So I reduce the system to one equivalent resistor. Then I use Ohm's law Vo = IoReq

So now I have this current that is coming out of the source terminal and going into R1. What is the next step here to get me closer to finding the current going into R4?

Do I need the voltage drop across R1 for anything? I was under the impression that I did not, since I know that the same current comes out of R1 and enters the 'block' of resistors.

Thanks for your help so far guys

EDIT: I just found this too; so I will checking this out now:http://www.uoregon.edu/~dparks/206/resistorvoltage/ [Broken]

Looks promising!

Last edited by a moderator: May 4, 2017
8. Apr 25, 2010

1.)OK, so I know that Io enters the 4-resistor loop. Lets call the equivalent resistor of that loop R25. Now I can find the Voltage drop across the equivalent resistor as
V25=IoR25. Okay then!

2.)Now I take R25 and break it back up into two parallel resistors (each of which is a sum of 2 resistors in series) that we will call R23 and R45.
So since these two resistors are in parallel, I belive I can say that each one of them takes half of V25 right?

3.)Use Ohm's law to find the current through each of R23 and R45 that we shall denote I23 and I45.

4.)Since R45 is comprised of 2 resistors in series, then R4 and R5 both have current of I45 running through them.

Any flaws in my logic here? Wouldn't be surprised!

9. Apr 25, 2010

### Integral

Staff Emeritus
Since you know the current through R1 you can apply Ohm's law to find the voltage drop. Use that to determine the voltage across the parallel block.

The current may or may not be equal through each of the branches of the parallel block. It depends upon the resistance of each branch.

10. Apr 25, 2010

### jdc15

Don't worry about it. This is the place to come for help right?

Anyways, you have to treat the circuit as a whole. These kind of questions assume that current is already flowing through the entire circuit. It doesn't "start" anywhere. Imagine a river flowing in a circle, with the battery at the top. At the first resistor, the voltage drops by a certain amount. It drops by V = IR. We represent this resistance as R1. I is currently unknown. Now it branches off into two paths. Now the current is less, but the sum of the currents going out equals the current going in. Now voltage drops across R2 and R3 in one pathway and R4 and R5 in the other pathway. At the junction, voltage does not drop (as per Kirchhoff loop rule).

When trying to do electricity and magnetism questions it's sometimes difficult to understand since it's very abstract and sometimes counter-intuitive. So, when dealing with a question like this, you simply have to memorize the rules and apply them. In this question specifically, there are a number of ways to solve the problem but most likely you'd be using Kirchhoff rules and Ohm's law. Here is a link relating to the current question: http://physics.bu.edu/py106/notes/Circuits.html" [Broken]

Good luck solving the problem.

Last edited by a moderator: May 4, 2017
11. Apr 25, 2010

### OmCheeto

wrong.

Though everything else looked ok.

12. Apr 25, 2010

Why is this wrong? Just curious. I did it this way and then I found the voltage drops across each resistor and they all summed up to the source voltage.

Resistors in parallel have the same voltage running through them yes?

13. Apr 25, 2010

### rock.freak667

If they are in parallel, then the two sets would have the same p.d. through them.

14. Apr 25, 2010

If 2 resistors Ra and Rb are in parallel and we know that the voltage drop across the equivalent resistor Rab is given as Vab, then can't we say that Va+Vb=Vab? And since they are in parallel Va=Vb and thus it follows that Va=Vb=Vab/2 ?

15. Apr 25, 2010

### OmCheeto

v23=v45=v25

It may just be that you are using layman terms to describe certain things which make people who speak "electric" cringe. It's like really bad grammar.

Resistors in parallel have the same voltage drop across them.
Resistors in series have the same current running through them.

Voltage drops, and current flows.

I'm sorry, your statement in neither right nor wrong, but makes no sense whatsoever.

v23=v45=v25

16. Apr 25, 2010

If 4 resistors are in a loop then they have an equivalent resistance. The voltage drop across the loop should add up to the individual voltage drops across each resistor. Is that statement not true? I suppose not, I suppose it is only true for those resistors in series.

So my previous post:

I should actually say that Va = Vb = Vab

PS I am sorry if I offended Maxwell with my blasphemous ramblings; I have no idea how anyone could have ever figured out what was talking about. Voltages running... what a jerk.

17. Apr 25, 2010

### OmCheeto

Yes.
Around the loop.
It is true. And the voltage drop around a loop of resistors will be zero.
It's only important to learn electrical grammar if you want to become an electrical engineer.

18. Apr 25, 2010

I am getting more confused now. This what I get for saying that I will help my girlfriend with her physics homework. I am going to ask this one more time and I am sooo sorry to do it, but I need to clear this up.

Look at the loop with resistors 2 through 5. I take R2 and R3 and create an equivalent R23; I do the same for R4 and R5 to crate R45. Now R23 and R45 are in parallel.

Please point out where I am wrong:

1.)Since R23 and R45 are in parallel they must have identical voltage drops across them.

2.)If I combine R23 and R45 into an equivalent resistor R25, then the voltage drop across R25 = R23 + R45.

According to what you have said OmCheeto, the latter is incorrect. But I also though that the drops across R23 and R45 had to sum to the drop across R25 (i.e. the sum around the loop). So I am screwing up somewhere.

19. Apr 25, 2010

### OmCheeto

Correct
You're using bad electrical grammar again. This makes no sense.
If you are trying to get the equivalent resistor R25 then you have to use the following equation: R25 = 1/(1/R23 + 1/R45)

Let's see if this helps.

Starting at node 1, going around the loop measuring the voltages across R2 then R3 then R5 then R4, returning to node 1, the voltages will add up to zero.
Starting at node 1, going across the loop to node 2, you will again measure voltages across all of the resistors. V2 + V3 = V23 and V4 + V5 = V45. You will also notice that V23 = V45. And as I mentioned before V23 = V45 = V25.

20. Apr 25, 2010

Hi OmCheeto! Thanks again for all of your help. You are right; that was terrible grammar!

It should have said that the voltage drop across R25 is given by V25 = V23 + V45, but I now see that even this is incorrect and may even constitute for, yet again, more bad grammar.

So now here we are at the end! I think I understand this all just enough to do a problem with some numerics; so, I will post back in a few with an actual problem to hopefully show that I am not a lost cause on this .

I do have one closing question though: We have from all of this that if two resistors are in parallel, then: i) they have equivalent voltage drops across them; ii)the voltage drop across each of the resistors is equal to the voltage drop across the equivalent resistor (of the two in parallel).

Now, the latter of these 2 statements is a little hard for me to get my head around (go figure ). Is there some rule or law that this should follow from? Should it be obvious to me?

I have no problem accepting that if an equivalent resistor comes from 2 resistors in series, then the current through each of the individual resistors is the same as the current through the equivalent.

But for some reason when I extend this idea to resistors in parallel and the voltage drops associated with them, my brain goes to pudding.

~Casey

21. Apr 25, 2010

### OmCheeto

hmmmm.... pudding eh?

According to Gustav Kirchhoff, the voltage around a closed loop must be equal to zero.

Starting at node 1 traveling to R23, we see that we'll drop a voltage +V23. Continuing on through node 2 and through R45 we see a voltage drop of -V45. According to Kirchhoff, V23 added to V45 must equal zero. And the only way this can happen is if they are equal in magnitude.

22. Apr 25, 2010

Hey OmCheeto. Ok, I see that but that still does not tell me why when I find R25 = (1/R23 +1/R45)-1 and the voltage drop V25 why V25 should be equal to V23 and V45.

But, you have me thinking now that if I apply Kirchoff's Rule to the new circuit that contains only R1 and R25 something may reveal itself to me. Not sure though.

23. Apr 25, 2010

### clombard1973

So you understand that an equivalent resistance with two resistors is in series with each other must really have the same current flowing through the two resistors, but you are having trouble understanding why two resistors in parallel have the same voltage dropped across them?

The previous reply is probably the best mathematical way to show why the voltage drop across each resistor in parallel with the other must be the same. You can also kind of "think" your way into an understanding of why it must be. If two resistors are in parallel, they are "seen" by the battery as one equivalent resistance, so the voltage dropped across the two parallel resistors is the same because the battery "sees" a resistance and simply drops a voltage across it. It happens to be two in parallel so the drop across both of them is the same, but the battery knows not that there are two resistors it is dropping some of its potential across, it simply "sees" a resistance and drops some voltage across it; since they are connected to each other on both ends, then the drop across the two of them must be the same.

The equivalent resistance of two resistors in parallel, say R1 and R2, can be found by 1/[(1/R1) + (1/R2)] which is also equal to simply R1R2/(R1 + R2). Once you have your equivalent resistance, you can use Ohms law for a circuit with two series resistor in it, those being the equivalent resistance of the parallel combination of the two series resistor sets, and the other resistor in series with that equivalent resistance is just R1.

Find the voltage dropped across the equivalent resistance and divide it by the series combination of R4 with R5 to find the current that flows through the two series resistors, as it will be the same. You just need to figure out how to properly combine R2 with R3, and R4 with R5, and then what the parallel combination of those two sums are to find the equivalent resistance.

Ohms law should help you from there by simply looking at the "new" circuit as two resistors in series, one being R1 with the equivalent resistance RE made from combing R2, R3, R4, and R5 in the correct manner to get the correct equivalent resistance.

If you can get that far, finding the voltage dropped across the equivalent resistance should be straight forward using Ohm's Law. Just remember that the voltage dropped across the equivalent resistance by combining R2, R3, R4, and R5 in the correct manner, is dropped really across both R2 + R3, which is the same drop as that across R4 + R5, so when you find the voltage, ask yourself, what is this voltage being dropped across in resistance such that you can find the current you are looking for.

If you simply find the voltage dropped across the equivalent resistance and divide it by that resistance you'll get the current that flows into both "legs" of the parallel combination of the resistors that are in series with each other, or in other words the current that is actually leaving the battery and some of it when it hits the node that splits into two different paths with two resistors in series in each path, will go down one path while the remaining current from the battery will go through the other path, exit from both "legs" and recombine to the full current that is coming from the battery.

Hope this helps.

Many Smiles,
Craig