What is total equivalent resistance and power dissipated?

[Apollinaria]

Another one.. I'm sorry but I will likely spend tomorrow and today here with you guys asking for help and mooching solutions. Same deal, badly explained unit. We spent 15 minutes on this. Attached questions as jpg...

Homework Statement

How do I find total equivalent resistance? What does that even mean? And power dissipated? What? :grumpy:

Homework Equations

No idea...

The Attempt at a Solution

I understand sometimes you add resistances and sometimes you add inverse resistances?

Can somebody please dumb this down into simple terms and try to explain step by step? Obviously I'm willing to work this through on paper with some help, guidance and feedback. I really want to understand the concept behind this.

 

[gneill]

The first problem ask to find the total equivalent resistance. That would be the net resistance "seen" by the battery. If you "cut" the circuit at the points a and b as shown in the diagram below, what is the net resistance between a and b? You'll want to simplify by combining parallel and serial resistor pairs until you're left with a single value for the total resistance.

Let's see how you do on that before moving on to the next part.

 

[Apollinaria]

Hi Gneill,

I'm not sure if I did this correctly but here...

 

[Apollinaria]

Nevermind, it wasn't.. Tried something else like this:

1/ (143/4300) + 10 + 10 and got 50.06. The correct answer.
Can you explain why there's a 1 over the parallel?

So the formula I have is

1/parallel = 1/R1 + 1/R2 ...

Using that, I get the whole value for the "parallel" portion of the formula. But to get the correct number I need to divide 1 by the number I got for the "parallel" part.

:uhh:

 

[gneill]

Nevermind, it wasn't.. Tried something else like this:

1/ (143/4300) + 10 + 10 and got 50.06. The correct answer.
Can you explain why there's a 1 over the parallel?

So the formula I have is

1/parallel = 1/R1 + 1/R2 ...

Using that, I get the whole value for the "parallel" portion of the formula. But to get the correct number I need to divide 1 by the number I got for the "parallel" part.

:uhh:

You didn't finish the calculation for the parallel resistance: you left off when you'd added 1/R1 + 1/R2, and failed to take the reciprocal of the sum.

A handy way to calculate parallel resistances when there are only two involved is $R_p = \frac{R1\times R2}{R1 + R2}$. The general formula of the reciprocal of the sum of reciprocals boils down to this when there are only two resistors involved.

 

[Apollinaria]

You didn't finish the calculation for the parallel resistance: you left off when you'd added 1/R1 + 1/R2, and failed to take the reciprocal of the sum.

A handy way to calculate parallel resistances when there are only two involved is $R_p = \frac{R1\times R2}{R1 + R2}$. The general formula of the reciprocal of the sum of reciprocals boils down to this when there are only two resistors involved.

I'm just trying to understand why I'm supposed to take the reciprocal of the sum... Why can't it be ready as in: Parallel = 1/R1 +1/R2... :rofl: ?

And that is an awesome formula. Now I just need to remember it :P

 

[gneill]

I'm just trying to understand why I'm supposed to take the reciprocal of the sum... Why can't it be ready as in: Parallel = 1/R1 +1/R2... :rofl: ?

It comes from the addition of conductances. Conductance is the reciprocal of resistance, so letting G stand for conductance, $G_T = G_1 + G_2 + G_3 ...$ when conductors are in parallel. Since resistance is the reciprocal of conductance, it stands to reason that $1/R_T = 1/R1 + 1/R2 + 1/R3 ...$

And that is an awesome formula. Now I just need to remember it :P

Yes, it comes in very handy.

 

[Apollinaria]

It comes from the addition of conductances. Conductance is the reciprocal of resistance, so letting G stand for conductance, $G_T = G_1 + G_2 + G_3 ...$ when conductors are in parallel. Since resistance is the reciprocal of conductance, it stands to reason that $1/R_T = 1/R1 + 1/R2 + 1/R3 ...$

Yes, it comes in very handy.

Thank you so much for explaining this. It's easier for me to remember something once I understand why it is that way.

 

[Apollinaria]

Oh my god, I completely forgot, power dissipated? What does this mean? How do I find it?
in the 33 ohm resistor...

 

[gneill]

Oh my god, I completely forgot, power dissipated? What does this mean? How do I find it?
in the 33 ohm resistor...

Look in your notes or text for power dissipated by a current flowing through a resistor. You should find several expressions which relate the current through, or the voltage drop across the resistor to the power dissipated.

 

[Apollinaria]

Look in your notes or text for power dissipated by a current flowing through a resistor. You should find several expressions which relate the current through, or the voltage drop across the resistor to the power dissipated.

I have...

V=IR
P=VI

12V/33= 0.3636A (I, current).
12V x 0.3636 = 4.36. Wrong.

 

[gneill]

I have...

V=IR
P=VI

12V/33= 0.3636A (I, current).
12V x 0.3636 = 4.36. Wrong.

You'll have to isolate the current flowing through the particular resistor, or the voltage across it. It's certainly not the total battery voltage across it nor is it the total current through it, since there are other current paths and potential drops in the circuit.

You may also find the formulas P = I²R and P = V²/R for the power dissipated in a resistor to be of use as you go along. Here I is the current flowing through the resistor, and V the potential drop across the resistor.

 

[Apollinaria]

You'll have to isolate the current flowing through the particular resistor, or the voltage across it. It's certainly not the total battery voltage across it nor is it the total current through it, since there are other current paths in the circuit.

You may also find the formulas P = I²R and P = V²/R for the power dissipated in a resistor to be of use as you go along. Here I is the current flowing through the resistor, and V the potential drop across the resistor.

Thats what I thought but how can I isolate it? Like I said, this unit wasn't clearly explained and we certainly never talked about isolating anything.

 

[gneill]

Thats what I thought but how can I isolate it? Like I said, this unit wasn't clearly explained and we certainly never talked about isolating anything.

You'll have to apply KVL and/or KCL (Kirchhoff's laws) to analyze the circuit.

If I may give a hint, you might find it helpful to first find the total current through the equivalent resistance, then treat the circuit like three resistances in series: the two 10 Ω resistors and the parallel resistance you found earlier (the 100 Ω in parallel with the 10 Ω + 33 Ω). The potential drop across the parallel resistance "block" can then be found, and thus the currents through the parallel branches...

 

[Apollinaria]

Just a hint?! LOL. I really cannot do this. Didn't learn/see KVL and KCL. How the heck do I find a potential drop? Are you tired yet? :/

R total = 20.03
V= 12V

12/20.03 = 0.599 A total.

Can you show me a solution?

 

[gneill]

Just a hint?! LOL. I really cannot do this. Didn't learn/see KVL and KCL. How the heck do I find a potential drop? Are you tired yet? :/

R total = 20.03

That's not the total resistance you found before
Better check!

The potential drop across a resistance is given by Ohm's law: V = I*R.

Can you show me a solution?

We can only provide hints and helpful (?) suggestions, not complete answers. Sorry, them's the rules

 

[Apollinaria]

That's not the total resistance you found before
Better check!

The potential drop across a resistance is given by Ohm's law: V = I*R.

We can only provide hints and helpful (?) suggestions, not complete answers. Sorry, them's the rules

Mostly helpful. I feel horrible having to interrogate you like this though. Failure at it's most pathetic. You're right. It's 50.06. I was looking at the wrong, partial solution I scribbled on the page.

V/R=I
12V/50.06ohms= 0.239 A

How do I get the 33 isolated?! :grumpy: I suspect having to substract stuff. From other stuff :rofl:

 

[gneill]

Mostly helpful. I feel horrible having to interrogate you like this though. Failure at it's most pathetic. You're right. It's 50.06. I was looking at the wrong, partial solution I scribbled on the page.

V/R=I
12V/50.06ohms= 0.239 A

How do I get the 33 isolated?! :grumpy: I suspect having to substract stuff. From other stuff :rofl:

Ha! Start by finding the potential drop across the "parallel block" resistance, R_p :

Once you have that you can narrow your focus to that portion of the circuit; two parallel branches with a known potential difference across them both.

 

[Apollinaria]

If V=IR and V is given as 12V,

then we already have the potential drop?

V/R = I, current
12/30.06 (for that block) = .399 A

Or,

12/43 = 0.279 A
12/100 = 0.12 A

for each branch of that block.:uhh:

 

[gneill]

If V=IR and V is given as 12V,

then we already have the potential drop?

That's the potential drop across the total resistance; it's the battery voltage across the net resistance of 50.06 Ω, yielding the total current ---> I = (12 V)/(50.06 Ω). The "parallel block" R_p will drop only a portion of it.

This total current I flows through a 10 Ω resistor, then through the R_p, then through another 10 Ω resistor. Since they're in series, they all have the same current passing through them. Each resistance will produce a voltage drop due to the current flowing through it (their sum should be equal to the 12V total).

You can work out the individual voltage drops across the resistances in this path using Ohm's law: V = I*R. What's the voltage drop across R_p?

Can you then find the current through the branch of R_p containing the 33 Ω resistor?

 

[Apollinaria]

That's the potential drop across the total resistance; it's the battery voltage across the net resistance of 50.06 Ω, yielding the total current ---> I = (12 V)/(50.06 Ω). The "parallel block" R_p will drop only a portion of it.

This total current I flows through a 10 Ω resistor, then through the R_p, then through another 10 Ω resistor. Since they're in series, they all have the same current passing through them. Each resistance will produce a voltage drop due to the current flowing through it (their sum should be equal to the 12V total).

You can work out the individual voltage drops across the resistances in this path using Ohm's law: V = I*R. What's the voltage drop across R_p?

Can you then find the current through the branch of R_p containing the 33 Ω resistor?

I got it! Part of it!
2.39V for each 10. and 7.18343V for the block entirely.

10.277V for the portion of the block with 33.
23.9V for the 100ohm branch of the block.

 

[gneill]

I got it! Part of it!
2.39V for each 10. and 7.18343V for the block entirely.

Yes, those look pretty good for those voltage drops. I think if you keep a few more decimal places in intermediate results you'll find that the drop across the parallel block is closer to 7.207 V.

10.277V for the portion of the block with 33.
23.9V for the 100ohm branch of the block.

Oops. No, You shouldn't ever find voltage drops larger than the total voltage being supplied. The total current divides in the parallel block; a portion goes through the 100 Ω resistor, and what's left goes through the 10 + 33 = 43 Ω branch.

This is where you make use of the voltage drop across the block; Parallel branches share the same potential difference across them. With about 7.2 V across the 43 Ω branch, what's the current there?

 

[Apollinaria]

Yes, those look pretty good for those voltage drops. I think if you keep a few more decimal places in intermediate results you'll find that the drop across the parallel block is closer to 7.207 V.

Oops. No, You shouldn't ever find voltage drops larger than the total voltage being supplied. The total current divides in the parallel block; a portion goes through the 100 Ω resistor, and what's left goes through the 10 + 33 = 43 Ω branch.

This is where you make use of the voltage drop across the block; Parallel branches share the same potential difference across them. With about 7.2 V across the 43 Ω branch, what's the current there?

This is truly mind blowing and my least favorite unit as of now.
The total current was 0.239 and it divides into the 43 and 100 branches. Unevenly.
Itot x Rblock = V block = 7.18 or 7.2V
7.2V/Rbranch = Ibranch
7.2/43 = .167 in the branch.

 

[gneill]

This is truly mind blowing and my least favorite unit as of now.
The total current was 0.239 and it divides into the 43 and 100 branches. Unevenly.
Itot x Rblock = V block = 7.18 or 7.2V
7.2V/Rbranch = Ibranch
7.2/43 = .167 in the branch.

Okay! Since you now know the current in the branch with the 33 Ω resistor you can find the power it dissipates using the formula mentioned before: P = I²R.

Since you explained that you haven't got a lot of circuit analysis rules and techniques in your "toolkit", the method I presented made use of what might be referred to as bare-bones or "brute force" techniques; Basic resistor combinations and Ohm's law to find potentials and currents at successive "levels" of the circuit. These can be quite effective with a bit of practice, and in some cases faster to use than more sophisticated methods.

 

[Apollinaria]

Success. .167² x 33. What boggles my mind is why I would use one of the resistors (the 33) because they both interfere with the current. Good god.
Edit: moreover, our formula sheet doesn't even have that formula. And I couldn't derive it.

 

[gneill]

Success. .167² x 33. What boggles my mind is why I would use one of the resistors (the 33) because they both interfere with the current. Good god.

Well, they did ask for the current dissipated by that particular resistor... no doubt because it was more difficult to "get at" than other choices.

Edit: moreover, our formula sheet doesn't even have that formula.

Could be they expected you to drill down a little further and determine both the current through and voltage across that resistor, then apply P = IV. Just one more step similar to all the rest. It can be handy to remember the P = I²R and P = V²/R forms.

 

[Apollinaria]

Well, they did ask for the current dissipated by that particular resistor... no doubt because it was more difficult to "get at" than other choices.


Could be they expected you to drill down a little further and determine both the current through and voltage across that resistor, then apply P = IV. Just one more step similar to all the rest. It can be handy to remember the P = I²R and P = V²/R forms.

Yes, they did, but what I mean is, from my lack of understanding of this unit, I would assume that the 10ohm resistor would mess with the 33ohm resistor power dissipated etc.. but you said they have the same current passing through them..
How would I do this the other way?

 

[gneill]

Yes, they did, but what I mean is, from my lack of understanding of this unit, I would assume that the 10ohm resistor would mess with the 33ohm resistor power dissipated etc.. but you said they have the same current passing through them..

The thing to remember is that components in series share the same current (same current flows through each) while components in parallel share the same potential drop. If you can reduce the parallel bit to a single resistance value, then you've effectively got a purely series circuit to deal with and can make use of the shared current property. That'll get you to the potential drop across the parallel bit, so you can focus your attention there where you can make use of the shared potential drop property.

So it's a matter of applying one tool after another until you get down to what's happening with the particular component you're interested in.

How would I do this the other way?

For a circuit like this I'd probably go about it the same way, first determining the equivalent resistance and total current. Then perhaps I'd make use of what's called the "current divider" rule to get the current in the parallel branch of interest rather than do the shared potential step. Both take about the same amount of time and effort; It didn't seem to be worthwhile to introduce another technique at the last stage of the game.

Other methods would involve writing equations using Kirchhoff's Laws (KVL, KCL) and solving them for the current of interest. In this case there would be a pair of simultaneous equations to solve; Perhaps a bit more effort, but it would solve for all the currents in one go and could be useful if I needed them for other parts of the problem.

 

[Apollinaria]

I learned more from you today than I did from my prof this entire semester. And I honestly don't know where I'd be if you'd hadn't shown me how to do these problems. Still pulling my hair most likely :rofl:

I'm going to try to find more similar problems and work on those as well because I need to solidify this very unstable base. Good for you for knowing so much so well :) Physics isn't my field... Thank you.

 

[gneill]

I learned more from you today than I did from my prof this entire semester. And I honestly don't know where I'd be if you'd hadn't shown me how to do these problems. Still pulling my hair most likely :rofl:

No worries. Here at Physics Forums we're always happy to help anyone who's making the effort. You've demonstrated tenacity and effort, and I think you'll do fine once you get comfortable with the concepts.

I'm going to try to find more similar problems and work on those as well because I need to solidify this very unstable base. Good for you for knowing so much so well :) Physics isn't my field... Thank you.

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