resistor question

[TheFirm]

Hi all,

I didn't know where to post this so I decide to post it here. But I'm drawing a blank and my brain is killing me trying to figure out 2 questions. Can someone please help me


thanks in advance. Any insights on how to solve it will be great

 

[Matthiasa]

The resistance is calculated using parallel and series resistance. The Voltage then comes from voltage division.
Redrawing it so that all the wires are vertical or horizontal might also make it easier for you to visualize it.

Oh and this probably should be in off topic. As hw is not suppose to be done here.

 

[PottedMeat]

lol @ 'Complex circuit.'

and yeah, just use the rules for series and parallel resistances and ohms law.

if that doesn't help then you're hopeless.

 

[TuxDave]

Simplify the resistors one at a time.
1) R1 and R2 are in series, substitute both with a single resistor of the same resistance
2) R4 and R5 are in parallel, same as above
... repeat until you end up with one resistor.

 

[bobdole369]

Not exactly complex. Tux has it right though.

 

[DominionSeraph]

Redraw it so your series and parallel circuits are more easily identifiable. (Trace your current paths)

R3 is in parallel with R1+R2
R4 is in parallel with R5
Those two networks are in series with each other.
R6 is in series with everything.
As they're giving you Ri, you need to add that in too.
Then it's ohm's law -- voltage divided by resistance. That gives you current, which you use for 3-51.

 

[Juncar]

Total resistance is 15 ohm.
Voltage across the equivalent resistor for R1 to R6 is 52 V, since the equivalent resistor is 13 ohm. Then use voltage division which is 60 V * (13 omh / 15 ohm). Are you studying electrical engineering? Since I did similar question as yours except for mine had like 30 resistors. It was an exam question.

 

[DominionSeraph]

Then use voltage division which is 60 V * (13 omh / 15 ohm).

I'm curious as to why you'd do it that way. 60/15*13 gives you total current as an intermediate step. 13/15 is just a particular resistance fraction.

 

[esun]

I'm curious as to why you'd do it that way. 60/15*13 gives you total current as an intermediate step. 13/15 is just a particular resistance fraction.

Very commonly used expression in EE is for a voltage divider.

V_X = V_DD (R_2) / (R_1 + R_2)

Where you have a source connected to two series resistors R_1 and R_2 with an intermediate unknown voltage V_X between them. Also commonly used is the corresponding expression for current division between parallel resistors. I typically viewed it as a voltage time a ratio of resistances rather than a current times a resistance (though they're obviously equivalent).

 

[DominionSeraph]

I suppose that would keep things simple when designing a voltage divider. Altering the ratio alters the voltage.
I wouldn't want to troubleshoot like that, though. As you change test points, you'd have to store the resistance value, subtract it from one side, add it to the other, then divide. Gah!

Current is just nice. Easy voltage drops, and thus power on demand, too.
R6 is a toasy little fellow at 80w.

 

[Juncar]

Well its a simplified form of the general voltage division rule, when you only have two resistors in series. Its a one liner and easy enough to do it in your head, but current works as well. In EE, you're taught different methods to solve the same circuit, its just that for most circuits, certain methods makes it easier to solve over the other methods. In this case, resistors in series with a voltage source makes voltage division the easier method. It also makes sense in circuit analysis because the R1 / (R1 + R2) would be considered your transfer function for the very simple circuit.

 

[TuxDave]

I suppose that would keep things simple when designing a voltage divider. Altering the ratio alters the voltage.
I wouldn't want to troubleshoot like that, though. As you change test points, you'd have to store the resistance value, subtract it from one side, add it to the other, then divide. Gah!

Current is just nice. Easy voltage drops, and thus power on demand, too.
R6 is a toasy little fellow at 80w.

What if you swap out one resistor to another value? The voltage divider you change the value of one of the R's in the equation and do the math again. Your method you recalculate the current based on the new equivalent resistance and then multiply it against the portion of resistance. A two step process.

Honestly, if you do the algebra you'll find that your method and the voltage divider method are identical. The best method in each case is whatever takes the least amount of work.

 

[LifeStealer]

You could also solve it using a delta-y transform That would make it much more fun!

 

[TheFirm]

thanks, I've been pretty busy with work and just got back today. I'm study for an exam I had thursday... thanks for the help. Are most of you guys EE majors? or have your degree?

 

[Zoomer]

3-52: Substitute Es with an AC source. Draw the frequency and phase bode plots.

Hint: This is very easy.

 

[johnnydrinkawater]

An easy way to decipher Req for weird looking resistor connections is to identify series combinations first (resistors sharing only one node), then to identify parallel combinations (resistors or combinations of resistors sharing two nodes).

 

[johnnydrinkawater]

3-52: Substitute Es with an AC source. Draw the frequency and phase bode plots.

Hint: This is very easy.

I think you mean the mag. & phase bode ^_^

 

[Juncar]

thanks, I've been pretty busy with work and just got back today. I'm study for an exam I had thursday... thanks for the help. Are most of you guys EE majors? or have your degree?

Yea I'm doing my majors for EE atm, currently third year. Can I ask what course the exam was for?

 

[ussfletcher]

That is a terrible schematic for an easy circuit. On first glance it makes it look more complex. Actually I was thinking it was some weirdly constructed wheatstone bridge when I scrolled past it to see the questions.

 

[bobsmith1492]

Looks like a Circuits 1 problem. I can do it in my head other than the 5||7.5 though... otherwise it's 12+(5||7.5) counting the battery's internal series resistance. It's not entirely clear whether or not that is part of the "total resistance" they ask for.

 

[Juncar]

Well if they give you the internal resistance, then 99% of the time, when they ask for total resistance, it includes that
Or else they wouldn't include it in the circuit.

 

[DominionSeraph]

What if you swap out one resistor to another value?

That would be 'designing a voltage divider'.

Honestly, if you do the algebra you'll find that your method and the voltage divider method are identical.

Yes, but they're not identical philosophically, which is why they're not mentally identical in either situation.
His way leverages a second-order property. Mine sticks to first-order properties.
Because his way connects things 'round the back way through a mathematical relationship, you can break the direct relationships to give you conceptual malleability. Good for design. The downside is that you have to maintain the connection between the derived function and reality or you set yourself up to make all sorts of mistakes. If you don't understand WHY the equation works, you could mistakenly use it in a situation where it doesn't apply.
For troubleshooting, you're only dealing with stark reality. The values are already engineered, so there's no need for the conceptual mutability gained by the additional distance of specialty equations. When the need arises you may generate such an equation on the spot from your understanding of electrical fundamentals and basic math, so there's no reason to keep a mental toolbox filled with such single-use tools.

Are most of you guys EE majors?

Ex-Navy Electronics Technician.

That is a terrible schematic for an easy circuit.

They give these examples on purpose so that students don't get locked into using visual pattern matching to identify series and parallel circuits.
"Series = straight line. Parallel = box." Sure, in school. In the first two weeks of it. After that, that identification system doesn't work so well.

 

[Juncar]

That is like saying 3 x 3 is really 3 + 3 + 3...makes no difference as long as you know the underlying principle.
Voltage division is useful for finding out the output across a component without having to calculate the equivalent impedance and the current running through the branch. Your method is based on ohm's law, and voltage division is derived from that. Its not that really that much more complicated though, and its frequently used so most EE who work with circuits should know it.

 

[TuxDave]

That would be 'designing a voltage divider'.

Well aren't you a smart ass.

Yes, but they're not identical philosophically, which is why they're not mentally identical in either situation.
His way leverages a second-order property. Mine sticks to first-order properties.
Because his way connects things 'round the back way through a mathematical relationship, you can break the direct relationships to give you conceptual malleability. Good for design. The downside is that you have to maintain the connection between the derived function and reality or you set yourself up to make all sorts of mistakes. If you don't understand WHY the equation works, you could mistakenly use it in a situation where it doesn't apply.
For troubleshooting, you're only dealing with stark reality. The values are already engineered, so there's no need for the conceptual mutability gained by the additional distance of specialty equations. When the need arises you may generate such an equation on the spot from your understanding of electrical fundamentals and basic math, so there's no reason to keep a mental toolbox filled with such single-use tools.

That's a difference of opinion. For any decent EE, a voltage divider is one algebra step away from V=IR and to most of them making that "second order property" is a no-brainer. I use (AB)/(A+B) proudly for parallel resistance. BIG DEAL. Oh no I'm abstracting so far away from the fundamentals.

You don't see me trying to rederive the resistance based on cross section area, mobility of carriers and length. That's the truly fundamental thing if you never want to lose sight of what the heck you're calculating. "single use tools". Yeah like calculating the small signal gain of a transistor is a nice single use tool but guess what, it's the only way we can do things to a first degree error without having a computer in your head. This is just algebra. Not rocket science.

And yes I'm an EE with a Masters working in the industry.

 

[DominionSeraph]

That is like saying 3 x 3 is really 3 + 3 + 3. makes no difference as long as you know the underlying principle.

No, 3*3 also equals 3+3+3, 3+3+3, 3+3+3, 3+3+3, and 3+3+3. So it works for all 6 combinations, which means there's no limitations on its equivalence.
This is not the case in R1/R2.

Here:

Billy can run 16 MPH.
Little Sally can run 8 MPH.

8/16 = 1/2. Sally runs at one-half Billy's speed.
8 MPH - 16 MPH = -8 MPH. Sally runs 8 MPH slower than Billy.

Billy hurt his knee and can only run 10MPH. How fast can Little Sally run?
One half? 10/2 = 5 MPH?
Eight less? 10-8 = 2 MPH?

No, Little Sally isn't the one hurt. Little Sally can still run 8 MPH.

"One-half" and "eight less" are mathematical description, not prescription.

Similar situation with R1/R2. The laws of nature prescribe the relationships described by Ohm's Law. Because of those laws, the ratio between series resistances ends up being proportional to the voltage drops. But R1/R2 itself has no central meaning. The extracted ratio means nothing to the universe as such. Throw two resistors on the ground and the universe isn't going to magically generate voltage just because the combination can be described as 13/15.

Because the foundation is solid, the derived equation is solid. But the equation isn't foundational. And this is in fact WHY it is so useful for circuit design. To design, you have to be able to play with values. But I, E, and R are solidly locked together. To have conceptual freedom, you need to break that block. R1/R2 gives you a place from which to apply leverage.

For any decent EE, a voltage divider is one algebra step away from V=IR and to most of them making that "second order property" is a no-brainer.

That doesn't make it in any way efficient.
He's in a class that's obviously currently on electronics fundamentals. There's no design there, unless there's some really odd trend going on in the electric heater field that I'm not aware of. So, he'll be on Ohm's Law, and they'll be doing a lot of, "I give you two legs; find the third." One scat, many questions regarding it. So why are you trying to defend a method that gives you trash as an intermediate step? Why defend a method where every voltage divider calculation requires you to start from scratch? There's twelve of them in that circuit alone. (ignoring the big one)

One value: 4A, does away with all that. There's no need to keep adding up all the resistances, as current is a function of resistance. And there's a good chance that one of the questions will ask for current anyway.

It's a pretty basic test-taking skill to predict what questions will be asked and then to work the current problem in such a way as to push in the direction of their answers.

You don't see me trying to rederive the resistance based on cross section area, mobility of carriers and length.

You seem to have a really hard time differentiating troubleshooting from engineering.

When I say, "X doesn't make sense in a troubleshooting capacity, but I see how it would be useful in design," and you come back with, "DON'T TELL ME IT ISN'T USEFUL BECAUSE SEE I USE IT IN DESIGN ALL THE TIME HAHA PROVED YOU WRONG N00B" ... yeeeeah... do you see how that doesn't quite marry?

Anyway, if you want to know resistance, just put a known voltage across it and read the current.