resistor question

[canis]

I am just kidding, but for the person who calculated 30 resistors on a exam, the room must have been noisy with calculator buttons being furiously pressed, or the question must have been worth big points where the most time is spent. Again, I am just kidding.

 

[Juncar]

Nah it was for a 1.5h mid-term, it had five questions and they were all worth the same amount. The resistor question involved source transformation and superposition, which simplifies it somewhat, but yea it was furious button mashing.

 

[TuxDave]

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)

/sigh. The method is not anything different than using adding resistances in series or using the resistances in parallel equation. Both of those are algebraically the same if you do it the slow way and start rederiving the current down each leg of parallel resistors to find the equivalent resistance. The voltage divider equation is the same class of equations. Very fundamental stuff and very efficient to use. I have a hunch that since you haven't seen the voltage divider equation you haven't developed the intuition on when to use it. I don't see a need for generating 12 voltage divider equations in the OP.

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.

While I'm 99% design, I would say the small time that I spend troubleshooting some poor undergraduate lab experiment, the voltage divider can easily show how an amplifier was losing its gain by the voltage division of the series resistance of a voltage source to the base biasing resistors (yeah we were using BJTs as the time). Not using a voltage divider equation is like not using a parallel resistance equation. It's a very basic tool and can come in handy.

As for your last comment. That's probably just the way you think. Why bother with theoreticals when you can just place a meter across and measure it. Unfortunately if you're already at that state of the product that means you already wasted the $$$ in making a broken product. But interestingly enough, applying a theoretical "test voltage" and calculating the theoretical "test current" is a method in EE. Just in more complicated cases than the above.

 

[DominionSeraph]

/sigh. The method is not anything different than using adding resistances in series or using the resistances in parallel equation. Both of those are algebraically the same if you do it the slow way and start rederiving the current down each leg of parallel resistors to find the equivalent resistance. The voltage divider equation is the same class of equations. Very fundamental stuff and very efficient to use.

You keep failing to understand that this isn't a math issue.

I have a hunch that since you haven't seen the voltage divider equation you haven't developed the intuition on when to use it.

I don't need to have seen it -- it's there in the numbers. But it's just not particularly useful in troubleshooting. It's much better backwards.

I don't see a need for generating 12 voltage divider equations in the OP.

Widen your view.

Lack of foresight leads to wasted effort.

While I'm 99% design, I would say the small time that I spend troubleshooting some poor undergraduate lab experiment, the voltage divider can easily show how an amplifier was losing its gain by the voltage division of the series resistance of a voltage source to the base biasing resistors (yeah we were using BJTs as the time).

For troubleshooting you should have checked the voltage, not calculated it. The voltage proportion gives you resistance proportion. Now you have everything, not just what's on paper.
Testing the hardware beats testing the schematic.

 

[TuxDave]

/gives up. So you measure the input voltage to the amp and see that the voltage is low. Why is it low? Where do you fix it? The resistors are all within spec. You tell me how do you fix it if you don't know how to find a cause that matches your measurements.

Oh...and my foresight based on the questions given shows me that I don't need to calculate 12 voltage dividers and hence, not waste effort. 😛

 

[DominionSeraph]

Do you find an actual voltage reading to be a magical entity completely disassociated from paper calculations?

If you figured out it was a biasing error from a voltage divider calculation, the same number taken off a direct reading shouldn't be giving you this much difficulty.

 

[Juncar]

Dominion every electrical circuit can be represented using math. It is what engineers use to model the circuit when they design those computer parts that you are using. Scratch that, most engineering models are based on some mathematical representation, anything from civil engineering to chemical engineering.

A quick hand analysis is useful for estimating the values that you expect to get from your measurement, so that you have something to compare to.

 

[TuxDave]

Do you find an actual voltage reading to be a magical entity completely disassociated from paper calculations?

If you figured out it was a biasing error from a voltage divider calculation, the same number taken off a direct reading shouldn't be giving you this much difficulty.

Troll much? If I thought that, why would I be arguing to use calculations to match your readings?

The biasing was actually correct. Just for whatever reason we were getting half the voltage we wanted at the input of the amplifier. Actually, along those lines, how would you even know that you weren't supposed to get half? You measured some voltage, how do you know that it's the wrong one?

A direct reading of the input waveform will just show you some value which you don't even know if it's correct. BJT is fine, resistors are fine, voltage supply is fine. It turns out that the amplifier chosen will never work with the source chosen to get what you expected because the input impedance of the entire amplifier is too low.

I can't believe you're still arguing this.