Blargh! Someone help me or this is going to drive me insane

Gregness · 12-02-2010, 10:57 PM

So, yesterday I had my laboratory section final in my basic circuit analysis class and I wanted to literaly break the planet in half when I was done with it. The theory part was easy enough, but I had the practical evaluation first which ALSO should have been easy except for one thing. The ENTIRE exam hinged on me being able to use a 10V source, and a 2K, 3.6K, 5.1K, and a 10K resistor (in Ohms) to design a Thevenin equivalent circuit with an equivalent resistance of 4.856K Ohms with an equivalent voltage of 6.62V (near there, anyway).

I don't know if I missed the lecture where they explained how to come up with an equivalence from nothing (as opposed to finding the equivalent of an existing circuit, which I can do) but what I came up with was this:

(the forward slashes are because the boards hate leading spaces apparantly).

/////////////////// ____2K_____
///////////////////|//////////////|
//////// ________|___5.1K___ |_____________ A
444444|//////////| /////////////|////////|
111111|//////////|___10K____|////////|
////////|//////////////////////////////////|
/////// +//////////////////////////////////|
////// 10V ////////////////////////////////|
//////// - ////////////////////////////////|
////////|//////////////////////////////////|
////////|__________3.6K____________|______B

As far as I can tell, this gives the correct equivalent resistance, but not the correct equivalent voltage. I got this far in about fifteen minutes, then just fucking stared at this thing for forty-five minutes running every other combination I could think of through in my head. I couldn't find another combination that even had the correct resistance much less voltage.

I could have done the rest of the practical part in my sleep. It was just stuff like calculating percent difference and power delivered and demonstrating the correct way to measure using a multimeter. But it all hinged on me being able to design the correct circuit!

I finally just asked the TA to let me build my incorrect circuit so that I could move on, but by that time he took my test, had me build it and demonstrate a few things, and then didn't let me finish the last question. And by this time I only had like 40 minutes left for the other half of the test so I had to rush it.

God, where's one of those FFffffffuuu- pics when I need one?

Anyone good with circuits wanna tell me where I screwed up?

Ryanderman · 12-03-2010, 09:50 AM

It's been a while since I've done circuits, but this is what I came up with:

........______3.6k___________10k______A
.......|.......................|
.......|..................___|___
...../.+.\...............|..........|
....|10V|.............2k........5.1k
.....\.-./................|______|
.......|.......................|
.......|______________|______________B
.......|
...Ground

Gregness · 12-03-2010, 02:09 PM

err, yeah but if you arrange it that way isn't the voltage at the other end of the 3.6k resistor 1.4366/5.0366 *10 or 2.85V

Isn't the voltage division equation Va = R2/(R2+R1)?

Vi______R1__Va____R2____

Sithdarth · 12-03-2010, 03:28 PM

That should work or at least be a close as you can get with what you where given. The Thevenin resistance is 4.875 Kohms and the voltage is 6.415 Volts.

Ryanderman · 12-03-2010, 03:44 PM

Arg! You're right! I screwed up my math.

Sith, wouldn't your Rth be just = 7.1K? To find Rth, don't you short the terminals, and find the Req? So the 3.6K and 10K wouldn't even see any current, because you have no resistor on the other path.

Sithdarth · 12-03-2010, 03:54 PM

Nope to find the Thevenin resistance you pretend the battery is a piece of wire and calculate the resistance left over. That puts the two sets of resistors in parallel. The resistance is calculated with the circuit still open. The voltage is calculated by shorting the terminals. In which case it is the current that would have been flowing through the circuit before it was shorted times the resistance of just the 2k and 5.1k resistors.

Which makes me notice I did screw up the value of the 5.1k resistor. Fixing that the equivalent resistance is 4.665 Ohms and the voltage is 6.570 Volts. Which are even closer.

Ryanderman · 12-03-2010, 03:59 PM

Ah. Knew it had been too long. What I was thinking of was finding the Norton equivalent current.

Grimpond · 12-03-2010, 04:16 PM

<3

synkr0nized · 12-03-2010, 04:34 PM

Quote:

Originally Posted by Gregness

(the forward slashes are because the boards hate leading spaces apparently).

This is just a general FYI for folks:

Unlike a word processor or other text editors, when HTML* is rendered by browsers white space beyond one character is not considered. In order to force multiple spaces, one usually makes use of the " " (non breaking space) special character. Unfortunately for users wishing to get blank space while posting on a forum, HTML is normally turned off or very limited; you won't be able to use   in your posts as an HTML character.

* even though you may use bulletin code, it all gets translated into HTML when viewed.
e: You'll notice that line breaks are the same in a web page (i.e. without a tag the whitespace means nothing), but on a forum your line breaks are properly parsed and given appropriate tags (<br>) when rendered.

An alternative that is viable here would be to use the CODE board tags. Here's an example:

Code:

This is the           code tag.
       It should leave my     extra       spaces in.
                                                  See?

Ryanderman · 12-03-2010, 04:57 PM

Because this is a fun time wasting problem, I think (if I've done the math correctly this time) I have another solution.

Code:

 
                            ___2K___
        ______5.1K_________|        |___A
       |              |    |__3.6K__|
       |              |            
     / + \            |
    | 10V |          10K
     \ - /            |
       |              |
       |______________|_________________B
       |
   Ground

That should give:
Rth=4.66 KOhms
Vth=6.62 Volts

EDIT 1: Hmmm. That code tag isn't as simple as I'd hoped. Fixing

EDIT Something-teen: There. Fixed. Looks all funky in the edit window though. I hope my math was right to make it worth it.

12-02-2010, 10:57 PM	#1
Gregness Never give up. Never give in. Join Date: Oct 2009 Location: Sacramento, CA Posts: 1,034	Blargh! Someone help me or this is going to drive me insane So, yesterday I had my laboratory section final in my basic circuit analysis class and I wanted to literaly break the planet in half when I was done with it. The theory part was easy enough, but I had the practical evaluation first which ALSO should have been easy except for one thing. The ENTIRE exam hinged on me being able to use a 10V source, and a 2K, 3.6K, 5.1K, and a 10K resistor (in Ohms) to design a Thevenin equivalent circuit with an equivalent resistance of 4.856K Ohms with an equivalent voltage of 6.62V (near there, anyway). I don't know if I missed the lecture where they explained how to come up with an equivalence from nothing (as opposed to finding the equivalent of an existing circuit, which I can do) but what I came up with was this: (the forward slashes are because the boards hate leading spaces apparantly). /////////////////// ____2K_____ ///////////////////\|//////////////\| //////// ________\|___5.1K___ \|_____________ A 444444\|//////////\| /////////////\|////////\| 111111\|//////////\|___10K____\|////////\| ////////\|//////////////////////////////////\| /////// +//////////////////////////////////\| ////// 10V ////////////////////////////////\| //////// - ////////////////////////////////\| ////////\|//////////////////////////////////\| ////////\|__________3.6K____________\|______B As far as I can tell, this gives the correct equivalent resistance, but not the correct equivalent voltage. I got this far in about fifteen minutes, then just fucking stared at this thing for forty-five minutes running every other combination I could think of through in my head. I couldn't find another combination that even had the correct resistance much less voltage. I could have done the rest of the practical part in my sleep. It was just stuff like calculating percent difference and power delivered and demonstrating the correct way to measure using a multimeter. But it all hinged on me being able to design the correct circuit! I finally just asked the TA to let me build my incorrect circuit so that I could move on, but by that time he took my test, had me build it and demonstrate a few things, and then didn't let me finish the last question. And by this time I only had like 40 minutes left for the other half of the test so I had to rush it. God, where's one of those FFffffffuuu- pics when I need one? Anyone good with circuits wanna tell me where I screwed up? Last edited by Gregness; 12-02-2010 at 11:02 PM.

12-03-2010, 09:50 AM	#2
Ryanderman Beard of Leadership Join Date: Apr 2006 Posts: 827	It's been a while since I've done circuits, but this is what I came up with: ........______3.6k___________10k______A .......\|.......................\| .......\|..................___\|___ ...../.+.\...............\|..........\| ....\|10V\|.............2k........5.1k .....\.-./................\|______\| .......\|.......................\| .......\|______________\|______________B .......\| ...Ground __________________ ~Your robot reminds me of you. You tell it to stop, it turns. You tell it to turn, it stops. You tell it to take out the trash, it watches reruns of Firefly.~

12-03-2010, 03:44 PM	#5
Ryanderman Beard of Leadership Join Date: Apr 2006 Posts: 827	Arg! You're right! I screwed up my math. Sith, wouldn't your Rth be just = 7.1K? To find Rth, don't you short the terminals, and find the Req? So the 3.6K and 10K wouldn't even see any current, because you have no resistor on the other path. __________________ ~Your robot reminds me of you. You tell it to stop, it turns. You tell it to turn, it stops. You tell it to take out the trash, it watches reruns of Firefly.~

12-03-2010, 03:59 PM	#7
Ryanderman Beard of Leadership Join Date: Apr 2006 Posts: 827	Ah. Knew it had been too long. What I was thinking of was finding the Norton equivalent current. __________________ ~Your robot reminds me of you. You tell it to stop, it turns. You tell it to turn, it stops. You tell it to take out the trash, it watches reruns of Firefly.~

12-03-2010, 04:57 PM	#10
Ryanderman Beard of Leadership Join Date: Apr 2006 Posts: 827	Because this is a fun time wasting problem, I think (if I've done the math correctly this time) I have another solution. Code: ___2K___ ______5.1K_________\| \|___A \| \| \|__3.6K__\| \| \| / + \ \| \| 10V \| 10K \ - / \| \| \| \|______________\|_________________B \| Ground That should give: Rth=4.66 KOhms Vth=6.62 Volts EDIT 1: Hmmm. That code tag isn't as simple as I'd hoped. Fixing EDIT Something-teen: There. Fixed. Looks all funky in the edit window though. I hope my math was right to make it worth it. __________________ ~Your robot reminds me of you. You tell it to stop, it turns. You tell it to turn, it stops. You tell it to take out the trash, it watches reruns of Firefly.~ Last edited by Ryanderman; 12-03-2010 at 05:05 PM.

12-03-2010, 02:09 PM	#3
Gregness Never give up. Never give in. Join Date: Oct 2009 Location: Sacramento, CA Posts: 1,034	err, yeah but if you arrange it that way isn't the voltage at the other end of the 3.6k resistor 1.4366/5.0366 *10 or 2.85V Isn't the voltage division equation Va = R2/(R2+R1)? Vi______R1__Va____R2____

12-03-2010, 03:28 PM	#4
Sithdarth Friendly Neighborhood Quantum Hobo Join Date: Mar 2004 Location: Outside the M-brane look'n in Posts: 5,403	That should work or at least be a close as you can get with what you where given. The Thevenin resistance is 4.875 Kohms and the voltage is 6.415 Volts.

12-03-2010, 03:54 PM	#6
Sithdarth Friendly Neighborhood Quantum Hobo Join Date: Mar 2004 Location: Outside the M-brane look'n in Posts: 5,403	Nope to find the Thevenin resistance you pretend the battery is a piece of wire and calculate the resistance left over. That puts the two sets of resistors in parallel. The resistance is calculated with the circuit still open. The voltage is calculated by shorting the terminals. In which case it is the current that would have been flowing through the circuit before it was shorted times the resistance of just the 2k and 5.1k resistors. Which makes me notice I did screw up the value of the 5.1k resistor. Fixing that the equivalent resistance is 4.665 Ohms and the voltage is 6.570 Volts. Which are even closer.