ECE Homework Help

iamtrout

Diamond Member
Nov 21, 2001
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What is going on here?

http://img100.imageshack.us/img100/7863/temp3lj.png

a). Draw an equivalent circuit clearly showing ?ground? (reference) node. (25 points)
b). Calculate current flowing through the 1K resistor and identify its direction (30 points)
c). Calculate voltage of the Signal out at device (45 points).
Help & Hints:

It?s sometimes acceptable to identify output terminals (such as Signal out at device) without explicitly referencing them to ground node. So, what that means in this case is that you should find a voltage between the node closest to the said output terminal and ground.

I'm not understanding the diagram at all. 1.5V is the ground reference node, but what's that 3.3V doing? It's not even in a loop, so does that mean that the circuit is equivalent to
http://img193.imageshack.us/img193/5766/temp20eb.png

We're currently doing nodal analysis but all of our examples have been nice little loops or multiloops. We've never had something where the circuit was not a loop (the signal out just kinda terminates).

TIA
 

dighn

Lifer
Aug 12, 2001
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well, you can assume that the ground of the 1.5 src nd the 3.3v src are connected. so yes you do have a loop
 

slpaulson

Diamond Member
Jun 5, 2000
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Draw your 3V source like your 1.5V source is in your edited pic.

I don't really remember what they mean by a reference node... wouldn't that be ground?
 

iamtrout

Diamond Member
Nov 21, 2001
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Originally posted by: JohnCU
Originally posted by: cRazYdood
Draw your 3V source like your 1.5V source is in your edited pic.


That's what I thought initially, but rereading his questions "a). Draw an equivalent circuit clearly showing ?ground? (reference) node."

"node" is singular. And he says later in the homework:

It is customary in real engineering circuit diagrams to identify ground-referenced voltage sources as (1.5V on top of the "T"). this kind of symbol describes a voltage source and you can replace it in an equivalent circuit as follows: (1.5V next to the circle with +/- and chassis ground symbol).

He's making it sound like 1.5V is the ONLY ground referenced voltage source, which is why I'm wary of labeling the 3.3V as also a ground referenced source.

But suppose the 3.3V is also a source, would the current arrows be something like this?

http://img299.imageshack.us/img299/32/untitled4ir.png

Aren't the arrows supposed to be backwards, though. But if they're backwards then how is signal going INTO the device?
 

slpaulson

Diamond Member
Jun 5, 2000
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Originally posted by: iamtrout
Originally posted by: JohnCU
Originally posted by: cRazYdood
Draw your 3V source like your 1.5V source is in your edited pic.


That's what I thought initially, but rereading his questions "a). Draw an equivalent circuit clearly showing ?ground? (reference) node."

"node" is singular. And he says later in the homework:

It is customary in real engineering circuit diagrams to identify ground-referenced voltage sources as (1.5V on top of the "T"). this kind of symbol describes a voltage source and you can replace it in an equivalent circuit as follows: (1.5V next to the circle with +/- and chassis ground symbol).

He's making it sound like 1.5V is the ONLY ground referenced voltage source, which is why I'm wary of labeling the 3.3V as also a ground referenced source.

But suppose the 3.3V is also a source, would the current arrows be something like this?

http://img299.imageshack.us/img299/32/untitled4ir.png

Aren't the arrows supposed to be backwards, though. But if they're backwards then how is signal going INTO the device?

It's generally assumed that no current flows into a probe, or in this case the output of the device.

Draw both the 3.3V and the 1.5V as you drew your 1.5V source in your other post. You'll see that they both have ground on one end. You can then connect the grounds together, and you have a loop.

Current ALWAYS flows from high voltage to low voltage.
Hopefully this helps.

Pic

You can simplify the 2 voltage sources into 1 source.

Edit: Actually get rid of the arrow between the negative terminals of the voltage sources.
 

iamtrout

Diamond Member
Nov 21, 2001
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Ahhh... I see. In your pic, shouldn't the arrows be in the opposite direction? The 3.3V (going from + to - ) flowing into the 1.5V ( - to + ), across the resistors, and then back into the + end of the 3.3V?

How would one get voltage across the signal out device? There's no resistance to plug into the V = IR equation... or current for that matter since you said current doesn't flow through a probe.
 

dighn

Lifer
Aug 12, 2001
22,820
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Originally posted by: iamtrout
Ahhh... I see. In your pic, shouldn't the arrows be in the opposite direction? The 3.3V (going from + to - ) flowing into the 1.5V ( - to + ), across the resistors, and then back into the + end of the 3.3V?

How would one get voltage across the signal out device? There's no resistance to plug into the V = IR equation... or current for that matter since you said current doesn't flow through a probe.

for a voltage source, the current goes from - to + inside but + to - outside. another way to look at it is that the 3.3 v's + is stronger than 1.5v's + so current flows from 3.3v's + terminal to 1.5V +'s terminal.

as for the voltage, it would be relative to the ground which is the same as the - terminals of the two voltage srouces. so you can add from the ground = 3.3v - 5k * curent_thru_5k

edit: made a sign error
 

slpaulson

Diamond Member
Jun 5, 2000
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In V = IR, V is the voltage drop across the resistor, not the voltage at an end of the resistor.

Edit: nevermind on the direction thing.
 

iamtrout

Diamond Member
Nov 21, 2001
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Ahhh.. it all makes perfect sense now...

Thanks a lot guys! I don't think I could ever have done this homework without your insight.

:beer: :heart:
 
Feb 19, 2001
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It doesn't matter what you set as positive, negative, you just need to be consistent in your sign convention.

Bleh. Nodal analysis. Mesh analysis. Fun.
 

slpaulson

Diamond Member
Jun 5, 2000
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Yeah, you could have had your current going in the direction you drew it, you would just have negative numbers.

Either way, I'm glad I was some help.