Electrical Engineering problem

[WisMan]

I've been working on some EE review problems and absolutely cannot get this problem to come out right.

Crappy MSPaint drawing of the problem

I'm fairly certain i need to add in a test source and find the Vx/Ix but i can't seem to get it to work out.

 

[TecHNooB]

I got R1 + (1+gR1) / R2. I hope i'm right!

 

[TecHNooB]

What's the answer? i wanna go to bed

 

[WisMan]

I don't have the answer, which was making this a lot more difficult.

That looks pretty close to what i would have expected it to look like though.

 

[TecHNooB]

Well here's what I did.

Set 1A as test current.

gV1 now becomes gR1.

Node Rule (subtract current going in):
-1 + Va/R2 - gR1 = 0

(Va - Vt)/R1 = -1

Subsitute Va with Vt or however you wanna do it. Vt / It just equals Vt because we set It to 1.

 

[Qacer]

R1 + R2

Since R1 and R2 are always constant, their values are not affected by the sources. From that perspective of Ri and using superposition (treat gVi as an open circuit), then your test source will always see R1 in series with R2. Your input resistance will just be R1 + R2. If you're looking at it from the perspective of gVi, then your input resistance would be just R2 assuming that the source connected to R1 is a voltage source.

[edit 2]
Hmm... Let me think a little bit more. I didn't actually consider what is actually involved in measuring the resistance. I think a multimeter sends out a test current (known value), measures the voltage across the resistance, and calculates the resistance using Ohm's law.

If we were to put a multimeter with this setup, then gVi will affect the voltage across R2 and thereby, affecting the overall voltage reading.

[edit 3]
Studies have shown that sleep helps the brain solve problems. In that case, I will sleep on it and hopefully, we'll have a clearer understanding later. If you have a multimeter and a voltage source handy, just play around and mimic that setup on a breadboard. Use a constant source in place of gV1 and measure the resistance with varying voltage source. See if it gives you different values.

 

[TecHNooB]

Goddamnit qacer, now I cant go to bed.

 

[PurdueRy]

I agree with Technoob except for one point. I think the answer should be Rt = R1+(gR1+1)R2

I am not sure why he's dividing by R2 as we're adding voltages to get this equation.

EDIT: Just checked it using his method and still got my answer. I think you might have a small error somewhere in your simplification Technoob.

 

[WisMan]

Originally posted by: PurdueRy
I agree with Technoob except for one point. I think the answer should be Rt = R1+(gR1+1)R2

I am not sure why he's dividing by R2 as we're adding voltages to get this equation.

EDIT: Just checked it using his method and still got my answer. I think you might have a small error somewhere in your simplification Technoob.

I just worked it and got an answer very similar to yours. I got Rin = R2+gR1R2-R1. Only different by a sign. So i just probably made an algebra error someone. I'm happy enough with that to finally go to bed.

Thanks for the help you guys!

 

[George P Burdell]

Stop helping n00bies with their homework!

 

[PurdueRy]

Originally posted by: George P Burdell
Stop helping n00bies with their homework!

If it wasn't for the walkthroughs on certain problems from friends, professors and family members I would have found learning some of this material much harder. Sometimes its hard to understand the method without seeing a good example or two.

 

[esun]

http://www.eudean.com/misc/problem.pdf

BTW, Technoob's answer is definitely wrong, since the units don't work out.

 

[TecHNooB]

Just woke up

Damn 7 30 classes.

Somehow, I went from Va/R2 -1 - gR1 = 0 to Va = (1 + gR1) / R2. Algebra ftl

 

[Qacer]

My original answer was wrong. The right answer is posted in this thread. I wasn't thinking properly.