Does current flow through R4 when calculating Vout?

[Femme_physics]

In trying to figure out what's Vout, do I discount R4? I don't believe any current will flow through there.

http://img825.imageshack.us/img825/7236/vinvout.jpg

 

[Phrak]

If the op amp is not saturated Va=Vb=Vout.

Set all these nodes equal and solve.

 

[I like Serena]

Nope.

Current will flow through R4.

Actually the main flow of current will be from the 6 V pole to the earth. ;)

 

[Carl Pugh]

Phrak is correct.

Sharp observation Phrak.

 

[Zryn]

In trying to figure out what's Vout, do I discount R4? I don't believe any current will flow through there.

Why do you believe that?

 

[Femme_physics]

Why do you believe that?

Because of this

I know that in the left circuit, current will not flow where the resistors are attached to the straight wire connecting A and B, but on the right one, current will flow where the resistors are attached to the straight wire x and y

http://img824.imageshack.us/img824/880/this1p.jpg

Why is it different here?

 

[tedbradly]

Because of this



Why is it different here?

Why would current not flow in the circuit to the left? As long as v_b not equal to the voltage at the node at the bottom of the pair of resistors, current will flow.

 

[Femme_physics]

I didn't say current will not flow, I said current will not flow in the resistor box of the circuit to the left. It will just flow from a to b not entering it. Why is it entering R4 in the original question, if R4 leads the current nowhere?

 

[Phrak]

Do want the answer or how to solve the problem?

To solve the problem you need

A) to setup the system of linear equations. Uses variables such as I1, I2, I3 and I4 for your unknow currents. You have two unkown voltages which are Va and Vb, and theses are equal. This are 5 unknows when Va=Vb, so you should have 5 equations.

or B) solve it heuristically. For this, note that the current flowing in R1 is equally divided into R2 and R3, and the current through R3 is equal to the current though R4.

 

[Phrak]

Phrak is correct.

Sharp observation Phrak.

You're a gentlemen, Carl Pugh, and appreciated. Thank you.

 

[Femme_physics]

Do want the answer or how to solve the problem?

Answer to what? To what I asked in the first post? I got it, actually (that it does flow), I'm just trying to understand it now.

A) to setup the system of linear equations. Uses variables such as I1, I2, I3 and I4 for your unknow currents. You have two unkown voltages which are Va and Vb, and theses are equal. This are 5 unknows when Va=Vb, so you should have 5 equations.

Thanks. That's then next step. I didn't mean to get there just yet (was trying to clear the confusion) but now I just might :)

You're a gentlemen, Carl Pugh, and appreciated. Thank you.

*rolls eyes* men...

 

[I like Serena]

I didn't say current will not flow, I said current will not flow in the resistor box of the circuit to the left. It will just flow from a to b not entering it. Why is it entering R4 in the original question, if R4 leads the current nowhere?

R4 does lead the current somewhere.

It leads it to earth, which is effectively the negative pole of a battery.
The 6 V pole is effectively the plus pole of a battery.

This is a case where the battery is not part of the circuit.
Here the battery is indeed an "external force" or something like that!

Once upon a time, I said that you'd get to that sooner or later!

 

[Phrak]

*rolls eyes* men...

Yeah, those guys. Combative, contensious, always in oneupmanship given half a chance. It's not just refreshing and encouraging when we are given hope that there can be a world without self predation but peace.

 

[Femme_physics]

R4 does lead the current somewhere.

It leads it to earth, which is effectively the negative pole of a battery.
The 6 V pole is effectively the plus pole of a battery.

This is a case where the battery is not part of the circuit.
Here the battery is indeed an "external force" or something like that!

Once upon a time, I said that you'd get to that sooner or later!

Hmm, so "Vout" isn't the negative pole of the battery?

 

[Ouabache]

Did your instructor discuss two golden rules when
analyzing basic op-amp circuits with external feedback?

Rule 1) the output attempts to do whatever is necessary to make the
voltage difference between the - and + inputs equal zero.

Rule 2) The - and + inputs to the op-amp draw no current.

ref: http://books.google.com/books?id=bk...resnum=3&ved=0CCcQ6AEwAg#v=onepage&q&f=false" by Horowitz and Hill .

You can then solve for Vout using:
KVL : "The algebraic sum of all voltages in a loop must equal zero"
KCL: "The algebraic sum of all currents entering and exiting a node must equal zero"

Hmm, so "Vout" isn't the negative pole of the battery?

Correct ! (and dooley noted ), and why is that so? Hint: the negative pole of battery is at ground potential.

So you have me curious, in your post #6, does "מעגל א" translate to Circle A?

 

[I like Serena]

Hmm, so "Vout" isn't the negative pole of the battery?

No. Vout is not a pole of a battery.
For all practical purposes Vout is a loose hanging end of wire.

We are interested in the voltage at Vout, but in the problems you're facing nothing will be connected to it.

 

[uart]

I just want to point out that I usually get my resistors discounted if I buy more than 100 of the same type at a time.

 

[Zryn]

Because of this



Why is it different here?

When you look at the picture on the left, the red voltage encompasses both terminals of the two resistors. If you consider in general that Vxy = Vx - Vy then you can see when both resistances are equal that the voltage across a component is zero, and thus by Ohms Law when V = 0 and R has a value, then I = 0.

When you look at the picture on the right, you can see that the same Vx - Vy will produce a non zero result as the two voltages must be different to each other, since there will be a drop across the component between them.

This is a good example of a Short Circuit (the circuit has been made much shorter by a connection somewhere that bypasses components).

 

[Femme_physics]

I was just told by my instructor that this problem is "off my league" and we'll only study it in the next semester. It turns out I solved all the problems possible for our first semester (correctly, thanks to this forum :) ) and in the last class I had nothing to do but roll my eyes and yell out all the answers to prove I got juice. Heh.

So you have me curious, in your post #6, does "מעגל א" translate to circle A?

Yes :)

Did your instructor discuss two golden rules when
analyzing basic op-amp circuits with external feedback?

Rule 1) the output attempts to do whatever is necessary to make the
voltage difference between the - and + inputs equal zero.

Rule 2) The - and + inputs to the op-amp draw no current.

ref: The Art of Electronics by Horowitz and Hill .

You can then solve for Vout using:
KVL : "The algebraic sum of all voltages in a loop must equal zero"
KCL: "The algebraic sum of all currents entering and exiting a node must equal zero"

Theoretically I can, but since this is 2nd semester stuff I'll leave it alone for now. :) Thanks for the awesome help effort though! Is this reference book any good? Should I use it to study electronics/circuits better? I'm looking for a good source.

and why is that so? Hint: the negative pole of battery is at ground potential.

Thanks for the answer :)

When you look at the picture on the left, the red voltage encompasses both terminals of the two resistors. If you consider in general that Vxy = Vx - Vy then you can see when both resistances are equal that the voltage across a component is zero, and thus by Ohms Law when V = 0 and R has a value, then I = 0.

When you look at the picture on the right, you can see that the same Vx - Vy will produce a non zero result as the two voltages must be different to each other, since there will be a drop across the component between them.

This is a good example of a Short Circuit (the circuit has been made much shorter by a connection somewhere that bypasses components).

Yes, I understand that :) But the original question seemed the same thing to me, but I guess I'm wrong since the ground point is in fact the other terminal of the battery! Anyway, I'm not sure if it's a good idea for now since this is next semester stuff. I appreciate the reply! :)

 

[uart]

I was just told by my instructor that this problem is "off my league" and we'll only study it in the next semester.

Well here's a quick solution just so you know how to do this type of problem.

Start by using the fact that $V_a = V_b = V_o$ as has already been pointed out above. To keep the number of variables to a minimum I will just use $V_a$ any time I get any of these three in an equation.

Also note that I will use "G's" to represent resistance reciprocals, eg $G_1 = 1/R_1$ etc.

Finally note that I will need to introduce one extra node voltage not currently labeled on the diagram. Let "x" represent the node at the junction of R1, R2 and R3, and $V_x$ its voltage.

KCL at node "x" gives :

$$G_1 (V_x - V_i) + G_2 (V_x-V_o) + G_3 (V_x - V_a) = 0$$

Rearranging and setting Vo = Va gives,

$$V_x (G_1 + G_2 + G_3) = G_1 V_i + (G_2 + G_3) V_a$$

And since all R's and hence all G's are equal value this reduces to,

$$V_x = \frac{1}{3} V_i + \frac{2}{3} V_a$$

Now we need just one more equation connecting Vx and Va and then we can eliminate Vx. From the simple "voltage divider" (formed by R3 and R4) we can write,

$$V_x = 2 V_a$$

Now we can just eliminate Vx between the previous two equations to get,$$2 V_a = \frac{1}{3} V_i + \frac{2}{3} V_a$$

$$\frac{4}{3} V_a = \frac{1}{3} V_i$$

$$V_a = \frac{1}{4} V_i$$

So given that $V_i = 6$ we get the output voltage is $V_o = V_a = 1.5$ volts.

 

[Ouabache]

ref: 'The Art of Electronics' by Horowitz and Hill

Is this reference book any good? Should I use it to study electronics/circuits better? I'm looking for a good source.

I like it and find this a very useful text for electronics.
We have a more detailed review of this book by one of our Engineering Mentors,
berkeman shared in https://www.physicsforums.com/showthread.php?t=174586".

 

[Femme_physics]

Well here's a quick solution just so you know how to do this type of problem.

Start by using the fact that $V_a = V_b = V_o$ as has already been pointed out above. To keep the number of variables to a minimum I will just use $V_a$ any time I get any of these three in an equation.

Also note that I will use "G's" to represent resistance reciprocals, eg $G_1 = 1/R_1$ etc.

Finally note that I will need to introduce one extra node voltage not currently labeled on the diagram. Let "x" represent the node at the junction of R1, R2 and R3, and $V_x$ its voltage.

KCL at node "x" gives :

$$G_1 (V_x - V_i) + G_2 (V_x-V_o) + G_3 (V_x - V_a) = 0$$

Rearranging and setting Vo = Va gives,

$$V_x (G_1 + G_2 + G_3) = G_1 V_i + (G_2 + G_3) V_a$$

And since all R's and hence all G's are equal value this reduces to,

$$V_x = \frac{1}{3} V_i + \frac{2}{3} V_a$$

Now we need just one more equation connecting Vx and Va and then we can eliminate Vx. From the simple "voltage divider" (formed by R3 and R4) we can write,

$$V_x = 2 V_a$$

Now we can just eliminate Vx between the previous two equations to get,


$$2 V_a = \frac{1}{3} V_i + \frac{2}{3} V_a$$

$$\frac{4}{3} V_a = \frac{1}{3} V_i$$

$$V_a = \frac{1}{4} V_i$$

So given that $V_i = 6$ we get the output voltage is $V_o = V_a = 1.5$ volts.

Thanks a lot! I'll review it Saturday when I'll have plenty of time :)

I like it and find this a very useful text for electronics.
We have a more detailed review of this book by one of our Engineering Mentors,
berkeman shared in this post.

Really appreciate it! :)