# Choosing of Resistance Value

#### Fuzz92

Joined Aug 5, 2015
8
Hi Guys,

First post on here so apologies if I have posted this in the wrong section or what-not but I need help with a general question I do not understand, not homework.

The question is to choose the value of the R resistor to set the node to 5V.

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#### ericgibbs

Joined Jan 29, 2010
9,343
hi fuzz,
If its homework, you are in the correct forum.

What do you think the resistance value is.? show us your attempt.

E

#### Fuzz92

Joined Aug 5, 2015
8
Hi Eric,

As I said, it's not homework but not sure where else I could have posted it!

Ok, so I tried to work out the KCL of the central node, wasn't sure whether to use the voltages of the top and bottom (20V, -10V) or use the difference between the top and bottom, and the central node (20V-5V, -10V,-5V). I guess my confusion is why the node has a voltage. If its the difference, I found the current passing through each resistance to be 3A either way out of the node (up and down), which would mean 0 current going left (is that correct?).

If not then the voltage drop across would equal a 4A current going up, 2A current going down and therefore a 2A current going left, meaning the Resistance should be 2V/2A = 1ohm?

I think it's just the 5V node tripping me up here, any help would be appreciated

#### Jony130

Joined Feb 17, 2009
5,127
I found the current passing through each resistance to be 3A either way out of the node (up and down), which would mean 0 current going left (is that correct?).
Yes you are right about "current going left". If so the value of a R resistor is equal to ?

#### Fuzz92

Joined Aug 5, 2015
8
I am right in that it is 0A? If so, then the Resistance would be infinite, no?

#### ericgibbs

Joined Jan 29, 2010
9,343
What is the voltage at the junction of the two, 5R resistors if you remove the unknown resistor.??

#### Fuzz92

Joined Aug 5, 2015
8
Voltage between -10V and 20V is 30, with equal resistance it would be the same voltage drop across both, 15V (correct?).

That would make the junction of the two, 5R resistors: 5V? Does that mean that the resistance of the wire on the left would have to act like an open circuit to keep the same voltage across the two, infinite resistance? What would the power be across the R resistor then, 0W?

Appreciate the help guys!

#### WBahn

Joined Mar 31, 2012
25,090
Hi Eric,

As I said, it's not homework but not sure where else I could have posted it!
Out of curiosity, if it's not homework, what's it for?

Ok, so I tried to work out the KCL of the central node, wasn't sure whether to use the voltages of the top and bottom (20V, -10V) or use the difference between the top and bottom, and the central node (20V-5V, -10V,-5V). I guess my confusion is why the node has a voltage.
The voltage AT or ON a node is merely the voltage difference between that node and some arbitrary reference node.

For applying Ohm's Law, you need the voltage ACROSS a resistor, which means the voltage difference between the two nodes it is connected to.

If its the difference, I found the current passing through each resistance to be 3A either way out of the node (up and down), which would mean 0 current going left (is that correct?).

If not then the voltage drop across would equal a 4A current going up, 2A current going down and therefore a 2A current going left, meaning the Resistance should be 2V/2A = 1ohm?

I think it's just the 5V node tripping me up here, any help would be appreciated
The 5V is merely telling you what the voltage is on that node, relative to whatever reference the voltages on the other nodes are referred to, when the correct valued resistor is used.

Problems always work out much better if you start with a clearly labeled diagram:

ASIDE: Note that the above image file is only 44 kB, less than 4% of the 1.2 MB monster you posted, despite the added annotations. Please make an effort to keep image sizes reasonable.

Now you can apply the rules and concepts in a clear and unambiguous manner.

Apply KCL at the central node:

$$I_1 \; + \; I_2 \; = \; I_3$$

Apply Ohm's Law for each resistor:

$$I_1 \; = \; \frac{(20V) \; - \; (5V)}{5 \Omega} I_2 \; = \; \frac{(-10V) \; - \; (5V)}{5 \Omega} I_3 \; = \; \frac{(5V) \; - \; (-2V)}{R}$$

By inspection, you can see that, if the unknown resistance were removed entirely, that the voltage at the central node would simply be half way between the 20 V an the -10 V, which would be 5V. So if there resistor IS there, it must not have any appreciable current flowing in it. While it would need to be infinite to have no current flowing in it, if it is 100x to 1000x or more compared to the other resistors, the effect of the small current that does flow will be negligible.

#### WBahn

Joined Mar 31, 2012
25,090
Voltage between -10V and 20V is 30, with equal resistance it would be the same voltage drop across both, 15V (correct?).

That would make the junction of the two, 5R resistors: 5V? Does that mean that the resistance of the wire on the left would have to act like an open circuit to keep the same voltage across the two, infinite resistance? What would the power be across the R resistor then, 0W?

Appreciate the help guys!
Yep.

#### WBahn

Joined Mar 31, 2012
25,090
Fuzz92 said:
If not then the voltage drop across would equal a 4A current going up, 2A current going down and therefore a 2A current going left, meaning the Resistance should be 2V/2A = 1ohm?
That sounds right to me.
Huh?

Have you LOOKED at the problem? Even if the current were correct, the voltage across the unknown resistor is 7V, not 2V.

#### tjohnson

Joined Dec 23, 2014
611
Huh?

Have you LOOKED at the problem? Even if the current were correct, the voltage across the unknown resistor is 7V, not 2V.
I did look at it, but rather quickly since I was on my phone at the time so I completely missed the 5V node. Sorry about that.

@WBahn: I'm curious how you were able to quote my post and leave the quote within it intact? Nested quotes don't seem to work for me.

#### WBahn

Joined Mar 31, 2012
25,090
I did look at it, but rather quickly since I was on my phone at the time so I completely missed the 5V node. Sorry about that.

@WBahn: I'm curious how you were able to quote my post and leave the quote within it intact? Nested quotes don't seem to work for me.
They don't work for me, either. I did a copy/paste of the embedded quote and entered the tags manually.

#### Fuzz92

Joined Aug 5, 2015
8
Thanks for the help guys, much appreciated!

WBahn you explained it perfectly, so thanks for that.

#### WBahn

Joined Mar 31, 2012
25,090
Thanks for the help guys, much appreciated!

WBahn you explained it perfectly, so thanks for that.
You're more than welcome, glad it was useful.

#### MrAl

Joined Jun 17, 2014
6,959
Hi,

Yes when you consider the difference voltage across the two 5 ohm resistors you get 3 amps, and 3 amps times 5 ohms is 15 volts, and 15 volts down from the top voltage is 5 volts, and 15 volts up from the bottom voltage is 5 volts, so the center node is at 5 volts already without suppling more current or taking more current away from that node.

With a -2 volt source on the left the only way this can happen is if the third resistor on the left is infinite in value, or just not connected into the circuit. If the source on the left was 5 volts, then we could make that resistor anything we wanted it to be because there would be no current flow, and that would mean there would be an infinite number of resistance values that work. Since it is not 5v on the left however, only an open circuit works for that resistor.

#### WBahn

Joined Mar 31, 2012
25,090
A good follow up exercise would be to solve the more general problem. Consider a circuit with the same topology but with unspecified values. You have Vcc (instead of the +20V supply), Vee (instead of the -20V supply), and Vbb (instead of the -2V supply). The resistors are Rc, Re, and Rb (which is which is hopefully obvious from the subscripts). Vcc, Vee, Rc, and Re are fixed. Your task is to find values of Rb and Vbb that will let you set the value of Vx (the voltage on the center node) to within some tolerance, Vdx, of a desired value.

This is more than just a made-up exercise -- it is pretty much exactly how we adjust the bias currents and voltages inside the IC chips we designed when testing them; the only significant difference is that either Rc or Re would actually be an on-chip transistor that was part of a current mirror circuit. In designing the test electronics, we needed to decide how much variability we needed and what kind of resolution we wanted. We could typically vary Vbb over a range of +/-12V, so the main decision was determining the value of Rb that would give us the desired effect.

#### Fuzz92

Joined Aug 5, 2015
8
That sounds interesting WBahn, do you have some questions you could give for me to solve, to check my understanding?

#### WBahn

Joined Mar 31, 2012
25,090
That sounds interesting WBahn, do you have some questions you could give for me to solve, to check my understanding?
I actually DID give it to you.

First, let's ignore the tolerance question and assume everything is ideal. Consider a circuit with the same topology but with unspecified values. You have Vcc (instead of the +20V supply), Vee (instead of the -20V supply), and Vbb (instead of the -2V supply). The resistors are Rc, Re, and Rb (which is which is hopefully obvious from the subscripts). Vcc, Vee, Rc, and Re are fixed. Let's further assume, at first, that Vbb is fixed. Your task is to find the value of Rb that will set the value of Vx to the desired value.

So what you want to do is analyze the circuit symbolically and come up with an equation for Rb in terms of all of the other parameters.

#### Fuzz92

Joined Aug 5, 2015
8
Still not sure if I get this yet.. let me know , so:

I1 = (Vcc - Vx)/Rc
I2 = (Vx - Vee)/Re

Using KCL on given numbers can give you how much current is flowing in the 'left' part of the circuit, I3.

From that

(Vx - Vbb)/I3 = Rb

#### WBahn

Joined Mar 31, 2012
25,090
Still not sure if I get this yet.. let me know , so:

I1 = (Vcc - Vx)/Rc
I2 = (Vx - Vee)/Re

Using KCL on given numbers can give you how much current is flowing in the 'left' part of the circuit, I3.

From that

(Vx - Vbb)/I3 = Rb
Close, but you've got to pay very close attention to sign and polarity issues -- these errors are very easy to make and very hard to track down. That is why using clearly annotated drawings are so important.

So the first question is: What are the definitions of I1, I2, and I3 that you are using? If they are the ones I provided in Post #8, then is your equation for I2 consistent with that definition?

Other than that, you basically have it and just need to bring everything together and give Rb as a function of Vx (and the other parameters).