Series-Parallel Circuits (R/2R Ladder Networks)

[Lay1]

Homework Statement

Determine Vout for the R/2R ladder network in the given figure for the following condition:
SW3 and SW4 to +12V, SW1 and SW2 to ground

Relevant Equations

V=I/R
I(x)=(Rt/Rx)*I(t)

This is the figure given by the question.

Here is the redraw of above circuit.

What I request to ask are:
1. Is my redraw is correct or not?
2. The Vout asked in the question is referred to the voltage after passing throgh R8?
Thank you for your answers.

 

[BvU]

Homework Statement: Determine Vout for the R/2R ladder network in the given figure for the following condition:

1. Is my redraw is correct or not?
2. The Vout asked in the question is referred to the voltage after passing throgh R8?
Thank you for your answers.

1. I think it is. But I don't find it very clarifying
2. Voltage does not really 'pass'

Your working isn't very clear to me either; perhaps you can type it ?
What is the $28.4\ k\Omega$ you calculate ?
And I don't see a $V_\text{out}$ in your calculations ?

I can't distinguish the other questions in the exercise (cropped off ) but for the one you picked, my approach would be to redraw with ground at the bottom and +12 V at the top. e.g.:

Then substitute the parallel resistances 1 through 5 with R_eq before looking at currents.

$\$

 

[gneill]

Hi Lay1.

I believe that your redrawing is correct, although I can't quite make out the resistor symbols. I think that R8 is the one shown here?

Not sure what you mean by "passing through R8" though.

There are several ways to redraw the circuit, here's an example that isn't far from the original. You should be able to reduce the indicated resistances to one value before proceeding:

Thus:

 

[ace1719]

Hi Lay1,

Your redraw is correct.

To answer your second question, Vout is the node that is common to R7 and R8.

Some of the other commenters have noted that voltage doesn't really "pass through". I'll try and clarify things for you a bit. Voltage is always measured between two points.

For example, the voltage across R8 is: V_r8 = (12 V - Vout). If you apply Ohm's law: 12 V - Vout = I_r8*R_r8.

While voltage is always measured between two points, you will often be asked (such as in this case) about the voltage at a particular point. The convention is that is is referenced to "ground", however the same technique for finding it can apply.

Vout - Vgnd = I_req*R_req, where I_req and R_req is the current passing through and the resistance of an equivalent resistance between whichever node you are interested in and ground. I will note that by convention, Vgnd = 0.

To me, this question feels as though it is trying to test your understanding of the definition of what it means for a resistance to be in parallel or series. Gneill alluded to this, and BvU drew out the circuit in a what that exposes this a bit more clearly. I would definitely try and find that equivalent resistance and then work from there.

 

[Lay1]

1. I think it is. But I don't find it very clarifying
2. Voltage does not really 'pass'

Your working isn't very clear to me either; perhaps you can type it ?
What is the $28.4\ k\Omega$ you calculate ?
And I don't see a $V_\text{out}$ in your calculations ?

I can't distinguish the other questions in the exercise (cropped off ) but for the one you picked, my approach would be to redraw with ground at the bottom and +12 V at the top. e.g.:
View attachment 325066

Then substitute the parallel resistances 1 through 5 with R_eq before looking at currents.

$\$

Thank you for your help. Have a nice day.

 

[Lay1]

Hi Lay1.

I believe that your redrawing is correct, although I can't quite make out the resistor symbols. I think that R8 is the one shown here?
View attachment 325067

Not sure what you mean by "passing through R8" though.

There are several ways to redraw the circuit, here's an example that isn't far from the original. You should be able to reduce the indicated resistances to one value before proceeding:

View attachment 325069

Thus:
View attachment 325070

I will note that down. Thank you for your explanation.

 

[Lay1]

Hi Lay1,

Your redraw is correct.

To answer your second question, Vout is the node that is common to R7 and R8.

Some of the other commenters have noted that voltage doesn't really "pass through". I'll try and clarify things for you a bit. Voltage is always measured between two points.

For example, the voltage across R8 is: V_r8 = (12 V - Vout). If you apply Ohm's law: 12 V - Vout = I_r8*R_r8.

While voltage is always measured between two points, you will often be asked (such as in this case) about the voltage at a particular point. The convention is that is is referenced to "ground", however the same technique for finding it can apply.

Vout - Vgnd = I_req*R_req, where I_req and R_req is the current passing through and the resistance of an equivalent resistance between whichever node you are interested in and ground. I will note that by convention, Vgnd = 0.

To me, this question feels as though it is trying to test your understanding of the definition of what it means for a resistance to be in parallel or series. Gneill alluded to this, and BvU drew out the circuit in a what that exposes this a bit more clearly. I would definitely try and find that equivalent resistance and then work from there.

Thank you for your thorough explanation. I will remember your guidance.

 

[berkeman]

BTW, the R-2R Ladder DAC is a handy tool to understand. Now that you've basically worked through this problem the hard way, here is a tutorial to help you understand their voltage output characteristics:

https://www.tek.com/en/blog/tutorial-digital-analog-conversion-r-2r-dac