What's the difference in resistance between A and B vs A and C?

[MermaidWonders]

View attachment 7919

I tried redrawing the above circuit so that it resembles more like the circuit diagrams we see from day to day. However, what confused me was when it asks for the resistance measured between A and B versus between A and C What's the difference between those two, and how would the calculations differ in each case?

 

[MermaidWonders]

I tried redrawing the above circuit so that it resembles more like the circuit diagrams we see from day to day. However, what confused was when it asks for the resistance measured between A and B versus between A and C? What's the difference between those two, and how would the calculations differ in each case?

As for part (a), I tried to find the ${R}_{eq}$ of the entire circuit since I thought you must go through the whole circuit before arriving at B from A, but I don't think that that would get me the answer. :( Now, I'm even more confused... :(

 

[MermaidWonders]

2nd question:

View attachment 7920

I know this question's kinda lame, but it's just that when I tried to solve it, I couldn't get the right answer. (I got 3.375 $\times$ $10^{-9}$ A.) :(
(Answer = 1.5 mA)

 

[I like Serena]

https://www.physicsforums.com/attachments/7919

I tried redrawing the above circuit so that it resembles more like the circuit diagrams we see from day to day. However, what confused was when it asks for the resistance measured between A and B versus between A and C? What's the difference between those two, and how would the calculations differ in each case?

Hi MermaidWonders! ;)

To find the resistance between A and B, we can rearrange the circuit as:
\begin{tikzpicture}
%preamble \usepackage{circuitikz}
\draw[color=black, thick]
(0,0) node

{A} to [R, *-] (2,0) to (2,1) to [R, -*] (4,1) to [R] (6,1) to (6,0) to [R, -*] (8,0) node

{B}
(2,0) to (2,-1) to [R] (6,-1) to (6,0);
\end{tikzpicture}

Can you merge the resistors in series together?
And calculate the resitance of parallel resistors?

For the resistance between A and C we need:
\begin{tikzpicture}
%preamble \usepackage{circuitikz}
\draw[color=black, thick]
(0,0) node

{A} to [R, *-] (2,0) to (2,-1) to [R] (4,-1) to [R] (6,-1) to (6,0) to [short, -*] (7,0) node

{C}
(2,0) to (2,1) to [R] (6,1) to (6,0)
(4,-1) to [R, -*] (4, -3) node[below] {B} ;
\end{tikzpicture}
The resistor to B is a 'loose end' here, which does not participate.
Can you find the resistance between A and C?​

 

[MermaidWonders]

Thanks for your reply! As for this question, I got the same re-drawn circuit diagram as the one you have here for part (a), but I'm totally lost as to how you got part (b) in order to get the resistance for part (b). Can you please explain?

 

[I like Serena]

Thanks for your reply! As for this question, I got the same re-drawn circuit diagram as the one you have here for part (a), but I'm totally lost as to how you got part (b) in order to get the resistance for part (b). Can you please explain?

Just by deforming the circuit.
An intermediate step is:
\begin{tikzpicture}
%preamble \usepackage{circuitikz}
\draw[color=black, thick]
(0,0) node

{A} to [R, *-] (2,0) to [R] (4,-2) to [R, -*] (6,0) node

{C}
(2,0) to [R] (6,0)
(4,-2) to [R, -*] (2, -3) node[below] {B} ;
\end{tikzpicture}
And I've separated C a bit with a separate connection, which is equivalent, since it behaves like a 'short'.​

 

[MermaidWonders]

Hi MermaidWonders! ;)

To find the resistance between A and B, we can rearrange the circuit as:
\begin{tikzpicture}
%preamble \usepackage{circuitikz}
\draw[color=black, thick]
(0,0) node

{A} to [R, *-] (2,0) to (2,1) to [R, -*] (4,1) to [R] (6,1) to (6,0) to [R, -*] (8,0) node

{B}
(2,0) to (2,-1) to [R] (6,-1) to (6,0);
\end{tikzpicture}

Can you merge the resistors in series together?
And calculate the resitance of parallel resistors?

For the resistance between A and C we need:
\begin{tikzpicture}
%preamble \usepackage{circuitikz}
\draw[color=black, thick]
(0,0) node

{A} to [R, *-] (2,0) to (2,-1) to [R] (4,-1) to [R] (6,-1) to (6,0) to [short, -*] (7,0) node

{C}
(2,0) to (2,1) to [R] (6,1) to (6,0)
(4,-1) to [R, -*] (4, -3) node[below] {B} ;
\end{tikzpicture}
The resistor to B is a 'loose end' here, which does not participate.
Can you find the resistance between A and C?​

Is the resistance between A and B equal to $\frac{7}{2R}$?​

 

[I like Serena]

Is the resistance between A and B equal to $\frac{7}{2R}$?

Not according to my calculations.
How did you get that?

 

[MermaidWonders]

Not according to my calculations.
How did you get that?

Oops, I just realized that I messed up... Let's try again.

Top 2 resistors of parallel portion of the circuit are in series with each other $\implies$ $R = R + R = 2R$

Bottom resistor of parallel portion of the circuit in parallel with top 2 resistors $\implies$ $\frac{1}{{R}_{eq}}$ = $\frac{1}{2R}$ + $\frac{1}{R}$ = $\frac{3}{2R}$ $\implies$ $\therefore$, ${R}_{eq}$ = $\frac{2R}{3}$

Now, this ${R}_{eq}$ will be in series with the remaining resistors $\implies$ ${R}_{total}$ = $\frac{8R}{3}$?

- - - Updated - - -

Just by deforming the circuit.
An intermediate step is:
\begin{tikzpicture}
%preamble \usepackage{circuitikz}
\draw[color=black, thick]
(0,0) node

{A} to [R, *-] (2,0) to [R] (4,-2) to [R, -*] (6,0) node

{C}
(2,0) to [R] (6,0)
(4,-2) to [R, -*] (2, -3) node[below] {B} ;
\end{tikzpicture}
And I've separated C a bit with a separate connection, which is equivalent, since it behaves like a 'short'.​

Oh, I see. Now, with this new diagram, how would one go about solving for the resistance between A and C?

P.S.: What's a 'short'? Do you mean the extra connection to C? If so, how come the 'short' won't affect C?​

 

[I like Serena]

Oops, I just realized that I messed up... Let's try again.

Top 2 resistors of parallel portion of the circuit are in series with each other $\implies$ $R = R + R = 2R$

Bottom resistor of parallel portion of the circuit in parallel with top 2 resistors $\implies$ $\frac{1}{{R}_{eq}}$ = $\frac{1}{2R}$ + $\frac{1}{R}$ = $\frac{3}{2R}$ $\implies$ $\therefore$, ${R}_{eq}$ = $\frac{2R}{3}$

Now, this ${R}_{eq}$ will be in series with the remaining resistors $\implies$ ${R}_{total}$ = $\frac{8R}{3}$?

Yep. That's correct. (Nod)

Oh, I see. Now, with this new diagram, how would one go about solving for the resistance between A and C?

P.S.: What's a 'short'? Do you mean the extra connection to C? If so, how come the 'short' won't affect C?

Just leave out the resistor to B and do the same thing:
\begin{tikzpicture}
%preamble \usepackage{circuitikz}
\draw[color=black, thick]
(0,0) node

{A} to [R, *-] (2,0) to (2,-1) to [R, -o] (4,-1) to [R] (6,-1) to (6,0) to [short, -*] (7,0) node

{C}
(2,0) to (2,1) to [R] (6,1) to (6,0);
\end{tikzpicture}
A 'short' means 'short circuit', which is a connection between 2 points with no (or negligible) resistance.
If that's an unintended connection, we'll get a bad result.
But in this case it's only to clarify the circuit, without actually making a connection to any other point in the diagram.
As such it has no impact whatsoever on the result.​

 

[MermaidWonders]

What happened to B now? Why is it gone?

 

[I like Serena]

What happened to B now? Why is it gone?

Since B is an open ended connection, no current can flow through B.
Therefore the resistor to B can be left out without changing the resistance between A and C.
And now we can calculate that resistance...

 

[MermaidWonders]

Since B is an open ended connection, no current can flow through B.
Therefore the resistor to B can be left out without changing the resistance between A and C.
And now we can calculate that resistance...

OK... So, to sum up, you would need 2 different circuit diagrams to refer to depending on whether you need the resistance between A and B or A and C, right, and for each case, you must alter the diagram such that you'll end up being able to calculate the equivalent resistance as you would for any circuit?

 

[I like Serena]

OK... So, to sum up, you would need 2 different circuit diagrams to refer to depending on whether you need the resistance between A and B or A and C, right, and for each case, you must alter the diagram such that you'll end up being able to calculate the equivalent resistance as you would for any circuit?

Yep. (Nod)

 

[MermaidWonders]

OK, thanks! :)

 

[I like Serena]

Oh, and about the second part... (Wait)

We know now that the resistance between A and B is $R_{AB}=\frac 83R$ with $R=1\text{ k}\Omega$ and $V=6\text{ V}$.
So the total current is:
$$I_{total}=\frac V{R_{AB}}=\frac {6\text{ V}}{\frac 83\cdot 1\text{ k}\Omega} = 2.25 \text{ mA}$$
Since the current is split between 1R along the vertical branch and 2R along the branch through C, it follows that $\frac 23$ of the current must flow through the vertical branch, which is:
$$I_{vertical} = \frac 23 \times 2.25 = 1.5\text{ mA}$$

 

[MermaidWonders]

I now understand where I went wrong in question #37. I didn't use $\frac{8R}{3}$ as the equivalent resistance since I messed up in the beginning... I finally got the right answer, although I do agree that your reasoning about the currents is much faster :) Many thanks again!