Question about resistors (parallel and series)

[x86]

Here is a picture of a circuit:

I have two questions:

(1) Are all resistors connected in one of two ways? I.e., resistors can be connected in parallel and in series, but is it possible for them to be connected as NOT parllel and NOT series? For instance, the two circled resistors don't seem to be connected in either parallel or series. That is, they're neither in series nor in parallel?

(2) Suppose we short the 12 V voltage source. Then the two resistors circled are in parallel.

If I want to simplify this circuit by combinding these two resistors, I find the equivalent resistance Req = 1/(1/4 + 1/4) = (1/(1/2)) = 2.

My question is, where do I put it? I have two resistors, what do I do with the second? Do make an open circuit from one of the resistors, and replace the other with Req? If I do this, there are two possibilities for the circuit. Are both of these two possible circuits equivalent?

 

[berkeman]

Parallel 2-port devices share 2 nodes. Series 2-port devices share one node...

 

[x86]

Parallel 2-port devices share 2 nodes. Series 2-port devices share one node...

So the resistors in the black circle at in series then? They have only one node in common, the bottom one.

 

[berkeman]

So the resistors in the black circle at in series then? They have only one node in common, the bottom one.

Correct. If you were not writing KCL equations to solve the circuit and instead wanted to use parallel and series combinations, you would combine the outside parallel combinations and then put them in series. Makes sense?

 

[x86]

Correct. If you were not writing KCL equations to solve the circuit and instead wanted to use parallel and series combinations, you would combine the outside parallel combinations and then put them in series. Makes sense?

Sort of. When I put them combine resistors in series, I am supposed to short one of the two resistors, and replace the other with the equivalent resistance correct? And in parallel, I open one of the two resistors. and replace the non-open resistor with the equivalent resistance?

 

[x86]

Sorry, I'm not understanding what you mean. When you have two impedances in zeries, you add them Ztot = Z1 + Z2 When you have two impedances in parallel, you add them with this:

1/Ztot = 1/[1/Z1 + 1/Z2]

I apologize for the poor explanation.

If we have two resistors R1 and R2 connected in series, and we replace them with Req then we are removing two resistors from our circuit and adding one resistor to our circuit. (Net loss of one resistor). Therefore, we collapse two elements into one element.

So since we had 2 elements in our first circuit, and 1 element in our second circuit, it raises the question: what do we replace the missing element with?

My question is as follows:
If we have two resistors in series, do we replace one of the resistors we removed with a short circuit?
If we have two resistors in parallel, do we replace one of the resistors we moved with an open circuit?

 

[berkeman]

I apologize for the poor explanation.

If we have two resistors R1 and R2 connected in series, and we replace them with Req then we are removing two resistors from our circuit and adding one resistor to our circuit. (Net loss of one resistor). Therefore, we collapse two elements into one element.

So since we had 2 elements in our first circuit, and 1 element in our second circuit, it raises the question: what do we replace the missing element with?

My question is as follows:
If we have two resistors in series, do we replace one of the resistors we removed with a short circuit?
If we have two resistors in parallel, do we replace one of the resistors we moved with an open circuit?

First, sorry about the quoted text -- I may not have used the correct equation.

Next, when simplifying circuits, you are replacing series or parallel components with their equivalent impedance. It's a 1-1 change. There is no need to do anything after you figure out the equivalent component to put between the nodes.

 

[x86]

First, sorry about the quoted text -- I may not have used the correct equation.

Next, when simplifying circuits, you are replacing series or parallel components with their equivalent impedance. It's a 1-1 change. There is no need to do anything after you figure out the equivalent component to put between the nodes.

I am a little confused. If you were to explain it to me like you would explain it to a computer, would you say that my above statements were correct?

We are going from 2 resistors to 1. Suppose we have two positions in our circuit which are that of our resistor elements (A and B). We calculate an equivalent resistance, Req. Clearly, in our simplified circuit we have a net loss of one resistor. This means we are removing the resistors from A and B, and then putting an equivallent resistor in A exclusive or B.

Let's assume we put Req in A. What happens to B? Of course, we have to remove the old resistor that was in B (because now Req is in A).

My question is as follows: If A and B are parallel, do we open B? If A and B are series, do we short B?

 

[Drakkith]

Let's assume we put Req in A. What happens to B? Of course, we have to remove the old resistor that was in B (because now Req is in A).

My question is as follows: If A and B are parallel, do we open B? If A and B are series, do we short B?

A and B no longer exist in the circuit. Instead you have a single resistor, C, with a resistance equal to R_eq.

Note that you usually don't physically remove resistors and replace them in this manner. If a circuit has two resistors instead of one, it's usually done for a reason. (Even it's just because of cost) Replacing resistors A and B with resistor C so that C has a resistance equal to R_eq is usually just a way to simplify a circuit diagram and any calculations. If a circuit component is composed of a dozen resistors in a complicated pattern, but you don't care about the specifics of what's going on in that component, then it's usually okay to just use the R_eq for that component instead of including all of those resistors in your analysis of the circuit. Especially if you're given R_eq for the component.

 

[milesyoung]

Are all resistors connected in one of two ways?

No, the series/parallel configuration is just very common.

For instance, the two circled resistors don't seem to be connected in either parallel or series. That is, they're neither in series nor in parallel?

Yes, they're not in series or parallel.

A series connection of two elements share a common node and are connected such that the same current flows through each one. In other words: Nothing else must be connected to their common node, since the current then branches off.

A parallel connection of two elements share two common nodes. A consequence of this is that they have the same voltage across them.

The two circled resistors share only one common node, but this node is also shared by the switch and capacitor, so the resistors are neither in series or parallel.

My question is, where do I put it?

You replace the parallel combination with a single equivalent resistance between their common nodes.

Do make an open circuit from one of the resistors, and replace the other with Req?

Schematically, it doesn't matter which resistor you replace or "open up", the result will be the same. You will have replaced a parallel combination with a single equivalent resistance between their common nodes.

If I do this, there are two possibilities for the circuit.

No, since we're assuming the 12 V source is shorted, there will just be one resistor connected between the same two nodes.