Calculate the equivalent resistance

[Andrei0408]

Homework Statement

I need to calculate the equivalent resistance in 4 different cases

Relevant Equations

1/Rep = 1/R1 + 1/R2 + ... + 1/Ri
Res = R1 + R2 + ... + Ri

R1 = 2 ohms ; R2 = 2 ohms; R3 = 1ohm
First figure:
I need to calculate the resistance between: a) A and C and b) A and B
a) R1 and R3 are in series => R13 = R1 + R3 = 3 ohms
R13 and R2 are in parallel => 1/Rp = 1/R13 + 1/R2 => Rp = 1.2 ohms
b) Here, between A and B there is only R1 so the equivalent resistance will be 1 ohm? (is this correct?)
Second figure:
I need to calculate the resistance between: a) A and C and b) B and C
a) R1 and R3 are in series => Rs = R1 + R3 = 3 ohms
But here, if we take another path: R1 and R2 parallel the result will be 1 ohm.
Which one is correct and why are they different?
b) Between B and C there is only R3 = 1 ohm.
But if we take R2 parallel with R3 => 0.66 ohms
Which one is correct?
Third figure:
I need the resistance between A and D:
R1 and R3 are in series => R1+R3 = 3 ohms
Fourth figure:
I need the resistance between A and D:
Here couldn't I take the same path as earlier and get 3 ohms or the path with R2 only for 2 ohms?
Could you please explain where I'm wrong and which one is the correct answer at each point? Thank you, in advance!

 

[BvU]

b) Here, between A and B there is only R1 so the equivalent resistance will be 1 ohm? (is this correct?)

One at a time. R1 = 2 $\Omega$, so where does your 1 $\Omega$ come from ?
And what about R2 and R3 ?

 

[haruspex]

Homework Statement:: I need to calculate the equivalent resistance in 4 different cases
Relevant Equations:: 1/Rep = 1/R1 + 1/R2 + ... + 1/Ri
Res = R1 + R2 + ... + Ri

b) Here, between A and B there is only R1 so the equivalent resistance will be 1 ohm? (is this correct?)

If there is a voltage difference between A and B, why would the current shun the R2-R3 path?

Second figure:
I need to calculate the resistance between: a) A and C and b) B and C
a) R1 and R3 are in series

No, you can only count resistors as being in series if there is no junction between them that could bring current in or out.

b) Between B and C there is only R3 = 1 ohm.
But if we take R2 parallel with R3 => 0.66 ohms

take the same path as earlier and get 3 ohms or the path with R2 only for 2 ohms?

Again, you must consider all paths current could take as carrying some. It is not a question of one or the other.

 

[Andrei0408]

One at a time. R1 = 2 $\Omega$, so where does your 1 $\Omega$ come from ?
And what about R2 and R3 ?

One at a time. R1 = 2 $\Omega$, so where does your 1 $\Omega$ come from ?
And what about R2 and R3 ?

Okay so, R1 and R2 are in parallel => R12 = 1 ohm. Will R12 be in parallel or in series with R3?
Also, is the first one correct because there is a junction between R1 and R3 so they can't be in series, right?

 

[BvU]

Do we have the figures mixed up ?

First figure:
I need to calculate the resistance between: a) A and C and b) A and B
a) R1 and R3 are in series => R13 = R1 + R3 = 3 ohms
R13 and R2 are in parallel => 1/Rp = 1/R13 + 1/R2 => Rp = 1.2 ohms
b) Here, between A and B there is only R1 so the equivalent resistance will be 1 ohm? (is this correct?)

​

a) correct !
b) No. FIrst: R1 is 2 $\Omega$. Second: there is another path

Okay so, R1 and R2 are in parallel => R12 = 1 ohm. Will R12 be in parallel or in series with R3?

IF this is about b) in the first picture, then: No !

Also, is the first one correct because there is a junction between R1 and R3 so they can't be in series, right?

the first one is correct. I don't understand why a junction (connection) means R1 and R3 can not be in series.

We need to first establish how you are looking at this.
Look at the paths the current can take:
Series means if it goes through the one, it has to go through the other.
Parallel means the current splits up and a part goes through the one, the remainder goes through the other.

First figure

a) between A and C: two paths $\Rightarrow$ parallel​

first path through (R1 and R3) $\Rightarrow$ series​

second path through R2​

​

b) between A and B: also two paths $\Rightarrow$ parallel​

first path through R1​

second path through R2 and R3 $\Rightarrow$ series​

​

$\$​

 

[Andrei0408]

If there is a voltage difference between A and B, why would the current shun the R2-R3 path?

No, you can only count resistors as being in series if there is no junction between them that could bring current in or out.Again, you must consider all paths current could take as carrying some. It is not a question of one or the other.
[/QUO

Do we have the figures mixed up ?

View attachment 278191​

a) correct !
b) No. FIrst: R1 is 2 $\Omega$. Second: there is another pathIF this is about b) in the first picture, then: No !

the first one is correct. I don't understand why a junction (connection) means R1 and R3 can not be in series.

We need to first establish how you are looking at this.
Look at the paths the current can take:
Series means if it goes through the one, it has to go through the other.
Parallel means the current splits up and a part goes through the one, the remainder goes through the other.

First figure

a) between A and C: two paths $\Rightarrow$ parallel​

first path through (R1 and R3) $\Rightarrow$ series​

second path through R2​

​

b) between A and B: also two paths $\Rightarrow$ parallel​

first path through R1​

second path through R2 and R3 $\Rightarrow$ series​

​

$\$​

Then isn't RAB going to be 1.2 ohms, just like the first one?

 

[BvU]

Then isn't RAB going to be 1.2 ohms, just like the first one?

Bingo !

 

[Andrei0408]

Bingo !

Thanks a lot, I think I understand now, I have 2 more questions if you don't mind: what is the maximum and the minimum value of equivalent resistance that I could obtain using all 3 resistors. (how should I make the drawings?)

 

[BvU]

You should be able to answer that one yourself now !
Make a few guesses

 

[Andrei0408]

You should be able to answer that one yourself now !
Make a few guesses

For the maximum we should have all the resistors connected in series, and for minimum they should be connected in parallel?

 

[BvU]

Bingo again. Calculate the equivalent resistance to convince yourself !

 

[haruspex]

I don't understand why a junction (connection) means R1 and R3 can not be in series.

Not sure which diagram that refers to, but @Andrei0408's remark follows from one of mine in post #3 in relation to screenshot 467 (figure 2?)
Andrei had started by adding R1 and R3 because they are in series, but the junction with R2 in between invalidates that. It is necessary to collapse the parallel R2 and R3 first.