What is the effective resistance between X and Y in this circuit?

[Kajan thana]

Homework Statement

Find the effective resistance between X and Y. I have attached the circuit to this post.

Homework Equations

V=IR

The Attempt at a Solution

I did 0.25+0.5+0.5=1.25 then flipped the fraction to get 0.8ohm. I considered as if two resistors are connected in series and rest are connected in parallel .

 

[gneill]

It's hard to tell from what you've posted which resistors you've considered to be in series and which in parallel. Here's your diagram with the resistors labeled:

Take us through your calculation step by step, identifying which resistors are involved.

 

[Kajan thana]

It's hard to tell from what you've posted which resistors you've considered to be in series and which in parallel. Here's your diagram with the resistors labeled:

View attachment 148776

Take us through your calculation step by step, identifying which resistors are involved.

Thank you for labelling it for me.

Basically I combined R1 and R2 to give me 4 ohms. Then treated (R1&R2), R3 and R4 as if they are connected in parallel to get the effective resistance 0.8 ohms.

 

[gneill]

Thank you for labelling it for me.

Basically I combined R1 and R2 to give me 4 ohms. Then treated (R1&R2), R3 and R4 as if they are connected in parallel to get the effective resistance 0.8 ohms.

Okay, well since X is the junction between R1 and R2 they can't be in series for any path that starts at X and ends elsewhere (at Y for example). Automatically there are two paths leaving point X, one through R1 and one through R2. The best you could hope for is that they are in parallel, but a quick look at the circuit tells you that they are not (they only share one node in common: X).

Take a look for other series or parallel opportunities to start the simplification. What can you find?

 

[Kajan thana]

Okay, well since X is the junction between R1 and R2 they can't be in series for any path that starts at X and ends elsewhere (at Y for example). Automatically there are two paths leaving point X, one through R1 and one through R2. The best you could hope for is that they are in parallel, but a quick look at the circuit tells you that they are not (they only share one node in common: X).

Take a look for other series or parallel opportunities to start the simplification. What can you find?

I got the answer as 1.2 ohms which matches the book's answer as well, But I don't understand it properly. Can you explain to me please? I basically took R3 and R4 as parallel and got the overall resistance 1 ohm, then combined that with R2 to give me 3ohms. Then I took the combined 3ohm and R2 to be parallel to give me the answer 1.2. The part that does not make sense to me is why are we taking R3 and R4 parallel, because they are coming from the same node isn't?

 

[Kajan thana]

since X is the junction between R1 and R2 they can't be in series for any path that starts at X

I don't understand when you said X is the junction..., can you please expand on it?
Thank you so much.

 

[gneill]

I got the answer as 1.2 ohms which matches the book's answer as well, But I don't understand it properly. Can you explain to me please? I basically took R3 and R4 as parallel and got the overall resistance 1 ohm, then combined that with R2 to give me 3ohms. Then I took the combined 3ohm and R2 to be parallel to give me the answer 1.2. The part that does not make sense to me is why are we taking R3 and R4 parallel, because they are coming from the same node isn't?

Your analysis is correct.

R3 and R4 are in parallel because their connections share two nodes. Anther way to tell that they are in parallel is that you can follow a closed path that passes through only through those two components:

I don't understand when you said X is the junction..., can you please expand on it?
Thank you so much.

The node X is where R1 and R2 are connected. Thus it is the junction between those two components.

You're looking for the resistance between nodes X and Y. So you need to consider them as starting and ending points, or the terminals of the network. Node X happens lie at the junction of two components, R1 and R2, so there are two paths leading from X into the network.

 

[Kajan thana]

Your analysis is correct.

R3 and R4 are in parallel because their connections share two nodes. Anther way to tell that they are in parallel is that you can follow a closed path that passes through only through those two components:

View attachment 149580The node X is where R1 and R2 are connected. Thus it is the junction between those two components.

You're looking for the resistance between nodes X and Y. So you need to consider them as starting and ending points, or the terminals of the network. Node X happens lie at the junction of two components, R1 and R2, so there are two paths leading from X into the network.

Thank you so much.
Make sense now