Calculating the resistance when the switch is off

kakadas

1. The problem statement, all variables and given/known data

EDIT: The chain is connected to a battery, thus the current in this chain is direct.



When the electric chain's switch (J) is on, resistance Rab (from dots A to B) is equal to 80 Ohms.

What is the electric chain's resistance when the switch (J) is off?

2. Relevant equations


3. The attempt at a solution

I thought that the resistors are connected in parallel , but my teacher said that when the switch if on resistors are connected in parallel and when it's off it's in series.

Is this true, cause I think it's parallel, because the wires split either way, no matter if the switch is on or not...

Don't know for sure what's the connection type in this chain, so...

gneill

When the switch J is closed, isn't there a direct path from A to B passing right through it? How can the resistance then be 80Ω? Can you confirm the question statement and diagram?

kakadas

Well the book says exactly that. In my view this is reasonable because only some electricity is flowing through the switch when it's on, not all of it...

Anyway, does anyone has the answer to my original questions?

gneill

No, an ideal switch and wires have zero resistance. The switch will form a short circuit, and ALL of the current will pass through it. So, either the text is wrong, or there is more to this problem than has been stated. Is the switch non-ideal with some resistance of its own? Could it be that the switch should really be oriented horizontally across the circuit rather than vertically?

NascentOxygen

No one knows, because you have got the diagram wrong, kakadas. As gneill observed.

CWatters

I agree with the other posters. The original question, the diagram and your teachers comment are all inconsistent with each other.

Can you scan that page of the book?

PS Are you sure the switch shouldn't be between the other two corners of the square?

kakadas

I'm 100% sure of that. I will upload a scan ASAP.

Oh I think I forgot to mention a crucial part of this chain (Or maybe not, I'm just learning..): It's connected to a battery, thus the current is direct.

EDIT:

Well maybe You're a bit confused because English is not my native language, so...
But I really think that I translated the original question correctly...

Here's the Photo of the chain from my book:



I really hope someone will answer my questions.

azizlwl

When the electric chain's switch (J) is on, resistance Rab (from dots A to B) is equal to 80 Ohms.

What is the electric chain's resistance when the switch (J) is off?

2. Relevant equations


3. The attempt at a solution

I thought that the resistors are connected in parallel , but my teacher said that when the switch if on resistors are connected in parallel and when it's off it's in series.
--------------------------
Parallel circuit
[itex]R_{total}=\frac{R_1 R_2 R_3}{R_1 + R_2 +R_3}[/itex]

Any factor in the numerator with a zero value will result in total zero resistance.

gneill

According to the diagram, R_AB must be zero when the switch (j) is closed (on).

Is it possible that there is a language confusion regarding the switch positions? "closed" or "on" means that the contacts are touching and current can flow. "Open" or "off" means that the contacts are not touching and current cannot flow --- open circuit.

By the way, out of curiosity, what is the original language of the text?

kakadas

That was exactly what i meant to say..
But does the current not flow when the switch is off? I think it still flows, because IMO the resistors are connected in parallel, so I think that the current flows either way, no matter in what state the switch is.

If I'm wrong, Could anyone prove to me that the current doesn't flow when the switch is off?

Regards, kakadas

gneill

Yes the current flows when the switch is off (open) --- It then must flow through the resistor network. But your problem statement contends that the resistance is 80 Ohms when the switch is closed (on). That is not possible with the given information.

CWatters

I agree. Looks like ON and OFF may have been confused but...

If it's 80 Ohms with the switch OFF (open) that would allow R to be calculated but when the switch is ON (conducting) the equivalent resistance is ZERO.

The teachers comment about it changing from serial to parallel depending on the switch position makes no sense.

kakadas

Why is that so, please explain in more detail?

BTW does it make a difference if the current is DC or AC? In my case it's DC.

CWatters

With the switch OFF you have two resistors in series, in parallel with, two resistors in series.

With the switch ON the resistors are irrelevant. All the current flows through the switch. You could replace the resistors with ones that were much bigger, much smaller, or a short circuit and it would make no difference. The equivalent resistance is zero Ohms.

Makes no difference if it's AC or DC.

kakadas

I Just want to clearinfy that By telling that the switch is ON I meant that it's CONDUCTING.
OFF means that it's open and NOT CONDUCTING.

azizlwl

'


As others said, there are errors in the data and in the drawing.
It is waste of time computing on it.

I guess the problem meant for Wheatstone Bridge.

http://en.wikipedia.org/wiki/Wheatstone_bridge

CWatters

That's what I had assumed as well.