Resistance of both parallel and series

[kalakuja]

I was digging through my old uni files while cleaning up computer and found a following question.

What is the total resistance in the following circuit:

.

My physics was just minor so i got no clue what to do or if it is a trick question. I remember that you used the parallel and series and added them together but what do i do when they seem like they are in both connections?

[Obby]

Zero. There are shorts going around both resistors.

[kalakuja]

Zero. There are shorts going around both resistors.

AA i see. The thing about the current choosing the "easiest" part ?
Hmm i remember counting somekind of a similar system, What if the there is batteries in the similar connection, What is the total voltage produced if those resistors were batteries (minus sides on the left side of the rectangle and positives on right) ?

[yurell]

AA i see. The thing about the current choosing the "easiest" part ?

So you don't need to remember separate laws, you can always treat a wire as a resistor with resistance 0. Then, using your law for resistance in parallel (1/R = 1/R1 + 1/R2 + ...), you can see the resistance is 0. As you know, you can treat a system of resistors as a single resistor with the total resistance, which means you have two resistors of resistance 0 in series, which results in a single resistor of resistance 0.

Or you could just look to see if a short-circuit exists, and then ignore anything it bypasses.

[nehpest]

Zero. There are shorts going around both resistors.

AA i see. The thing about the current choosing the "easiest" part ?
Hmm i remember counting somekind of a similar system, What if the there is batteries in the similar connection, What is the total voltage produced if those resistors were batteries (minus sides on the left side of the rectangle and positives on right) ?

If you have batteries instead of resistors, you will have a degenerate case. The batteries are both shorted, so you'll discharge a very large current through the shorts and probably destroy the batteries. Probably best not to try that at home

[yurell]

If you have batteries instead of resistors, you will have a degenerate case. The batteries are both shorted, so you'll discharge a very large current through the shorts and probably destroy the batteries. Probably best not to try that at home

Tested that as part of a project at school. AAA batteries are virtually dead after an hour, with Duracell beating Energiser, and some cheap Chinese crap actually outlasting them both by a long margin (we set some potential difference between the two ends to be the 'dead' margin, can't remember what it was). But by God they get hot ... we had to keep them cool to make sure they wouldn't explode.

In hindsight, I honestly can't believe the teacher let us do it.

[jmorgan3]

Tested that as part of a project at school. AAA batteries are virtually dead after an hour, with Duracell beating Energiser, and some cheap Chinese crap actually outlasting them both by a long margin

This might have been from high internal resistance rather than high energy content.

[kalakuja]

Thank you

[nehpest]

Tested that as part of a project at school. AAA batteries are virtually dead after an hour, with Duracell beating Energiser, and some cheap Chinese crap actually outlasting them both by a long margin

This might have been from high internal resistance rather than high energy content.

True. Yurell, did you perchance measure current during discharge?

[yurell]

To be honest, I can't remember, and I don't seem to have it saved in my high school folder (I only really started keeping digital copies in Form 5). From the price of the battery, though (I do remember it was the cheapest available), I'd imagine that it's due to high internal resistance, but we'll never know unless someone wants to science it again!

[Adam Preston]

If there are 2 parallel resistors I find this equation much easier :

R1*R2/ R1+R2

Saves me a heck of time when answering questions which concern as I said only 2 parallel resistors, with series resistors you simply add them up. If you had more than 2 parallel resistors then yes as stated before you would need to use the formula 1/R = 1/R1 + 1/R2 + ...

[eSOANEM]

Perhaps, kalakuja, as a more interesting (and non-zero) problem, you might like to consider the same problem but with another resistor of resistance R placed between the two shorts.

If you had more than 2 parallel resistors then yes as stated before you would need to use the formula 1/R = 1/R1 + 1/R2 + ...

Nope. The total resistance would still be the ratio of the products of the resistances to the sum of the partial products (where by "the" partial products, I refer only to those with n-1 distinct resistor's resistance in them). The reason it doesn't look quite the same is that there is are degenerate cases when n<3.

Anyway, the OP was asking about cases where there are resistor networks containing parallel and series sections with connections between them.

[thoughtfully]

Here's how to think about this sort of problem in general. It's true enough that the shorts make the resistance zero, but what if they were resistors? Or maybe they are half-ohm resistors, and the wires in the shorts are really long, or really fine, making your abstraction leaky, or you need to show the math in a rigorous way.

What is going on is there are a pair of parallel resistors (if you consider the shorts as zero ohm resistors) in series. The circuits I've drawn and the original are all equivalent. The total resistance is the sum of the parallel resistances, which is 0+0:

Code: Select all

R*0   R+0
--- + ---
R+0   R+0

[shawnhcorey]

The circuit is called a bridge: http://en.wikipedia.org/wiki/Bridge_circuit

You solve it with Kirkoff's circuit laws: http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws

[eSOANEM]

If you're trying to solve one though, there are often much easier ways than sledgehammering through with Kirchoff.

For one thing, dealing with any symmetry first is always a good idea. It's possible to reduce a lot of problems to simple combinations of the standard results for the addition of series and parallel resistors (for example the tetrahedron of equal resistors reduces to two 2R resistors in parallel with a 1R when the symmetries are properly noted).

[Afif_D]

I did not any of the answers that other users gave
but
I feel, the resistance will be R/2.
Both of them are parallel to each other, their ends are connected to the same point(the diagram is cleverly drawn). So the potential across both are equivalent.

[jaap]

I did not any of the answers that other users gave
but
I feel, the resistance will be R/2.
Both of them are parallel to each other, their ends are connected to the same point(the diagram is cleverly drawn). So the potential across both are equivalent.

That would only be the answer if you cut the connection in the centre of the diagram.

[eSOANEM]

I did not any of the answers that other users gave
but
I feel, the resistance will be R/2.
Both of them are parallel to each other, their ends are connected to the same point(the diagram is cleverly drawn). So the potential across both are equivalent.

As it is, there is a route through with 0 resistance (following the two shorts). As the current splits between routes in inverse proportion to the resistance, 100% of the current follows this route so the resistance is 0.