nav[aria-label="Primary Navigation"] { padding: 0; & ul { list-style: none; width: 100%; display: flex; flex-direction: row; justify-content: start; align-items: start; gap: 30px; padding: 0; & li { margin: 0; } & ul li { list-style: none; } } }

Child Item

How to work out total resistance

phantom_lord

Any help much appriciated...

Find more posts tagged with

Quick Links

Comments

Capt'n Midnight

Is there an electrical connection at the centre ?

Since no one is going to do your homework for you, what have you tried so far ?

phantom_lord

http://img496.imageshack.us/img496/6278/schematic7mt.jpg that's another way of looking at it. It's not homework just a questions i'm working out. I've worked out several answers but they're all incorrect and I was just wondering if anyone had any ideas of how to solve it.

*edit*
really i'm just wondering what way does the current flow (2nd diagram) when the three branches are connected, and what effect this has on the resistance, I'm not asking someone to calculate t, more how to go about calculating it.

Teg Veece

MultiSim is giving me a total resistance value of 8 Ohms. I'm not too sure how you'd go about solving it though. Maybe if you break it up into different current loops which would leave you with about 9 simulataneous equations and as many variables for currents. I'll let you try that...

CathalMc

Actually, this is a far simpler problem than it might appear. The naive answer is to evaluate the zillion variable with Kirchoffs, but you can simplify the answer down by identifying symmetries.

Firstly, that the currents (Ia) are equal is apparent from vertical and horizontal symmetry. It is also quite intuitive. Similarly, it is apparent that the currents marked Ib are equal.

Now if you examine currents passing through the green dotted line, it is now apparent that the currents marked Ic must be equal. This is already enough to vastly simplify Kirchoffs equations.

But now consider the central node. On the branch going from input to output carrying Ib, we have determined that the current is constant throughout: ie: no current is added from the triangular side networks. (Note, this is true of all external bias conditions.) Now, we can redraw the network with the Ib branch disconnected from everything else (ie: a parallel branch with two 10 Ohm resistors). Similarly, we can disconnect the two triangular parts from each other, because each side is sourcing and sinking 1 Ic -> No current is exchanged. (This may be seem slightly irregular, but consider the voltages of the separated branches where they were connected: it is obvious they are all Vapplied/2. In a sense there is no potential difference to cause current to flow when they are connected; so the connection is irrelevent)

So finally, you have 3 parallel branches, of trivial difficulty.
The resistence of the left or right branch is 10 + 10||(10 + 10) + 10 = 80/3, and the resistence of the middle branch is 10 + 10 = 20. So the total resistence is (80/3)||20||(80/3) = 8 Ohms

Stoner

Lads is it just me is anyone having difficulty even seeing a voltage or current source on this diagram.
That seems like a question for Saint Flood from Kevin Street the Saint of electrical field theory. I feel guilty because I can't answer that question anymore or maybe its because I'm not bothered anymore but its deffo guilt.

Well answered CathalMc,

I watch too much TV

CathalMc

Teg Veece wrote:

MultiSim is giving me a total resistance value of 8 Ohms...

Coulda sworn you said 10 Ohms before I posted my answer!

ozt9vdujny3srf

Stoner wrote:

Lads is it just me is anyone having difficulty even seeing a voltage or current source on this diagram.
That seems like a question for Saint Flood from Kevin Street the Saint of electrical field theory. I feel guilty because I can't answer that question anymore or maybe its because I'm not bothered anymore but its deffo guilt.

Well answered CathalMc,

I watch too much TV

I'm not sure what voltage or current source really has to do with total resistance in fairness.

Stoner

The question brought me back a few years. I was applauding the answer

in fairness

ozt9vdujny3srf

Still doesnt mean a voltage or current source has anything to do with the question.

meh.

Stoner

True,
I was impressed with an answer and you seem to want to gain some level of superiority from the fact that my initial response had an error in it
What has been gained for your comments? somebody else answered the question

ozt9vdujny3srf

That meh was to point out that i wasnt interested in a stupid spiralling off topic bicker. I don't want to gain any levely of superiority, i was just pointing out a reduntant comment that could only confuse things.

Anyways, I agree, the answer was a good one.

In fairness

phantom_lord

Yep it was....

Can anymore point me in the direction of solving this one? I'll just not totally sure of what approach to take.

Stoner

1 Ohm in MHO

from a to b, the path is two resistors in series in parallel with two resistors in series. This would be the lowest resistance between points a and b.
having said that it seems a little too easy compare to the other question

phantom_lord

R + ((r + R) parallel R) = r
where R=1 and r= total resistance
then r/2=Ans

=> 1 + (r+1)||R = r
=> 1 + [{(r+1)*(1)}/r+1+1] = r
=> 1 + [r+1/r+2] = r
=> r + 2 + r + 1 =r2 + 2r .... where r2=r squared
=> 3 = r2 +2r - 2r
=> r = sqrt3

=> r = 1.732050808
Which means Ans = r/2 = .866025403 Ohms

CathalMc

phantom_lord wrote:

Which means Ans = r/2 = .866025403 Ohms

I got the same, albeit a slightly longer route it seems.

Could you explain how you formulated R + ((r + R) parallel R) = r?

Incidently, there's a nice related problem to this one, and that is a grid of squares, like in this question, but also extending in the vertical direction. So the entire plane is covered in squares of resistors...

phantom_lord

CathalMc wrote:

I got the same, albeit a slightly longer route it seems.

Could you explain how you formulated R + ((r + R) parallel R) = r?

Incidently, there's a nice related problem to this one, and that is a grid of squares, like in this question, but also extending in the vertical direction. So the entire plane is covered in squares of resistors...

Will do, I just don't have time right now, I'll post it later today.

phantom_lord

There are two properties of resistors that are useful in this problem.
1. Resistors in series add linearly. i.e. R(total) = R(1)+R(2)+R(3)+...
2. Resistors in parallel add inversely. i.e. 1/R(total) = 1/R(1) + 1/R(2)
+ 1/R(3) +...,
which can be simplified to R(total) = (R(1)*R(2))/(R(1)+R(2)) for the case
of only two resistors.

If it were a finite 'row' then we would be able to use the two properties
above calulate the resistance. Taking one box we have three
resistors in series in parallel with one other (Taking the last box, the
upper,
right and lower resistors are in series, which are in parallel to the
resistor on the left of the box). We can add the three in series together
to
obtain a resistor of value 3 in parallel with a resistor of value 1. Adding
these:

3*1/(3+1) = 3/4

We get another box, only now the right side has value 3/4 instead of 1. We
can repeat this process numerous times, getting values like 11/15 and 41/56,
but this won't help in finding the correct answer. The fact that this is
an
repetitive process will.

The general equation for the next resistor value in the series is given by:

(2+x)*1/(2+x+1) = (2+x)/(3+x)

taking x to be the previous vertical resistor and the values of the rest
to
be 1. This is used to consider the original infinite chain of
resistors.

Another matter that needs attention is whether this infinite chain of
resistors
has some limit.
This series is monotonic down, (the next value is
two plus the last value divided by two plus the last value plus one: f(x+1)
= (2+f(x)) / (2+f(x)+1))

So if it has a limit, it will be lower than any of the values computed from
a finite chain.
If all the horizontal resistors are removed.

Following the same logic as before in in computing g(x+1), we get:
g(x+1) = g(x) / (g(x) + 1), which is also monotonic down for the same
reasons
so if both series start with the same f(x), i.e. f(x) = g(x) and the first
series is larger than the second:
f(x) / (f(x) + 1) < (2 + f(x)) / (3 + f(x)), so by cross multiplying
3*f(x) + f(x)*f(x) < 2 + 3*f(x) + f(x)*f(x)
0 < 2, which is true

After one iteration, the first series will be larger than the second one.
Since the first series
now uses a larger value than the second series, the gap between the two can
only become larger.
Then the first series is always greater than the second, so if there is a
limit for the second series, this will be a lower bound to the first.

One divided by the sum of any number of resistors in parallel is equal to
the sum of one divided by each of the resistors.
Taking the second series, all it is is an infinite number of identical
resistors summed in parallel. Each has a value of 1, so the sum of an
infinite
number of ones divided by 1 approaches infinity, as it does for any other
finite value for the resistor. And if one divided by the equivalent
resistor approaches infinity, the equivalent resistor approaches zero, which
is a finite limit. Thus since the original resistor chain is monotonic down
and bounded from below, it then must approach some finite value
as more of the infinite chain of resistors is considered. This valus must
be found.

I have an equation to find any resistor value granted we know the previous
one. So the resistance between A and B, P(AB) with
respect to the resistance across the resistor immediately to the right of
it,
P(A'B').

P(AB) = (2+P(A'B'))/(3+P(A'B'))

Now consider P(A'B'): There is still an infinite chain of resistors
extending from it, so, P(AB) = P(A'B') when the entire chain is considered.

Substituting P(AB) = P(A'B') = r into the equation we get:

r = (2+r)/(3+r), thus:
3*r+r^2 = 2+r, and
r^2 + 2*r -2 = 0

Solving with the quadratic equation we get:

r = (-2 (+or-) sqrt (4 + 4*2*1))/2
r = (-2 (+or-) 2*sqrt(3))/2, and finally
r = -1 (+or-) sqrt(3)

since you cannot have negative resistances the (-)
solution is eliminated , leaving

r = sqrt(3) - 1

That plus the resistance for the resistors between a and b is the answer

CathalMc

Kudos to you sir, I'm impressed. I found the usage of g(x) unneccessary because the natural bounds of f(x+1) are obvious. Having said that, our approaches are identical all but in description.

Having worked out the same recursion equation from inspection:
f(x+1) = [f(x) + 2]/[f(x) + 3]
two simple observations are apparent, that any one f(x+1) iteration is neccessarily bounded by 2/3 and 1 for f(x) positive. (consider f(x) -> 0,inf).

Subtracting f(x) from both sides gives the deltaf(x+1), and simplifying gives

deltaf(x+1) = [2-f(x)^2 - 2f(x)]/[3 + f(x)] = [(f(x) + a)(f(x) + b)]/[f(x)+3]

where {a,b} = -1 +-sqrt(3)

Under further inspection, we can see that deltaf is positive and monotonically decreasing for 0 < f(x) < (sqrt(3)-1), and negative and monotonically decreasing for f(x) > (sqrt(3)-1), consistent with convergence on sqrt(3)-1 for f(0) in the range (0,+inf).

DMT

Superposition, Thevenin's theorem and Norton's Theorem should solve all of these if my memory serves me correctly.

Mr T

Thevenin's Theorem, Nortons Theorem and Kirchoffs laws thats all you need