Calculate the equivalent resistance from point #A# to #B#?
I haven't done this in a while, so check my work. I found it easier to redraw the circuit like this:
The first one is a re-draw to make it look easier to interpret.
The second one uses Kirchoff's current law at a junction, i.e.
#sum_i^N I_(i n,i) = sum_j^N I_(out,j)#
Since the central junction has three wires going in and out, it should be the same value in as the same value out. Thus, scrapping the junction doubles the resistances there (to
The third one is just a re-draw of the second to clarify that there are parallel circuits to consider. After this, I would simplify from the inside out using the following relations:
#R_(series) = sum_i^N R_i# #" "bb((1))#
#R_(parall el) = sum_i^N 1/(R_i)# #" "bb((2))#
- The top middle using
#1/R_(t o p) = 1/(R) + 1/(2R) = 3/(2R) => R_(t op) = 2/3 R#
The middle is now
#2R#, after scrapping the junction.
The bottom middle using
#(2)#is identical to the top middle and gives #R_(b o t t om) = 2/3 R#.
This then gives:
Simplifying the series circuits using
#R_(t o p)' = R_(b o t t o m)' = R + 2/3 R + R = 8/3 R#
And finally, this can be treated as a pair of parallel circuits using
#1/(R_(t o p)'') = 1/(8/3 R) + 1/(2R) = 3/(8R) + (4)/(8R) = 7/(8R)#
The top two then simplify to become a resistance of
#=> R_(t o p)'' = 8/7 R#
And the result parallel with the bottom circuit, we then get, using
#1/(R') = 1/(R_(t o p)'') + 1/(R_(b o t t o m)')#
#= 1/(8/7 R) + 1/(8/3 R)#
#= 7/(8R) + 3/(8R)#
So, the equivalent resistance is
#color(blue)(R' = 4/5 R)#
As the above given network of 12 resistors each of resistance
The current through
The equivalent resistance of this network will be same as that of the given network and now we can easily calculate it
So as a whole we have two
In this new symmetric circuit the point
The equivalent resistance now can be easily calculated.
And this becomes
If we see the given circuit net work composed of 12 resistors each of resistance
Now if we can find the equivalent resistance of one half shown above . then twice of that resistance will be the net equivalent resistance of the given circuit.
Th equivalent resistance of
The Fig-3 represents this situation . In this figure there exists three branches in parallel where two branches consist of two resistors
And thus we have the following circuit (Fig-4). Its equivalent resistance becomes
Hence the net total equivalent resistance of the given circuit (Fig-1) will be
SOLUTION-IV (Leaving hints only)
#color(magenta)" Step -I" to#Folding of the network axially
#color(magenta)"Step -II" to#Converting central #Delta to Y#.
#color(magenta)"Step -III" to#Removal of vertical central connector as pd across it is zero.
#color(magenta)"Step -IV" to#Calculation of equivalent resistance.
To simplify the given circuit for the sake of calculation of equivalent resistance If the wires are detached cutting off the axial wire and the network takes the following look then the equivalent resistance of this network will not be the same as above.
In this circuit from symmetric point of view potential of
This situation is similar to the folding of the network along its axis
The equivalent resistance of this circuit will be
Numbering the independent loops from left-top
Solving this system we get