# Calculate the equivalent resistance from point A to B?

## Jun 30, 2017

I got $\frac{4}{5} R$. Compare with dk-ch's method (first two solutions).

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.

${\sum}_{i}^{N} {I}_{i n , i} = {\sum}_{j}^{N} {I}_{o u t , j}$

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 $2 R$) when we combine into three parallel resistors.

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:

${R}_{s e r i e s} = {\sum}_{i}^{N} {R}_{i}$ $\text{ } \boldsymbol{\left(1\right)}$

${R}_{p a r a l l e l} = {\sum}_{i}^{N} \frac{1}{{R}_{i}}$ $\text{ } \boldsymbol{\left(2\right)}$

• The top middle using $\left(2\right)$ gives

$\frac{1}{R} _ \left(t o p\right) = \frac{1}{R} + \frac{1}{2 R} = \frac{3}{2 R} \implies {R}_{t o p} = \frac{2}{3} R$

• The middle is now $2 R$, after scrapping the junction.

• The bottom middle using $\left(2\right)$ is identical to the top middle and gives ${R}_{b o t t o m} = \frac{2}{3} R$.

This then gives: Simplifying the series circuits using $\left(1\right)$ gives:

${R}_{t o p} ' = {R}_{b o t t o m} ' = R + \frac{2}{3} R + R = \frac{8}{3} R$ And finally, this can be treated as a pair of parallel circuits using $\left(2\right)$:

$\frac{1}{{R}_{t o p} ' '} = \frac{1}{\frac{8}{3} R} + \frac{1}{2 R} = \frac{3}{8 R} + \frac{4}{8 R} = \frac{7}{8 R}$

The top two then simplify to become a resistance of

$\implies {R}_{t o p} ' ' = \frac{8}{7} R$

And the result parallel with the bottom circuit, we then get, using $\left(2\right)$,

$\frac{1}{R '} = \frac{1}{{R}_{t o p} ' '} + \frac{1}{{R}_{b o t t o m} '}$

$= \frac{1}{\frac{8}{7} R} + \frac{1}{\frac{8}{3} R}$

$= \frac{7}{8 R} + \frac{3}{8 R}$

$= \frac{10}{8 R}$

So, the equivalent resistance is

$\textcolor{b l u e}{R ' = \frac{4}{5} R}$

Jul 1, 2017

SOLUTION-I As the above given network of 12 resistors each of resistance $R \Omega$ is symmetric in respect of distribution of resistance. the flow of current in it also follows a symmetry.
The current through $B O \mathmr{and} O E$ Path is same as the current through $F O \mathmr{and} O C$ Path. So we can consider that the flow is similar to $B O \mathmr{and} O C$path as well as $F O \mathmr{and} O E$ Path. For this consideration the two paths can be separated from the central node as $B {O}_{1} C \mathmr{and} F {O}_{2} E$ and the network then looks as shown below . The equivalent resistance of this network will be same as that of the given network and now we can easily calculate it

Across $B C$ two resistors $R | | 2 R$ Giving an equivalent resistance $\frac{1}{\frac{1}{R} + \frac{1}{2 R}} = \frac{2 R}{3} \Omega$ and this combination being connected in series with $2 R \Omega$ the total resistance becomes $\left(2 R + \frac{2 R}{3}\right) \Omega = \frac{8 R}{3} \Omega$

So as a whole we have two $\frac{8 R}{3} \Omega$ and One $2 R \Omega$ in parallel combination and we get net equivalent resistance

$\frac{1}{\frac{3}{8 R} + \frac{3}{8 R} + \frac{1}{2 R}} = \frac{8 R}{10} = 0.8 R \Omega$

SOLUTION-II

Applying $\Delta \to Y$ conversion on two $\Delta$ circuits at the middle of the net work we get the following circuit. In this new symmetric circuit the point $X \mathmr{and} Y$ will have same potential so if the connection XY be removed the network will have same equivalent resistance and the network acquires the following look. The equivalent resistance now can be easily calculated.

And this becomes

$\frac{1}{\frac{1}{2 R + \frac{2 R}{3}} + \frac{1}{2 R + \frac{2 R}{3}} + \frac{1}{2 R}} = \frac{1}{\frac{3}{8 R} + \frac{3}{8 R} + \frac{1}{2 R}} = \frac{8 R}{10} = 0.8 R \Omega$

SOLUTION-III If we see the given circuit net work composed of 12 resistors each of resistance $R \Omega$, we see there exists a vertical line (XZ) of symmetry (shown in blue line in the figure). This line of symmetry is perpendicular to the axis of symmetry that joins the terminal of source of emf and cuts the network at equipotential points $\textcolor{red}{X , Y , Z}$. So if these three points be connected by a conductor there will be no change in main current i.e equivalent resistance remains unchanged. Hence equivalent resistance of the whole network is just twice of any half. 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 $\frac{R}{2} \Omega$connected in parallel combination with $R$ will be $= \frac{1}{\frac{2}{R} + \frac{1}{R}} \Omega = \frac{R}{3} \Omega$.

The Fig-3 represents this situation . In this figure there exists three branches in parallel where two branches consist of two resistors $R \mathmr{and} \frac{R}{3}$ connected in series,making total $\frac{4 R}{3} \Omega$ And thus we have the following circuit (Fig-4). Its equivalent resistance becomes

$= \frac{1}{\frac{3}{4 R} + \frac{3}{4 R} + \frac{1}{R}} \Omega = \frac{4 R}{10} \Omega = 0.4 R \Omega$ Hence the net total equivalent resistance of the given circuit (Fig-1) will be $\textcolor{red}{2 \times 0.4 R \Omega = 0.8 R \Omega}$

SOLUTION-IV (Leaving hints only)

• $\textcolor{m a \ge n t a}{\text{ Step -I}} \to$ Folding of the network axially • $\textcolor{m a \ge n t a}{\text{Step -II}} \to$Converting central $\Delta \to Y$. • $\textcolor{m a \ge n t a}{\text{Step -III}} \to$Removal of vertical central connector as pd across it is zero. • $\textcolor{m a \ge n t a}{\text{Step -IV}} \to$Calculation of equivalent resistance.

Equivalent resistance $= \frac{1}{\frac{3}{4 R} + \frac{1}{2 R}} = \frac{4 R}{5} = 0.8 R \Omega$

$\textcolor{red}{\text{ONE CAUTION}}$

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 $B$ is same as $F$ and potential of $C$ is same as $E$ So if $B \mathmr{and} F$ and $C \mathmr{and} E$ are connected by conductors then there will be no change in net equivalent resistance .
This situation is similar to the folding of the network along its axis $A D$ and it takes following look after folding. The equivalent resistance of this circuit will be

$= \frac{1}{\frac{2}{R} + \frac{2}{3 R}} + \frac{R}{2} + \frac{1}{\frac{2}{R} + \frac{2}{3 R}}$

$= \frac{3 R}{8} + \frac{R}{2} + \frac{3 R}{8} = \frac{10 R}{8}$

$= \frac{5 R}{4} \Omega \textcolor{red}{\to \text{not representative one of the original network}}$

Jul 6, 2017

$\left(\frac{4}{5}\right) r$

#### Explanation:

Numbering the independent loops from left-top $\to$ right-top and then right-bottom $\to$ left-bottom from ${i}_{1}$ to ${i}_{6}$ with ${i}_{7}$ in the big loop including the tension source $E$ we have

{(3ri_1-ri_2-ri_6=0), (3ri_2-ri_1-ri_3=0), (3ri_3-ri_2-ri_4=0), (3ri_4-ri_3-ri_5-ri_7=0), (3ri_5-ri_4-ri_6-ri_7=0), (3ri_6-ri_5-ri_7-ri_1=0), (3ri_7-ri_4-ri_5-ri_6=E):}

or

((3, -1, 0, 0, 0, -1, 0),(-1, 3, -1, 0, 0, 0, 0),(0, -1, 3, -1, 0, 0, 0),(0, 0, -1, 3, -1, 0, -1),(0, 0, 0, -1, 3, -1, -1),(-1, 0, 0, 0, -1, 3, -1),(0, 0, 0, -1, -1, -1, 3))((i_1),(i_2),(i_3),(i_4),(i_5),(i_6),(i_7))r = ((0),(0),(0),(0),(0),(0),(E))

Solving this system we get

$\left\{\begin{matrix}{i}_{1} = \left(\frac{3}{8}\right) \frac{E}{r} \\ {i}_{2} = \left(\frac{1}{4}\right) \frac{E}{r} \\ {i}_{3} = \left(\frac{3}{8}\right) \frac{E}{r} \\ {i}_{4} = \left(\frac{7}{8}\right) \frac{E}{r} \\ {i}_{5} = \frac{R}{r} \\ {i}_{6} = \left(\frac{7}{8}\right) \frac{E}{r} \\ {i}_{7} = \left(\frac{5}{4}\right) \frac{E}{r}\end{matrix}\right.$

so $R = \frac{E}{i} _ 7 = \left(\frac{4}{5}\right) r$