Question

Forty identical wires, each of resistance $R$ form a square net consisting of $16$ meshes. The resistance between the diagonal points of the network is $\frac{\alpha }{\beta }R$ then the value of a $\alpha -2\beta $ is

Answer 1

Above question can be seen through below figure,



From the axis symmetry about $AB$ , we have following mesh available,


Current distribution through the network is shown in below figure,


From the analysis, we can see that top resistances have same current $I_{5}\&I_{6}$ , so they are connected in series, and the case is shown in below figure,


Now connecting some of them as parallel and in series as shown in above network, we have following network,


Above network can be drawn as below figure,


Above network can be drawn as below figure,


Equivalent resistance for second part can be written as,
$R_{e q}{ }^{\prime}=\frac{V-0}{i_{1}+i_{2}} $
$\Rightarrow R_{e q}{ }^{\prime}=\frac{V-0}{\frac{(V-x)}{\frac{R}{2}}+\frac{(V-y)}{R}} $
$\Rightarrow R_{e q}{ }^{\prime}=\frac{(V-0) R}{2(V-x)+(V-y)} $
$\Rightarrow R_{e q}{ }^{\prime}=\frac{(V-0) R}{2\left(V-\left(\frac{37 V}{50}\right)\right)+\left(V-\frac{16 V}{25}\right)} $
$\Rightarrow R_{e q}{ }^{\prime}=\frac{(V-0) R}{2 V-\left(\frac{37 V}{25}\right)+V-\frac{16 V}{25}} $
$\Rightarrow R_{e q}{ }^{\prime}=\frac{(V-0) R}{3 V-\frac{53 V}{25}} $
$\Rightarrow R_{e q}{ }^{\prime}=\frac{25 V R}{75 V-53 V} $
$\Rightarrow R_{e q}{ }^{\prime}=\frac{25 R}{22}$
Now equivalent resistance of total network is given by,
$R_{e q}=R+R_{e q}{ }^{\prime} $
$\Rightarrow R_{e q}=R+\frac{25 R}{22}$
$\Rightarrow R_{e q}=\frac{22 R+25 R}{22}=\frac{47 R}{22}$
From the given question and comparing with, $R_{e q}$, we have
$\alpha=47, \beta=22$
So $\alpha-2 \beta=47-(2 \times 22)=3$