Seven resistors, each of 1Ω resistance are connected as shown in figu…

Let a cell of emf ε be connected between A and B. The currents through the various arms will be as shown in the figure.

Applying Kirchhoff’s loop law in closed loop ACEA, we get

$-I_1-I_2+\left(I-I_1\right)=0$

$I=2 I_1+I_2$                    (i)

Again applying Kirchhoff's loop law in closed CEDC, we get

$-I_2-I_2+\left(I_1-I_2\right)=0$      or      $3 I_2=I_1$

Putting this value in (i), we get

$I=2 I_1+\frac{I_1}{3}$      or      $I_1=\left(\frac{3}{7}\right) I$

Again applying Kirchhoff's loop law in closed loop AEBA, we get

$-\left(I-I_1\right)-\left(I-I_1\right)+\varepsilon=0$

$\varepsilon=2\left(I-I_1\right)$

$\varepsilon=2\left(I-\frac{3 I}{7}\right)=\frac{8 I}{7}$

If R is the effective resistance between A and B, then

ε = IR

So IR = $\frac{8 I}{7}$       or       $R=\frac{8}{7} \Omega$