In the circuit shown in the figure, the power dissipated in the circuit is $P_0$  if an ideal cell is connected across A and B same power is dissipated in the circuit if the same cell is connected across C and D . when the cell was connected across A and D or across B  and C, the power dissipated in the circuit is found to be $3P_0$  the power dissipated in the circuit if the cell is  connected across A and C is  $\left(\frac{15+6\sqrt{x}}{16+6\sqrt{x}}\right)p_0$ the  value of x is 

Equivalent resistance across A – B is given by
$\frac{1}{R_{AB}}=\frac{1}{R_1}+\frac{1}{R_2+R_3+R_4}$       …(1)
Resistance across C – D is   $\frac{1}{R_{CD}}=\frac{1}{R_3}+\frac{1}{R_2+R_1+R_4}$    …(2)
Same power will be dissipated in two cases if $R_{AB}=R_{CD}\Rightarrow R_1=R_3$
Similarly, one can prove that  $R_2=R_4.$
When cell is connected across B-C (or A - D) the equivalent resistance is 

Given  $\frac{1}{R_{BC}}=\frac{1}{R_2}+\frac{1}{R_1+R_2+R_4}$

$\frac{{\epsilon }^2}{R_{AB}}=P_{0}and\frac{{\epsilon }^2}{R_{BC}}=3P_0$      $\therefore \frac{1}{R_{BC}}=\frac{3}{R_{AB}}$

$$\begin{gathered} \Rightarrow \frac{1}{R_2}+\frac{1}{2R_1+R_2}=3[\frac{1}{R_1}+\frac{1}{R_1+2R_2}] \\ \Rightarrow \frac{2R_1+2R_2}{2R_1{R}_2+R_2^2}=3\left(\frac{2R_1+2R_2}{R_1^2+2R_1{R}_2}\right) \\ \Rightarrow R_1^2+2R_1{R}_2=3R_2^2+6R_1{R}_2 \Rightarrow R_1^2-4R_2\cdot R_1-3R_2^2=0 \\ \Rightarrow R_1=\frac{4R_2\pm \sqrt{16R_2^2+12R_2^2}}{2} \\ {R}_1=2R_2+\sqrt{7}{R}_2=(2+\sqrt{7})R_2 \\ \therefore P_0=\frac{{\epsilon }^2}{P_{AB}}={\epsilon }^2[\frac{1}{(2+\sqrt{7})R_2}+\frac{1}{(4+\sqrt{7})R_2}] \\ =\frac{{\epsilon }^2}{R_2}=\left[\frac{6+2\sqrt{7}}{(2+6\sqrt{7})}\right]=\frac{2{\epsilon }^2}{R_2}\frac{(3+\sqrt{7})}{(15+6\sqrt{7})} \end{gathered}$$

 Resistance across A – C is  $\frac{1}{R_{AC}}=\frac{1}{R_1+R_2}+\frac{1}{R_3+R_4}=\frac{1}{R_1+R_2}+\frac{1}{R_1+R_2}$

$$\begin{gathered} =\frac{2}{R_1+R_2}=\frac{2}{(3+\sqrt{7})R_2} \\ {P}_{AC}=\frac{{\epsilon }^2}{R_{AC}}=\frac{2{\epsilon }^2}{(3+\sqrt{7})R_2}=P_0\frac{(15+6\sqrt{7})}{(3+\sqrt{7})(3+\sqrt{7})}=\left(\frac{15+6\sqrt{7}}{16+6\sqrt{7}}\right)P_0 \end{gathered}$$