In the electrical network, shown in the diagram, $R_1=5\Omega ,R_2=2\Omega ,R_3=3\Omega$ and $E_1=2E_2=10V.$ Sources have negligible internal resistance. For this network, 

(A) Let say current through R₃ is x towards right and current through R₂ is y towards left.

Using Kirchhoff Loop Law,

$x=\frac{45}{31}; y=-\frac{10}{31}$

Current through R₃ is $x+y=\frac{35}{31}$

The power generated in $R_1$ is $i^{2}R=\frac{6125}{961}$ J/s.

(D) The ratio of the powers generated in $R_2$ and $R_3$ is ${\left(\frac{{\displaystyle \raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$31$}\right.}}{{\displaystyle \raisebox{1ex}{$45$}\!\left/ \!\raisebox{-1ex}{$31$}\right.}}\right)}^2\times \frac{2}{3}=\frac{8}{243}$.

(B) The sources of e.m.f $E_1$ and $E_2$ were physically inter changed, without changing the polarities.

Let say current through R₃ is x towards right and current through R₂ is y towards left.

Using Kirchhoff Loop Law,

$x=-\frac{15}{31}; y=\frac{55}{31}$

Current through R₃ is $x+y=\frac{40}{31}$

The power generated in $R_1$ is $i^{2}R=\frac{8000}{961}$ J/s

(C) The polarities of the sources $E_1$ and $E_2$ were both reversed.

Let say current through R₃ is x towards right and current through R₂ is y towards left.

Using Kirchhoff Loop Law,

$x=-\frac{45}{31}; y=\frac{10}{31}$

Current through R₃ is $x+y=-\frac{35}{31}$

The power generated in $R_1$ is $i^{2}R=\frac{6125}{961}$ J/s.