Given that current through CA = 1 A, current through C'A' = 2 (a) Now if A is connected to A' and B is connected B'. Find currents through CA and C'A', respectively.

${\mathrm{E}}_1=(2+4)\times 1=6\mathrm{V}, {\mathrm{E}}_2=(2+4)\times 2=12\mathrm{V}$

There will be no current in AB and A' B' , because all these points are at the same potential. From D to A':

$\begin{array}{l} {\mathrm{V}}_{\mathrm{D}}+6-2{\mathrm{I}}_1={\mathrm{V}}_{\mathrm{A}},\text{ but }{\mathrm{V}}_{\mathrm{D}}={\mathrm{V}}_{{\mathrm{A}}^{\mathrm{\prime }}}=0\\ \Rightarrow {\mathrm{I}}_1=3\mathrm{A}\end{array}$

For outermost loop: $6-12=2{\mathrm{I}}_1+4\left({\mathrm{I}}_1-{\mathrm{I}}_2\right)\text{ or }{\mathrm{I}}_2=6\mathrm{A}$

Current through ${\mathrm{C}}^{\mathrm{\prime }}{\mathrm{A}}^{\mathrm{\prime }}={\mathrm{I}}_1-{\mathrm{I}}_2=3-6=-3\mathrm{A}$