Equivalent Resistance Calculation:

\[ \varepsilon = \frac{5}{2}IR \]

\[ R_{eq} = \frac{\varepsilon}{3I} = \frac{5}{6}R \]

For \( R = 1 \Omega \), \( R_{eq} = \frac{5}{6} \Omega \)

The network is not reducible to simple series and parallel combinations of resistors. We exploit the symmetry in the problem:

The paths AA', AD and AB are symmetrically placed. Current in each must be the same (I).

At corners A', B and D, the incoming current I splits equally into two outgoing branches (I/2).

Applying Kirchhoff's second rule to loop ABCC'EA:

\[ -IR - \frac{1}{2}IR - IR + \varepsilon = 0 \]

\[ \varepsilon = \frac{5}{2}IR \]

Equivalent resistance:

\[ R_{eq} = \frac{\varepsilon}{3I} = \frac{5}{6}R \]

For \( R = 1 \Omega \), \( R_{eq} = \frac{5}{6} \Omega \)

For \( \varepsilon = 10 \) V, total current \( 3I = \frac{10}{5/6} = 12 \) A

Thus, \( I = 4 \) A and branch currents are as shown.