Determine the current in each branch of the network shown in the Figure.

(A) Let the currents in the various branches be as shown in the figure.
Applying Kirchhoff's voltage law to the closed loop $ABDA$:
$10 I_{2} + 5 I_{4} - 5 I_{3} = 0 \implies 2 I_{2} + I_{4} - I_{3} = 0 \implies I_{3} = 2 I_{2} + I_{4} \dots (i)$
Applying Kirchhoff's voltage law to the closed loop $BCDB$:
$5(I_{2} - I_{4}) - 10(I_{3} + I_{4}) - 5 I_{4} = 0 \implies 5 I_{2} - 10 I_{3} - 20 I_{4} = 0 \implies I_{2} = 2 I_{3} + 4 I_{4} \dots (ii)$
Applying Kirchhoff's voltage law to the outer loop $ABCFEA$:
$-10 + 10 I_{1} + 10 I_{2} + 5(I_{2} - I_{4}) = 0 \implies 10 I_{1} + 15 I_{2} - 5 I_{4} = 10 \implies 2 I_{1} + 3 I_{2} - I_{4} = 2 \dots (iii)$
Since $I_{1} = I_{2} + I_{3}$,substituting into $(iii)$:
$2(I_{2} + I_{3}) + 3 I_{2} - I_{4} = 2 \implies 5 I_{2} + 2 I_{3} - I_{4} = 2 \dots (iv)$
Solving the system of linear equations $(i), (ii),$ and $(iv)$:
From $(i)$ and $(ii)$,$I_{2} = 2(2 I_{2} + I_{4}) + 4 I_{4} = 4 I_{2} + 6 I_{4} \implies -3 I_{2} = 6 I_{4} \implies I_{2} = -2 I_{4}$.
Substituting $I_{2} = -2 I_{4}$ into $(i)$: $I_{3} = 2(-2 I_{4}) + I_{4} = -3 I_{4}$.
Substituting $I_{2}$ and $I_{3}$ into $(iv)$: $5(-2 I_{4}) + 2(-3 I_{4}) - I_{4} = 2 \implies -10 I_{4} - 6 I_{4} - I_{4} = 2 \implies -17 I_{4} = 2 \implies I_{4} = -2/17 \text{ A}$.
Thus,$I_{2} = -2(-2/17) = 4/17 \text{ A}$,$I_{3} = -3(-2/17) = 6/17 \text{ A}$.
Branch currents:
$I_{AB} = I_{2} = 4/17 \text{ A}$
$I_{BC} = I_{2} - I_{4} = 4/17 - (-2/17) = 6/17 \text{ A}$
$I_{AD} = I_{3} = 6/17 \text{ A}$
$I_{CD} = I_{3} + I_{4} = 6/17 - 2/17 = 4/17 \text{ A}$
$I_{BD} = I_{4} = -2/17 \text{ A}$
$I_{total} = I_{1} = I_{2} + I_{3} = 4/17 + 6/17 = 10/17 \text{ A}$.