Bridge Current

The resultant resistance in series with the 10 V voltage source $R_{AB} = (1||2)+(3||4) =$ $\dfrac {1\times 2}{1+2} + \dfrac {3 \times 4}{3+4} =$ $\dfrac 23 + \dfrac {12}7 = \dfrac {50}{21} \ \Omega$ .

The current through the voltage source $I = \dfrac {10}{\frac {50}{21}} = 4.2$ A.

By current division, the currents through the 1 $\Omega$ and 2 $\Omega$ are $I_{AC} = \dfrac 23 \times 4.2 = 2.8$ A and $I_{AD} = \dfrac 13 \times 4.2 = 1.4$ A.

Similarly, $I_{CB} = \dfrac 47 \times 4.2 = 2.4$ A and $I_{DB} = \dfrac 37 \times 4.2 = 1.8$ A.

Therefore, $I_{CD} = I_{AC} - I_{CB} = 2.8 - 2.4 = \boxed{0.4}$ A.

Note that $I_{AD} + I_{CD} = I_{DB}$ , which is true as $1.4 + 0.4 = 1.8$ A.