Currents flowing in each of the following circuits A and B respectively are

In circuit A, both (p-n) junction diode act as forward biasing.
Hence, current flows in circuit A.
Total resistance R is given by

$$\begin{gathered} \frac{1}{\mathrm{R}}=\frac{1}{4}+\frac{1}{4} \\ \Rightarrow \mathrm{R} = 2 \mathrm{\Omega } \end{gathered}$$

According to Ohm's law: 

$\begin{array}{rcl}\mathrm{V}& =& {\mathrm{I}}_{\mathrm{A}}\mathrm{R}\\ & \Rightarrow & 8={\mathrm{I}}_{\mathrm{A}}\times 2\\ & \Rightarrow & {\mathrm{I}}_{\mathrm{A}} = 4 \mathrm{A}\end{array}$

In circuit B, lower p-n-junction diode is reverse biased. Hence, no current will flow but upper diode is forward biased so current can flow through it.

According to Ohm's law: 

$\begin{array}{rcl}\mathrm{V}& =& {\mathrm{I}}_{\mathrm{B}}\mathrm{R}\\ & \Rightarrow & 8={\mathrm{I}}_{\mathrm{B}}\times 4\\ & \Rightarrow & {\mathrm{I}}_{\mathrm{B}} = 2 \mathrm{A}\end{array}$