Explanation

(c) In circuit $A$, both $(p - n)$ junction diode act as forward biased. Hence, current flows in circuit $A$,
Total resistance $R$ is given by
$\frac{1}{R} = \frac{1}{4} + \frac{1}{4} \Rightarrow \frac{1}{R} = \frac{2}{4} \Rightarrow R = 2\Omega$
According to Ohm’s law
$V + I_{A}R\ \Rightarrow 8 = I_{A} \times 2 \Rightarrow I_{A} = 4\ A$
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
$V = I_{B}R \Rightarrow 8 = I_{B} \times 4 \Rightarrow I_{B} = 2\ A$