a
(a)Since diode in upper branch is forward biased and in lower branch is reversed biased. So current through circuit \(i = \frac{V}{{R + {r_d}}}\); here \({r_d}\)= diode resistance in forward biasing \(= 0\)
==> \(i = \frac{V}{R} = \frac{2}{{10}} = 0.2A\).