b
(b)Consider the case when \( Ge\)  and \(Si \) diodes are connected as show in the given figure.
Equivalent voltage drop across the combination \(Ge\) and \(Si\) diode \(= 0.3 \,V\)
==> Current \(i = \frac{{12 - 0.3}}{{5\,k\Omega }} = 2.34\,mA\)
 Out put voltage \(V_0 = Ri = 5 k \Omega ×2.34 \,mA = 11.7 \,V\)
Now consider the case when diode connection are reversed. In this case voltage drop across the diode's combination \(= 0.7 \,V\)
==> Current \(i = \frac{{12 - 0.7}}{{5\,k\Omega }} = 2.26\,mA\)
 \({V_0} = iR = 2.26\,mA \times 5\,k\Omega = 11.3\,V\)
Hence charge in the value of \(V_0 = 11.7 -11.3 = 0.4 V\)