In the given circuit, a silicon diode ( D_1 ) and a germanium diode...

In the given circuit, the silicon diode ($D_1$) and the germanium diode ($D_2$) are connected in parallel.

The barrier potential of the germanium diode ($0.3 \text{ V}$) is lower than that of the silicon diode ($0.7 \text{ V}$).

When the voltage is applied, the germanium diode turns on first and clamps the voltage across the parallel combination to $0.3 \text{ V}$.

Since the voltage across the silicon diode is only $0.3 \text{ V}$, which is less than its barrier potential of $0.7 \text{ V}$, it remains in the non-conducting state. Therefore, the current passing through the silicon diode is $0 \text{ mA}$.

The total current supplied by the battery flows entirely through the germanium diode. Applying Kirchhoff's voltage law to the circuit:

$V - I R - V_{\text{Ge}} = 0$

$10.3 - I(2 \times 10^3) - 0.3 = 0$

$I = \dfrac{10.3 - 0.3}{2 \times 10^3} = \dfrac{10}{2 \times 10^3} = 5 \times 10^{-3} \text{ A} = 5 \text{ mA}$

Thus, the total current is $5 \text{ mA}$ and the current through the silicon diode is $0 \text{ mA}$.

Answer: $5 \text{ mA}$ and $0 \text{ mA}$