The circuit shown below contains two ideal diodes, each with a forward resistance of $$50\Omega$$. If the battery voltage is 6 V, the current through the $$100\Omega$$ resistance (in Amperes) is:

Biasing condition from battery connections: $$D_1 = \text{Forward-biased}, \quad D_2 = \text{Reverse-biased}$$

Equivalent resistances of the branches: $$R_{D1} = 50\ \Omega, \quad R_{D2} = \infty$$

Total resistance of the active circuit network: $$R_{\text{total}} = R_{D1} + 150 + 100 = 50 + 150 + 100 = 300\ \Omega$$

Current passing through the circuit and the $$100\ \Omega$$ resistor: $$I = \frac{V}{R_{\text{total}}} = \frac{6}{300} = 0.02\text{ A}$$