Figure shows a circuit in which three identical diodes are used. Each diode has forward resistance of 20$$\Omega$$ and infinite backward resistance. Resistors $$R_1 = R_2 = R_3 = 50\Omega$$. Battery voltage is 6 V. The current through $$R_3$$ is :

Since $$D_3$$ has infinite backward resistance, the entire middle branch acts as an open circuit and no current flows through it.

$$R_{\text{total}} = R_3 + r_{D1} + R_1$$

$$R_{\text{total}} = 50\ \Omega + 20\ \Omega + 50\ \Omega = 120\ \Omega$$

The current $$I$$ through $$R_3$$ is given by:

$$I = \frac{V}{R_{\text{total}}}$$

$$I = \frac{6}{120}\text{ A}$$

$$I = 50\text{ mA}$$

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