A circuit is arranged as shown in figure. The output voltage $$V_0$$ is equal to _________ V.

$$\text{Anodes of } D_1 \text{ and } D_2 \text{ are directly tied to ground } (0\text{ V}).$$

$$\text{Since cathodes connect via } R \text{ to } +5\text{ V, both diodes are reverse biased} \implies I_{D1} = I_{D2} = 0$$

$$V_B = 0\text{ V} \implies V_{BE} = 0\text{ V} < 0.7\text{ V}$$

$$\text{Transistor is in cut-off} \implies I_C = 0$$

$$V_o = 5 - I_C \cdot R = 5 - 0 = 5\text{ V}$$