Calculate the following when D2 is shorted-out: (i) Current through the resistor R, (ii) Voltage drop across diode D1, (iii) Voltage drop across resistor R, (iv) Current through resistor R.

Since the diodes are identical and $V_{D1} = 0.8 \, \text{V}$, the voltage drop across D2 is also $0.8 \, \text{V}$.

$$V_{diodes} = V_{D1} + V_{D2}$$ $$V_{diodes} = 0.8 \, \text{V} + 0.8 \, \text{V}$$ $$\boxed{\text{(i) } V_{diodes} = \textbf{1.6 V}}$$

Using Kirchhoff's Voltage Law (KVL) for the series circuit:

$$V_{supply} = V_{diodes} + V_R$$ $$5 \, \text{V} = 1.6 \, \text{V} + V_R$$ $$V_R = 5 \, \text{V} - 1.6 \, \text{V}$$ $$\boxed{\text{(ii) } V_R = \textbf{3.4 V}}$$

Using KVL for the new series circuit:

$$V_{supply} = V_{D1} + V_{D2} + V_R$$ $$5 \, \text{V} = 0.8 \, \text{V} + 0 \, \text{V} + V_R$$ $$V_R = 5 \, \text{V} - 0.8 \, \text{V}$$ $$\boxed{\text{(iii) } V_R = \textbf{4.2 V}}$$

Using Ohm's Law:

$$I_R = \frac{V_R}{R}$$ $$I_R = \frac{4.2 \, \text{V}}{1000 \, \Omega}$$ $$I_R = 0.0042 \, \text{A}$$ $$\boxed{\text{(i) } I_R = \textbf{4.2 mA}}$$ $$\boxed{\text{(iv) } I_R = \textbf{4.2 mA}}$$

The voltage drop across D1 remains the same as it is still forward-biased.

$$\boxed{\text{(ii) } V_{D1} = \textbf{0.8 V}}$$