Model each conducting diode as a constant 0.7 V drop (both are marked 0.7 V in the figure), then apply KVL and KCL.

Step 1 — the source loop through both diodes. Going round the 20 V source, D₁, D₂ and the 5.6 kΩ resistor:

\(20 = V_{D1} + V_{D2} + I_2 (5.6\ \text{k}\Omega)\)

\(20 = 0.7 + 0.7 + I_2(5600) \Rightarrow I_2 = \dfrac{18.6}{5600}\)

\(I_2 = 3.32\ \text{mA}\)

This I₂ is the total current the source pushes into the junction node.

Step 2 — the current in the 3.3 kΩ branch. The 3.3 kΩ resistor is connected across the same two nodes as D₂, so it sees the diode's forward drop:

\(I_1 = \dfrac{V_{D2}}{3.3\ \text{k}\Omega} = \dfrac{0.7}{3300} = 0.212\ \text{mA}\)

Step 3 — apply KCL at the junction. The incoming current I₂ splits between the resistor branch and D₂:

\(I_{D2} = I_2 - I_1 = 3.32 - 0.21\)

\(I_{D2} = 3.11\ \text{mA}\)

Why the method works. Because a conducting silicon diode clamps its terminals to about 0.7 V regardless of current, D₂ fixes the node voltage; everything else then follows from Ohm's law and current division. Note that the 3.3 kΩ branch takes only 0.21 mA — a diode in parallel with a resistor carries the lion's share of the current, which is exactly how clamping and protection circuits work.

Check. 3.11 mA + 0.21 mA = 3.32 mA = I₂, so KCL balances.

Hence, the current through D₂ is 3.11 mA.