Circuit. A single series loop: an 8 V d.c. source, a forward-biased silicon diode, and a 2.2 kΩ resistor. Because it is a series loop, the same current I flows through all three elements.

Model the diode. Use the piecewise-linear (constant-drop) model: once a conducting silicon diode is forward biased, its terminal voltage stays nearly fixed at the cut-in value

\(V_D \approx 0.7\ \text{V}\ (\text{Si}),\qquad V_D \approx 0.3\ \text{V}\ (\text{Ge})\)

Check the state first: the 8 V source pushes current into the anode, and 8 V is far above 0.7 V, so the diode is indeed ON and the model applies.

Apply KVL around the loop.

\(V_S = V_D + I R\)

Solve for the current.

\(I = \dfrac{V_S - V_D}{R}\)

Substitute the values \(V_S = 8\ \text{V}\), \(V_D = 0.7\ \text{V}\), \(R = 2.2\ \text{k}\Omega = 2200\ \Omega\):

\(I = \dfrac{8 - 0.7}{2200} = \dfrac{7.3}{2200}\)

\(I = 3.318 \times 10^{-3}\ \text{A} = 3.32\ \text{mA}\)

Where the other numbers come from — this is the useful lesson of the question. Each distractor corresponds to a different assumed diode drop:

Ideal diode (V_D = 0): \(8/2200 = 3.63\ \text{mA}\). Germanium diode (V_D = 0.3 V): \(7.7/2200 = 3.5\ \text{mA}\). A 1 V drop: \(7/2200 = 3.18\ \text{mA}\). Only the correct silicon value of 0.7 V gives 3.32 mA.

Takeaway. In any series diode–resistor problem, first confirm the diode is forward biased, then subtract its constant drop from the source voltage before dividing by the resistance; the material (Si or Ge) fixes which drop you subtract.

Hence, the current flowing in the circuit is 3.32 mA.