Small-Signal Input Resistance Seen by the Source: For the given common-emitter amplifier (β = 100, gm = 0.3861 A/V, rπ = 259 Ω, RB = 93 kΩ, RC = 250 Ω, RL = 1 kΩ, Rs = 1 kΩ; coupling capacitor C1 is effectively short and bypass capacitor C2 = 4.7 mF is a short at signal), compute the resistance seen by Vs.

Introduction / Context:
In small-signal BJT amplifier design, the source “sees” the Thevenin input resistance formed by the bias network and the transistor's input resistance (rπ). Correctly evaluating this resistance is essential for matching, signal attenuation estimates, and gain calculations.

Given Data / Assumptions:

• Common-emitter stage, emitter is AC-grounded by a large bypass capacitor (C2 ≈ short at signal).
• Coupling capacitor C1 is large (short at signal), so it does not add reactance.
• Transistor parameters: β = 100, gm = 0.3861 A/V, rπ = 259 Ω (consistent with rπ ≈ β/gm).
• Bias resistor to base: RB = 93 kΩ.
• Source resistance: Rs = 1 kΩ (series with Vs).
• We ignore ro (given r0 = ∞).

Concept / Approach:

The small-signal resistance seen by Vs equals Rs in series with the amplifier's input resistance Rin. With the emitter AC-grounded, Rin at the base is RB ∥ rπ (looking into the base). The collector network does not reflect to the input significantly in this idealized case (no feedback through ro).

Step-by-Step Solution:

Verification / Alternative check:

Check consistency: If C2 were open (no bypass), emitter degeneration would increase input resistance; here it is bypassed, so Rin is dominated by rπ, as used.

Why Other Options Are Wrong:

• 258 Ω: Omits Rs; this is only the approximate RB ∥ rπ.
• 93 kΩ: Ignores the strong shunting effect of rπ.
• ∞: Would require the input to be open-circuit, which is not the case.
• 1518 Ω: Not supported by the parallel/series combination given the numbers.

Common Pitfalls:

• Forgetting to include source resistance Rs.
• Using RB alone instead of RB ∥ rπ for the base input.

Final Answer:

1258 Ω