Step 1: An npn transistor in common-emitter mode has these output readings at I_B = 40 μA: V_CE = 1 V → I_C = 3.9 mA; V_CE = 5 V → I_C = 4.1 mA; V_CE = 10 V → I_C = 4.2 mA. Find β and the output resistance. Step 2: β = I_C/I_B in the active region: at V_CE = 5 V, β = 4.1 mA/40 μA = 102.5. Step 3: Output resistance r_o = ΔV_CE/ΔI_C in the active region: between 5 V and 10 V, ΔV = 5 V, ΔI_C = 0.1 mA. Step 4: r_o = 5/0.0001 = 50,000 Ω = 50 kΩ — the curves are nearly flat, as expected. Step 5: Input characteristic check: V_BE ≈ 0.7 V for I_B = 40 μA, like a forward diode (base–emitter is a PN junction). Result: β ≈ 100; r_o ≈ 50 kΩ. The flat output curves show the transistor acts as a current source controlled by I_B.