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For the network shown, determine z₁ , z₀ and A_v

z_i = 1.067 Ω; z₀ = 3 kΩ; A_v = 280

z_i = 1.067 Ω; z₀ = 3 kΩ; A_v = - 280.37

z_i = 937 Ω; z₀ = 3 Ω; A_v = 281

z_i = 937 Ω; z₀ = 33 kΩ; A_v = - 280.37

Answer : Option B

Explanation :

= 24.04 μA

I_E = (1 + β)I_B = 2.428 mA;

= 10.7 Ω

(i) z_i = R_B||Br_e =

= 0.937 K = 1.067 K

(ii) Z₀ = R_C = 3 kΩ

(iii) = -280.37

∴ Z_i = 1.067 KΩ;

Z₀ = 3 KΩ;

A_v = -280.37 .

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