The correct option is (c) z22 = 30 Ω
The explanation: z11 = \(\frac{V_1}{I_1} = \frac{(20+5)I_1}{I_1}\) = 25Ω
V0 = \(\frac{20}{25}\)V1 = 20 I1
-V0 – 4I2 + V2 = 0
Or, V2 = V0 + 4I1 = 20I1 + 4I1 = 24 I1
Or, z21 = \(\frac{V_2}{I_1}\) = 24 Ω
V2 = (10+20) I2 = 30 I2
Or, z22 = \(\frac{V_2}{I_1}\) = 30 Ω
V1 = 20I2
Or, z12 = \(\frac{V_1}{I_2}\) = 20 Ω
∴ [z] = [25:20; 24:30] Ω.