The correct choice is (c) z22 = -1.775 – j5.739 Ω
Easiest explanation: z1 = \(\frac{12(j10)}{12+j10-j5} = \frac{j120}{12+j5}\)
z2 = \(\frac{j60}{12+j5}\)
z3 = \(\frac{50}{12+j5}\)
z12 = z21 = z2 = \(\frac{(-j60)(12-j5)}{144+25}\) = -1.775 – j4.26
z11 = z1 + z12 = \(\frac{(j120)(12-j5)}{144+25}\) + z12 = 1.775 + j4.26
z22 = z3 + z21 = \(\frac{(50)(12-j5)}{144+25}\) + z21 = 1.7758 – j5.739
∴ [z] = [1.775 + j4.26; -1.775 – j4.26; -1.775 – j4.26; 1.775 – j5.739] Ω.
