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For the circuit given below, the value of the z12 parameter is ___________(a) z12 = 1 Ω(b) z12 = 4 Ω(c) z12 = 1.667 Ω(d) z12 = 2.33 ΩThe question was posed to me in an online quiz.My query is from Advanced Problems on Two Port Network in chapter Two-Port Networks of Network Theory |
Answer» RIGHT CHOICE is (a) z12 = 1 Ω To explain I would say: z11 = \(\FRAC{V_1}{I_1}\)= 1 + 6 || (4+2) = 4Ω I0 =\(\frac{1}{2} I_1 \) V2 = 2I0 = I1 z21 = \(\frac{V_2}{I_1}\) = 1Ω z22 = \(\frac{V_2}{I_2}\)= 2 || (4+6) = 1.667Ω So, I’0 = \(\frac{2}{2+10}I_2 = \frac{1}{6}I_2 \) V1 = 6I’0 = I2 z12 = \(\frac{V_1}{I_2}\)= 1Ω Hence, [z] = [4:1; 1:1.667] Ω. |
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