Question

For a T-network if the Short circuit admittance parameters are given as y11, y21, y12, y22, then y11 in terms of Hybrid parameters can be expressed as ________ (a) y11 = \left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right) (b) y11 = \frac{h_{21}}{h_{11}} (c) y11 = –\frac{h_{12}}{h_{11}} (d) y11 = \frac{1}{h_{11}} I have been asked this question during an online exam. My query is from Hybrid (h) Parameter topic in section Two-Port Networks of Network Theory Select the correct answer from above options network theory questions and answers, network theory questions pdf, network theory question bank, network theory gate questions and answers pdf, mcq on network theory pdf, gate network theory questions and solutions, network theory mcq Test , ies network theory questions, Network theory Questions for GATE EC Exam, Network Theory MCQ (Multiple Choice Questions)

Answer 1

The correct answer is (d) y11 = \frac{1}{h_{11}} Easy explanation: We know that the short circuit admittance parameters can be expressed in terms of voltages and currents as, I1 = y11 V1 + y12 V2 ……… (1) I2 = y21 V1 + y22 V2 ………. (2) And the Hybrid parameters can be expressed in terms of voltages and currents as, V1 = h11 I1 + h12 V2 ………. (3) I2 = h21 I1 + h22 V2 ……….. (4) Now, (3) and (4) can be rewritten as, I1 = \frac{V_1}{h_{11}} – \frac{h_{12} V_2}{h_{11}} ………. (5) And I2 = \frac{h_{21} V_1}{h_{11}} + \left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right) V_2 ………. (6) ∴ Comparing (1), (2) and (5), (6), we get, y11 = \frac{1}{h_{11}} y12 = –\frac{h_{12}}{h_{11}} y21 = \frac{h_{21}}{h_{11}} y22 = \left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right) .