Question

Determine the admittance parallel combination of the last elements after replacing with (3s+4)/2s(s+4) is? (a) (4s^2-19s+4)/(6s-8) (b) (4s^2+19s-4)/(6s+8) (c) (4s^2+19s-4)/(6s-8) (d) (4s^2+19s+4)/(6s+8) This question was posed to me in an international level competition. Enquiry is from Series and Parallel Combination of Elements topic in division S-Domain Analysis 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) (4s^2+19s+4)/(6s+8) Easiest explanation: The term admittance is defined as the inverse of the term impedance. As the impedance is Z2(s) = 1/2s+1/(s+4)=(3s+4)/2s(s+4), the admittance parallel combination of the last elements is Y2(s) = 1/2+2s(s+4)/(3s+4)=(4s^2+19s+4)/(6s+8).