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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

Answer»

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 isZ2(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).



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