Question

In the circuit shown below we get a nodal equation as (1/3+1/j4-1/j6)Va—(-1/j6)Vb=x. Find the value of ‘x^‘‘. (a) (5∠0^o)/3 (b) – (5∠0^o)/3 (c) (10∠0^o)/3 (d) – (10∠0^o)/3 I had been asked this question in an internship interview. This question is from Nodal Analysis in chapter Steady State AC 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

Correct choice is (c) (10∠0^o)/3 For explanation: The general equations are YaaVa+YabVb = I1, YbaVa+YbbVb = I2. We get Yaa=1/3+1/j4+1/(-j6) and the self admittance at node a is the sum of admittances connected to node a. Yab=-(1/(-j6)). I1 = (10∠0^o)/3=x.