

InterviewSolution
Saved Bookmarks
1. |
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)/3I had been asked this question in an internship interview.This question is from Nodal Analysis in chapter Steady State AC Analysis of Network Theory |
Answer» 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. |
|