Question

A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’ is ___________ (a) 6 Ω and 1.333 A (b) 6 Ω and 0.833 A (c) 32 Ω and 0.156 A (d) 32 Ω and 0.25 A The question was asked during an interview for a job. This interesting question is from Norton’s Theorem Involving Dependent and Independent Sources topic in section Useful Theorems in Circuit 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 option is (b) 6 Ω and 0.833 A To elaborate: We, draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’. Vxx’ = VN = \(\displaystyle\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}\) = 5V ∴ RN = 8 || (16 + 8) = \(\frac{8×24}{8+24}\) = 6 Ω ∴ \(I_N = \frac{V_N}{R_N} = \frac{5}{6}\) = 0.833 A.