Question

In the circuit given below, the value of V in terms of VS and IS using Superposition Theorem is ___________ (a) V = \frac{2}{7} I_S – \frac{4}{7} V_S (b) V = \frac{2}{7} V_S + \frac{4}{7} I_S (c) V = \frac{2}{7} V_S – \frac{4}{7} I_S (d) V = \frac{2}{7} I_S – \frac{4}{7} V_S I have been asked this question during an online interview. This interesting question is from Advanced Problems on Superposition Theorem topic in division 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

The correct choice is (c) V = \frac{2}{7} V_S – \frac{4}{7} I_S Explanation: By superposition, the current I is given by I = \frac{V_S}{5} − \frac{2}{5} × I_S − \frac{3}{5} × 3i This can be solved for I to obtain, I = \frac{V_S}{14} – \frac{I_S}{7} Now, by Superposition Theorem, V = \frac{V_S}{5} − \frac{2}{5} × I_S − \frac{3}{5} × 3i Or, V = \frac{V_S}{5} − \frac{2}{5} × I_S − \frac{3}{5} × 3(\frac{V_S}{14} – \frac{I_S}{7}) ∴ V = \frac{2}{7} V_S – \frac{4}{7} I_S.