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

In the circuit given below, the value of V in terms of VS and IS using

1.	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
Answer» 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 = \(\frac{2}{7} V_S – \frac{4}{7} I_S\).

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