In the circuit given below, the voltage across the 2Ω resistor is ________(a) 3.41 V(b) -3.41 V(c) 3.8 V(d) -3.8 VI have been asked this question at a job interview.This is a very interesting question from Advanced Problems on Network Theory topic in portion Network Theory and Complex Circuit Diagram of Network Theory

In the circuit given below, the voltage across the 2Ω resistor is __

1.	In the circuit given below, the voltage across the 2Ω resistor is ________(a) 3.41 V(b) -3.41 V(c) 3.8 V(d) -3.8 VI have been asked this question at a job interview.This is a very interesting question from Advanced Problems on Network Theory topic in portion Network Theory and Complex Circuit Diagram of Network Theory
Answer» The correct answer is (b) -3.41 V The best explanation: APPLYING KCL to node A, \(\FRAC{V_A-10}{10} + \frac{V_A}{20} + \frac{V_A}{7}\) = 0 Or, VA (0.1 + 0.05 + 0.143) = 1 Or, VA = 3.41 V The voltage across the 2 Ω resistor due to 10 V source is V2 = \(\frac{V_A}{7} × 2\) = 0.97 V V2Ω due to 20 V source, \(\frac{V_A}{10} + \frac{V_A}{20} + \frac{V_A-20}{7}\) = 0 Or, 0.1 VA + 0.05VA + 0.143VA = 2.86 ∴ VA = \(\frac{2.86}{0.293}\) = 9.76 V V2Ω = \(\frac{V_A-20}{7}\) × 2 = -2.92 V The CURRENT in 2 Ω resistor = 2 × \(\frac{5}{5+8.67}\) = \(\frac{10}{13.67}\) = 0.73 A The voltage across the 2 Ω resistor = 0.73 × 2 = 1.46 V V2Ω = 0.97 – 2.92 -1.46 = -3.41 V.

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