Question

Initial voltage on capacitor VO as marked |VO| = 5 V, VS = 8 u (t), where u (t) is the unit step. The voltage marked V at t=0^+ is _____________ (a) 1 V (b) -1 V (c) \frac{13}{3} V (d) –\frac{13}{3} V This question was posed to me during an online exam. My question comes from Dot Convention in Magnetically Coupled Circuits in chapter Resonance & Magnetically Coupled Circuit 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 (c) \frac{13}{3} V Easiest explanation: Applying voltage divider method, we get, I = \frac{V}{R_{EQ}} = \frac{8}{1+1||1} = \frac{8}{1+\frac{1}{2}} = \frac{16}{3} A I1 = \frac{16}{3} × \frac{1}{2} = \frac{8}{3} A And I’_2 = \frac{V}{R_{EQ}} = \frac{5}{1+1||1} = \frac{5}{1+\frac{1}{2}} = \frac{10}{3} A Now, I’_1 = I’_2 × \frac{1}{1+1} = \frac{10}{3} × \frac{1}{2} = \frac{5}{3} A Hence, the net current in 1Ω resistance = I1 + I’_1 = \frac{8}{3} + \frac{5}{3} = \frac{13}{3} A ∴ Voltage drop across 1Ω = \frac{13}{3} × 1 = \frac{13}{3} V.