Question

A periodic voltage v (t) = 1 + 4 sin ωt + 2 cos ωt is applied across a 1Ω resistance. The power dissipated is ____________ (a) 1 W (b) 11 W (c) 21 W (d) 24.5 W The question was posed to me in a job interview. The question is from Relation between Hybrid Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters in section Two-Port Networks 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

Right option is (b) 11 W For explanation I would say: Given that, v (t) = 1 + 4 sin ωt + 2 cos ωt So, Power is given by, Power, P = \frac{1^2}{1} + \frac{\frac{4^2}{\sqrt{2}}}{1} + \frac{\frac{2^2}{\sqrt{2}}}{1} = 11 W.