Question

In the figure given below, the power loss in 1 Ω resistor using Norton’s Theorem is ________ (a) 9.76 W (b) 9.26 W (c) 10.76 W (d) 11.70 W I have been asked this question during an online exam. I need to ask this question from Norton’s Theorem Involving Dependent and Independent Sources 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

Correct option is (b) 9.26 W Easy explanation: Let us remove the 1 Ω resistor and short x-y. At Node 1, assuming node potential to be V, \(\frac{V-10}{5}\) + ISC = 5 But ISC = \(\frac{V}{2}\) ∴ \(\frac{V-10}{5} + \frac{V}{2}\) = 5 Or, 0.7 V = 7 That is, V= 10 V ∴ ISC = \(\frac{V}{2}\) = 5 A To find Rint, all constant sources are deactivated. Rint = \(\frac{(5+2)×2}{5+2+2} = \frac{14}{9}\) = 1.56 Ω Rint = 1.56 Ω; ISC = IN = 5A Here, I = IN \(\frac{R_{int}}{R_{int}+1} = 5 × \frac{1.56}{1.56+1}\) = 3.04 A ∴ Power loss = (3.04)^2 × 1 = 9.26 W.