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 WI 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

In the figure given below, the power loss in 1 Ω resistor using Nort

1.	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 WI 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
Answer» 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.

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