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For the circuit shown in the figure below, the Norton Resistance looking into X-Y is __________(a) 2 Ω(b) \(\frac{2}{3}\)(c) \(\frac{5}{3}\)(d) 2 ΩI got this question in an international level competition.My enquiry is from Norton’s Theorem Involving Dependent and Independent Sources topic in division Useful Theorems in Circuit Analysis of Network Theory

Answer»

The correct CHOICE is (d) 2

The explanation is: \(R_N = \frac{V_{OC}}{I_{SC}}\)

VN = VOC

Applying KCL at node A, \(\frac{2I-V_N}{1} + 2 = I + \frac{V_N}{2}\)

Or, I = \(\frac{V_N}{1}\)

PUTTING, 2VN – VN + 2 = VN + \(\frac{V_N}{2}\)

Or, VN = 4 V.

∴ RN = 4/2 = 2Ω.



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