A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’ is ___________(a) 6 Ω and 1.333 A(b) 6 Ω and 0.833 A(c) 32 Ω and 0.156 A(d) 32 Ω and 0.25 AI had been asked this question during an online interview.Question is from Advanced Problems Involving Complex Circuit Diagram topic in chapter Network Theory and Complex Circuit Diagram of Network Theory

A circuit is given in the figure below. The Norton equivalent as viewe

1.	A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’ is ___________(a) 6 Ω and 1.333 A(b) 6 Ω and 0.833 A(c) 32 Ω and 0.156 A(d) 32 Ω and 0.25 AI had been asked this question during an online interview.Question is from Advanced Problems Involving Complex Circuit Diagram topic in chapter Network Theory and Complex Circuit Diagram of Network Theory
Answer» The correct choice is (b) 6 Ω and 0.833 A The explanation is: We, draw the Norton equivalent of the left SIDE of xx’ and SOURCE transformed right side of yy’. Vxx’ = VN = \(\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}\) = 5V ∴ RN = 8 \|\| (16 + 8) = \(\frac{8×24}{8+24}\) = 6 Ω ∴ IN = \(\frac{V_N}{R_N}= \frac{5}{6}\) = 0.833 A.

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