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 AThe question was asked during an interview for a job.This interesting question is from Norton’s Theorem Involving Dependent and Independent Sources topic in section Useful Theorems in Circuit Analysis 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 AThe question was asked during an interview for a job.This interesting question is from Norton’s Theorem Involving Dependent and Independent Sources topic in section Useful Theorems in Circuit Analysis of Network Theory
Answer» Correct option is (b) 6 Ω and 0.833 A To ELABORATE: We, draw the NORTON equivalent of the left side of xx’ and source TRANSFORMED right side of YY’. Vxx’ = VN = \(\displaystyle\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}\) = 5V ∴ RN = 8 \|\| (16 + 8) = \(\frac{8×24}{8+24}\) = 6 Ω ∴ \(I_N = \frac{V_N}{R_N} = \frac{5}{6}\) = 0.833 A.

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