Question

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 A

The 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 1

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.