Question

For the circuit given in figure below, the Norton equivalent as viewed from terminals y and y’ is _________

(a) 32 Ω and 0.25 A

(b) 32 Ω and 0.125 A

(c) 6 Ω and 0.833 A

(d) 6 Ω and 1.167 A

This question was posed to me during an online interview.

My question is based upon Norton’s Theorem Involving Dependent and Independent Sources topic in portion Useful Theorems in Circuit Analysis of Network Theory

Answer 1

The correct option is (d) 6 Ω and 1.167 A

For explanation: We draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’.

Norton equivalent as seen from terminal yy’ is

Vyy’ = VN =\(\displaystyle\frac{\frac{4}{24} + \frac{8}{8}}{\frac{1}{24} + \frac{1}{8}}\) = 5V

= \(\frac{0.167+1}{0.04167+0.125}\) = 7 V

∴ RN = (8 + 16) || 8

= \(\frac{24×8}{24+8}\) = 6 Ω

∴ IN = \(\frac{V_N}{R_N} = \frac{7}{6}\) = 1.167 A.