Question

The Norton equivalent impedance Z between the nodes P and Q in the following circuit is __________

(a) 1

(b) 1 + s + \(\frac{1}{s}\)

(c) 2 + s + \(\frac{1}{s}\)

(d) 3 + s + \(\frac{1}{s}\)

I had been asked this question at a job interview.

Question is taken from Norton’s Theorem Involving Dependent and Independent Sources in section Useful Theorems in Circuit Analysis of Network Theory

Answer 1

The correct answer is (a) 1

The explanation is: To calculate the Norton resistance, all the current sources get open-circuited and voltage sources get short-circuited.

∴ RN = (\(\frac{1}{s}\)  + 1) || (1+s)

= \(\frac{\left(\frac{1}{s} + 1\right)×(1+s)}{\left(\frac{1}{s} + 1\right)+(1+s)}\)

= \(\frac{\frac{1}{s}+1+1+s}{\frac{1}{s}+1+1+s}\) = 1

So, RN = 1.