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

The Norton equivalent impedance Z between the nodes P and Q in the fol

1.	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» 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.

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