Home › Network Theory › A one-port network has driving-point impedance Z(s)=(s^2+4s+3)/(s^2+2s+2). At omega=1 rad/s, what is the real part of Z(j omega)?

MCQ Practice Question

A one-port network has driving-point impedance Z(s)=(s^2+4s+3)/(s^2+2s+2). At omega=1 rad/s, what is the real part of Z(j omega)?

A1 ohm
B2 ohm
C3 ohm
D4 ohm

Show AnswerHide AnswerReport an error

Correct Answer: B. 2 ohm

Explanation: At s=j, the numerator is (j)^2+4j+3=2+4j and the denominator is (j)^2+2j+2=1+2j. Thus Z(j)=(2+4j)/(1+2j)=2 ohm, so its real part is 2 ohm.

Reference: Standard network theory: driving-point impedance evaluation

Continue practicing: Network Theory MCQs or browse all MCQs.