Question

In the circuit shown in the figure, the angular frequancy $\omega$ (in rad/s), at which the Norton equivalent impendance as seen from terminals b-b' is purely resistive, is

• A. 2

Answer 1

The correct option is A 2

$Finding{Z}_N$ :

$Z_{bb}^{{}^{\prime }}=\frac{1\times j0.5\omega }{1+j0.5\omega }+\frac{1}{j\omega }=\frac{j\omega }{2+j\omega }+\frac{1}{j\omega }$

$or{Z}_{bb}^{{}^{\prime }}=\frac{2-{\omega }^2+j\omega }{2j\omega -{\omega }^2}$

Rationalizing equation (i), we get,

$Z_{bb}^{{}^{\prime }}=\frac{(2-{\omega }^2)+j\omega }{2j\omega -{\omega }^2}\times \frac{-{\omega }^2-j2\omega }{-{\omega }^2-j2\omega }$

$=\frac{2{\omega }^2+{\omega }^4+2{\omega }^2}{{\omega }^4+2{\omega }^2}+\frac{({\omega }^3-4\omega )}{{\omega }^4+2{\omega }^2}$

In order to have a purely resistive impendance $Z_{bb}^{{}^{\prime }}$, the imaginary part of equation (ii) will be equaled to zero.

$\therefore \frac{-4\omega +{\omega }^3}{{\omega }^4+2{\omega }^2}=0$

$or{\omega }^3=4\omega$

$or\omega =\sqrt{4}=2rad/sec$