Question

If anyone knows this answer please reply me​

Answer 1

The $\sf{j5\ \Omega}$ inductor of and the rightmost $\sf{-j2\ \Omega}$ capacitor are in series, so their equivalent impedance,

$\sf{=j5\ \Omega-j2\ \Omega}$

$\sf{=j3\,\Omega}$

This combination is in parallel with the other $\sf{-j2\ \Omega}$ capacitor. So their equivalent impedance,

$\sf{=\dfrac{j3\ \Omega\times-j2\ \Omega}{j3\ \Omega-j2\ \Omega}}$

$\sf{=\dfrac{6}{j}\ \Omega}$

$\sf{=\dfrac{j6}{-1}\ \Omega\quad\left[\because\ j^2=-1\right]}$

$\sf{=-j6\ \Omega}$

This combination is in series with the $\sf{4\ \Omega}$ resistor.

So the equivalent impedance of the circuit is,

$\sf{\longrightarrow\underline{\underline{Z=(4-j6)\ \Omega}}}$

$\sf{\longrightarrow Z=\sqrt{4^2+(-6)^2}\angle\tan^{-1}\left(\dfrac{-6}{4}\right)\ \Omega}$

$\sf{\longrightarrow Z=2\sqrt{13}\angle-\tan^{-1}\left(\dfrac{3}{2}\right)\ \Omega}$

$\sf{\longrightarrow\underline{\underline{Z=7.211\angle-56.31^o\ \Omega}}}$

And the power factor,

$\sf{\longrightarrow PF=\cos\left(-\tan^{-1}\left(\dfrac{3}{2}\right)\right)}$

$\sf{\longrightarrow PF=\dfrac{2}{2\sqrt{13}}}$

$\sf{\longrightarrow\underline{\underline{PF=0.277}}}$