For the circuit of Fig. 4.81. Find the Thevenin equivalent to the left of terminals ‘ab’. Find ν ( t )  and also find the complex power consumed by 0.8 H inductor.

Transforming to frequency domain

$\begin{array}{cc}{\displaystyle }& {\displaystyle 5\sqrt{2}\cos 5t5\mathrm{\angle }{0}^{\circ }V,\omega =5 \mathrm{r}\mathrm{a}\mathrm{d}\mathrm{/}\mathrm{s}}\\ {\displaystyle }& {\displaystyle 2\mathrm{\Omega }\to 2\mathrm{\Omega }}\\ {\displaystyle }& {\displaystyle \frac{\mathrm{}}{}}\end{array}$$0.8\mathrm{H}\to j5\times 0.8=+j4\mathrm{\Omega }$

The corresponding circuit is drawn in Fig. 4.81 (a).

Open circuit voltage, disconnect inductor $j4\mathrm{\Omega }$ as in Fig. 4.81 (b).

$V_{oC}$VoC​ is same as voltage across $2\mathrm{\Omega }$2Ω as $-j 1\mathrm{\Omega }$ does not carry any current. Therefore

$\begin{array}{cc}{\displaystyle V_{OC}=V_{TH}}& {\displaystyle =\left(\frac{2}{2+j5}\right)5\mathrm{\angle }{0}^{\circ }}\\ {\displaystyle }& {\displaystyle =1.857\mathrm{\angle }-68.2^{\circ }\mathrm{V}}\end{array}$

Thevenin impedance

Short circuit $5\mathrm{\angle }{0}^{\circ }\mathrm{V }$as in Fig. 4.81 (c)

Impedance seen from ‘ $ab$ab ‘

$\begin{array}{cc}{\displaystyle {\bar{Z}}_{TH}}& {\displaystyle =-j1+\frac{j5\times 2}{2+j5}\frac{\mathrm{}}{}}\\ {\displaystyle }& {\displaystyle =-j1+1.857\mathrm{\angle }21.8^{\circ }}\\ {\displaystyle }& {\displaystyle =1.724-j0.310\mathrm{\Omega }}\end{array}$

Complex power drawn by $0.8\mathrm{H}$

The Thevenin equivalent with load 0.8 is drawn in Fig. 4.81 (d). From the figure

Current, $\bar{I}=\frac{1.875\mathrm{\angle }-68.2^{\circ }}{1.724-j0.310+j4}$∘​

$\begin{array}{cc}{\displaystyle \bar{I}}& {\displaystyle =\frac{1.875\mathrm{\angle }-68.2^{\circ }}{5.737\mathrm{\angle }3.1^{\circ }}=0.327\mathrm{\angle }-71.3^{\circ }\mathrm{A}}\\ {\displaystyle \bar{V}}& {\displaystyle =\bar{I}\mathrm{.}j0.8=0.327\mathrm{\angle }-71.3^{\circ }\times j4}\\ {\displaystyle }& {\displaystyle =1.308\mathrm{\angle }18.7^{\circ }\mathrm{V}}\end{array}$

Then

$\nu \left(t\right)=1.308\sqrt{}2\cos (5+18.7^{\circ })v\mathrm{}$Complex power consumed by $0.8 \mathrm{H}$0.8 H

$\bar{S}=j(0.327{)}^2\times 4=0+j0.428 \mathrm{V}\mathrm{A}$

or

$S=0.428 VARs\; (inductive)\;$Note + sign 

Also by relationship

$\begin{array}{cc}{\displaystyle \bar{S}}& {\displaystyle =\bar{V}{\bar{I}}^{\ast }=1.308\mathrm{\angle }18.7^{\circ }\times 0.327\mathrm{\angle }71.3^{\circ }}\\ {\displaystyle }& {\displaystyle =0.482\mathrm{\angle }90^{\circ }=0+j0.482 \mathrm{V}\mathrm{A}}\end{array}$