# 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.

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Transforming to frequency domain

$\begin{array}{cc}{\displaystyle}& {\displaystyle 5\sqrt{2}\mathrm{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 $-j1\mathrm{\Omega}$ 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 ‘

$\begin{array}{cc}{\displaystyle {\stackrel{\u02c9}{Z}}_{TH}}& {\displaystyle =-j1+\frac{j5\times 2}{2+j5}\frac{\mathrm{}}{}}\\ {\displaystyle}& {\displaystyle =-j1+1.857\mathrm{\angle}21.{8}^{\circ}}\\ {\displaystyle}& {\displaystyle =1.724-j0.310\mathrm{\Omega}}\end{array}$

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, $\stackrel{\u02c9}{I}=\frac{1.875\mathrm{\angle}-68.{2}^{\circ}}{1.724-j0.310+j4}$∘

$\begin{array}{cc}{\displaystyle \stackrel{\u02c9}{I}}& {\displaystyle =\frac{1.875\mathrm{\angle}-68.{2}^{\circ}}{5.737\mathrm{\angle}3.{1}^{\circ}}=0.327\mathrm{\angle}-71.{3}^{\circ}\mathrm{A}}\\ {\displaystyle \stackrel{\u02c9}{V}}& {\displaystyle =\stackrel{\u02c9}{I}\mathrm{.}j0.8=0.327\mathrm{\angle}-71.{3}^{\circ}\times j4}\\ {\displaystyle}& {\displaystyle =1.308\mathrm{\angle}18.{7}^{\circ}\mathrm{V}}\end{array}$

Then

$\nu \left(t\right)=1.308\sqrt{}2\mathrm{cos}(5+18.{7}^{\circ})v\mathrm{}$Complex power consumed by $0.8\mathrm{H}$0.8 H

$\stackrel{\u02c9}{S}=j(0.327{)}^{2}\times 4=0+j0.428\mathrm{V}\mathrm{A}$

or

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

Also by relationship

$\begin{array}{cc}{\displaystyle \stackrel{\u02c9}{S}}& {\displaystyle =\stackrel{\u02c9}{V}{\stackrel{\u02c9}{I}}^{\ast}=1.308\mathrm{\angle}18.{7}^{\circ}\times 0.327\mathrm{\angle}71.{3}^{\circ}}\\ {\displaystyle}& {\displaystyle =0.482\mathrm{\angle}9{0}^{\circ}=0+j0.482\mathrm{V}\mathrm{A}}\end{array}$