With reference to the network shown in Fig.

10.19, find the input impedance Zin that would be measured between terminals: (a) a and g; (b) b and g; (c) a and b

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Step 1The circuit diagram is,\omega=1000 \mathrm{rad} / \mathrm{s}Determining the impedance of the element,\begin{aligned} Z_{\text {5mH }} & =j \omega L \\ & =j(1000)\left(5 \times 10^{-3}\right) \\ & =j 5 \Omega \\ Z_{20 \mathrm{mH}} & =j \omega L \\ & =j(1000)\left(20 \times 10^{-3}\right) \\ & =j 20 \Omega\end{aligned}\begin{aligned} Z_{200 \mu F} & =-\frac{j}{\omega C} \\ & =-\frac{j}{1000 \times 200 \times 10^{-6}} \\ & =-j 5 \Omega \\ Z_{100 \mu F} & =-\frac{j}{\omega C} \\ & =-\frac{j}{1000 \times 100 \times 10^{-6}} \\ & =-j 10 \Omega\end{aligned}Step 2The equivalent circuit diagram is,(a).Determining the input impedance across a and g terminals,\begin{aligned} Z_{\text {in }} & =(j 5)\|(10)\|[(-j 5)+[(-j 10) \|(j 20)]] \\ & =(j 5)\|(10)\|\left[(-j 5)+\frac{(-j 10)(j 20)}{(-j 10)+(j 20)}\right] \\ & =(j 5)\|(10)\|[(-j 5)+(-j 20)] \\ & =(j 5)\|(10)\|(-j 25) \\ & =\left[\frac{(j 5)(10)}{(j 5)+(10)}\right] \|(-j 25) \\ & =\left(\frac{50 \angle 90^{\circ}}{11.18 \angle 26.56^{\circ}}\right) \|(-j 25) \\ & =\left(4.47 \angle 63.44^{\circ}\right) \|(-j 25) \\ & =(2+j 4) \|(-j 25)\end{aligned}          \begin{array}{l}=\frac{(2+j 4)(-j 25)}{(2+j 4)+(-j 25)} \\ =\frac{-j 50+100}{2-j 21} \\ =\frac{111.8 \angle-26.56^{\circ}}{21.095 \angle-84.56} \\ =5.3 \angle 58^{\circ} \\ =2.81+j 4.49 \Omega\end{array}Step 3(b).Determining the input impedance across terminals b and g,\begin{aligned} Z_{\text {in }} & =[((j 5) \|(10))+(-j 5)]\|(-j 10)\|(j 20) \\ & =\left[\frac{50 \angle 90^{\circ}}{11.18 \angle 26.56}+(-j 5)\right] \|\left[\frac{200}{j 10}\right] \\ & =\left[\left(4.47 \angle 63.44^{\circ}\right)-j 5\right] \|(-j 20) \\ & =(2+4 j-5 j) \|(-j 20) \\ & =\frac{(2-j)(-j 20)}{2-j-j 20} \\ & =\frac{-j 40-20}{2-j 21}\end{aligned}        \begin{array}{l}=\frac{44.72 \angle 243.435^{\circ}}{21.095 \angle-84.56^{\circ}} \\ =2.12 \angle-32^{\circ} \\ =1.798-1.123 j \Omega\end{array}Step 4(c).Determining the input impedance across terminals a and b,j 5 \Omega and 10 \Omega are connected in parallel,\begin{aligned}Z_{1} & =\frac{(j 5)(10)}{j 5+10} \\& =2+j 4 \Omega\end{aligned}-j 10 \Omega and j 20 \Omega are connected in parallel,\begin{aligned}Z_{2} & =\frac{(-j 10)(j 20)}{j 20-j 10} \\& =\frac{200}{j 10} \\& =-j 20 \Omega\end{aligned}Step 5The modified circuit is,\begin{aligned} Z_{i n} & =(2+j 4-j 20) \|(-j 5) \\ & =(2-j 16) \|(-j 5) \\ & =\frac{(2-j 16)(-j 5)}{2-j 16-j 5} \\ & =\frac{-j 10-80}{2-j 21} \\ & =\frac{80.62 \angle 187.125^{\circ}}{21.095 \angle-84.56^{\circ}} \\ & =3.82 \angle-88.315^{\circ} \\ & =0.1123-j 3.82 \Omega\end{aligned} ...