Question 2.Q.63: In the circuit of Fig. 1, compute I x . Demonstrate the r......
In the circuit of Fig. 1, compute \overline{I}_ x . Demonstrate the reciprocity theorem by interchanging the positions of the source and \overline{I}_ x.
Case i : To solve \overline{I}_ x in the given circuit Let us assume three mesh currents \overline{I}_{1},\overline{I}_{2} and \overline{I}_{3} as shown in Fig. 2. Now, the response, \overline{I}_{x}=\overline{I}_{3}.
With reference to Fig. 2, the mesh basis matrix equation is,
\begin{bmatrix} {8 + j4}&{{-j4}}&0\\ {{-j4}}&{{8 + j4 − j8}}&-(-j8) \\ 0 &-(-j8)& 3-j8\end{bmatrix} \begin{bmatrix}\overline{I} _{1} \\ \overline{I}_{2} \\ \overline{I}_{3}\end{bmatrix} =\begin{bmatrix} 50 \angle 30^\circ \\ 0 \\ 0\end{bmatrix}
Here,50\angle30^{\circ}=50\cos30^{\circ}+j50\sin30^{\circ}=43.3013+j25V
\begin{bmatrix} {8 + j4}&{{-j4}}&0\\ {{-j4}}&{{8 – j4 }}& j8 \\ 0 & j8 & 3-j8\end{bmatrix} \begin{bmatrix}\overline{I} _{1} \\ \overline{I}_{2} \\ \overline{I}_{3}\end{bmatrix} =\begin{bmatrix} 43.3013+j25 \\ 0 \\ 0\end{bmatrix} ………….(1)
\Delta=\begin{vmatrix}{{8 + j4}}&{{-j4}} & 0\\ {{-j4}}&8 – j4&j8 \\0&j8&3-j8 \end{vmatrix} =(8+\mathrm{j}4)\times\bigl[(8-\mathrm{j}4)\times(3-\mathrm{j}8)-(\mathrm{j}8)^{2}\bigr]-(-\mathrm{j}4)\times\bigl[-\mathrm{j}4\times(3-\mathrm{j}8)-0\bigr]+0=752-j384+48-j128=800-j512
\Delta_3=\begin{vmatrix}{{8 + j4}}&{{-j4}} & 43.3013+j25\\ {{-j4}}&8 – j4&0 \\0 & j8 & 0 \end{vmatrix} =0-0+(43.3013+j25)\times\lbrack-\mathrm{j}4\times\mathrm{j}8-0\rbrack=1385.6416 + j 800
∴ The response {\bar{I}}_{x}\;=\;{\bar{I}}_{3}\;=\;{\frac{\Delta_{3}}{\Delta}}=\frac{1385.6416+j800}{800-j512}
=\;0.7747\;+j1.4958\;A=\;1.6845∠62.6^{\circ}\;ACase ii : To demonstrate the reciprocity theorem by interchanging the positions of source and response
Let us interchange the position of source and response as shown in Fig. 3. Let us assume mesh currents as shown in Fig. 3.
Now, the response, {\overline{I}}_{{x}}\;=\;{{{\overline{I}}}}_{a}\;.
With reference to Fig. 3, the mesh basis matrix equation is,
\begin{bmatrix} {8 + j4}&{{-j4}}&0\\ {{-j4}}&{{8 + j4 − j8}}&-(-j8) \\ 0 &-(-j8)& 3-j8\end{bmatrix} \begin{bmatrix}\overline{I} _{a} \\ \overline{I}_{b} \\ \overline{I}_{c}\end{bmatrix} =\begin{bmatrix} 0\\ 0 \\ 50 \angle 30^\circ \end{bmatrix}
\begin{bmatrix} {8 + j4}&{{-j4}}&0\\ {{-j4}}&{{8 – j4}}&j8 \\ 0 &j8& 3-j8\end{bmatrix} \begin{bmatrix}\overline{I} _{a} \\ \overline{I}_{b} \\ \overline{I}_{c}\end{bmatrix} =\begin{bmatrix} 0\\ 0 \\ 43. 3013+j25 \end{bmatrix} ……………(2)
On comparing equations (1) and (2), we can say that the value of ∆ remains the same in both the cases.
∴ ∆ = 800 − j512
\Delta_a=\begin{vmatrix}{0}&{{-j4}} & 0\\ {{0}}&8 – j4&j8 \\43.3013 +j25&j8&3-j8 \end{vmatrix}= 0-(-\mathrm{j}4)\times\bigl\lbrack0-(43.3013+\mathrm{i}25)\times\bigr\rbrack8\bigr\rbrack+0=\,1385.6416+j800
∴ The response,{\bar{I}}_{x}\;=\;{\bar{I}}_{a}\;=\;{\frac{\Delta_{a}}{\Delta}}=\frac{1385.6416+j800}{800-j512}
=\;0.7747\;+j1.4958\;A=\;1.6845∠62.6^{\circ}\;AIt is observed that the response remains the same after interchanging the positions of source and response, which demonstrates the validity of the reciprocity theorem.
