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Chapter 2: Q2.5-28Q (page 93)

28. Show that if three shunt circuits (with resistances \({R_{\bf{1}}},{R_{\bf{2}}},\,{R_{\bf{3}}}\)) are connected in series, the resulting network has the same transfer matrix as a single shunt circuit. Find a formula for the resistance in that circuit.

Short Answer

Expert verified

The resulting network is itself a single shunt circuit with resistance \({1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}} + {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} + {1 \mathord{\left/

{\vphantom {1 {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}}\).

Step by step solution

01

Write the transfer matrices

The transfer matrices of the shunt circuit with resistances \({R_1},{R_2},\)and \({R_3}\) are \(\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}&1\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{aligned}} \right),\)and \(\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}}}&1\end{aligned}} \right)\), respectively.

02

Find the transfer matrix for the resulting network

\(\begin{aligned}{c}\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}}}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}&1\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}} - {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}} - {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}&1\end{aligned}} \right)\\\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}}}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}&1\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0\\{ - \left( {{1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}} + {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} + {1 \mathord{\left/

{\vphantom {1 {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}}} \right)}&1\end{aligned}} \right)\end{aligned}\)

03

Conclusion

Hence, the resulting network is itself a single shunt circuit with resistance \({1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}} + {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} + {1 \mathord{\left/

{\vphantom {1 {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}}\).

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