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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

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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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Most popular questions from this chapter

In exercise 11 and 12, mark each statement True or False.Justify each answer.

a. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}&{{A_{\bf{2}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), with \({A_{\bf{1}}}\) and \({A_{\bf{2}}}\) the same sizes as \({B_{\bf{1}}}\) and \({B_{\bf{2}}}\), respectively then \(A + B = \left[ {\begin{array}{*{20}{c}}{{A_1} + {B_1}}&{{A_{\bf{2}}} + {B_{\bf{2}}}}\end{array}} \right]\).

b. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_1}}\\{{B_{\bf{2}}}}\end{array}} \right]\), then the partitions of \(A\) and \(B\) are comfortable for block multiplication.

1. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{6}}\\{\bf{5}}&{\bf{4}}\end{aligned}} \right)\).

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{\bf{5}}\\{ - {\bf{3}}}&{\bf{1}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{5}}}\\{\bf{3}}&k\end{aligned}} \right)\). What value(s) of \(k\), if any will make \(AB = BA\)?

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

36. Write the command(s) that will create a \(6 \times 4\) matrix with random entries. In what range of numbers do the entries lie? Tell how to create a \(3 \times 3\) matrix with random integer entries between \( - {\bf{9}}\) and 9. (Hint:If xis a random number such that 0 < x < 1, then \( - 9.5 < 19\left( {x - .5} \right) < 9.5\).

If \(A = \left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\) and \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\), determine the first and second column of B.
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