Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 2: Q2.5-27Q (page 93)

27. Design two different ladder networks that each output 9 volts and 4 amps when the input is 12 volts and 6 amps.

Short Answer

Expert verified

The first ladder network is

Here, \[{v_1} = 12,{i_1} = 6,\] and \[{v_3} = 9,{i_3} = 4\].

The second ladder network is

Here, \[{v_1} = 12,{i_1} = 6,\] and \[{v_3} = 9,{i_3} = 4\].

Step by step solution

01

Write the transfer matrices

The transfer matrices of the series circuit with resistance \[{R_1}\] and shunt circuit with resistance \[{R_2}\] are \[\left[ {\begin{array}{*{20}{c}}1&{ - {R_1}}\\0&1\end{array}} \right]\], and \[\left[ {\begin{array}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

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

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{array}} \right]\], respectively.

02

Design the first ladder network

Consider the first ladder network as aseries circuitwith resistance \[{R_1}\], followed by a shunt circuitwith resistance \[{R_2}\]. Therefore, the transfer matrix for this network is

\[\left[ {\begin{array}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

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

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - {R_1}}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - {R_1}}\\{ - {1 \mathord{\left/

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

\kern-\nulldelimiterspace} {{R_2}}}}&{1 + {{{R_1}} \mathord{\left/

{\vphantom {{{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}\end{array}} \right]\].

To produce an output of 9 volts and 4 amps for an input of 12 volts and 6 amps, the transfer matrix must satisfy the equation:

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}1&{ - {R_1}}\\{ - {1 \mathord{\left/

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

\kern-\nulldelimiterspace} {{R_2}}}}&{1 + {{{R_1}} \mathord{\left/

{\vphantom {{{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{12}\\6\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\4\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{12 - 6{R_1}}\\{ - {{12} \mathord{\left/

{\vphantom {{12} {{R_2} + 6 + {{6{R_1}} \mathord{\left/

{\vphantom {{6{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}}} \right.

\kern-\nulldelimiterspace} {{R_2} + 6 + {{6{R_1}} \mathord{\left/

{\vphantom {{6{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\4\end{array}} \right]\end{array}\]

This implies that

\[\begin{array}{c}12 - 6{R_1} = 9\\6{R_1} = 3\\{R_1} = {1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}\end{array}\]

And

\[\begin{array}{c} - {{12} \mathord{\left/

{\vphantom {{12} {{R_2} + 6 + {{6{R_1}} \mathord{\left/

{\vphantom {{6{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}}} \right.

\kern-\nulldelimiterspace} {{R_2} + 6 + {{6{R_1}} \mathord{\left/

{\vphantom {{6{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}} = 4\\ - 12 + 6{R_2} + 6\left( {{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}} \right) = 4{R_2}\\2{R_2} = 12 - 3\\{R_2} = {9 \mathord{\left/

{\vphantom {9 2}} \right.

\kern-\nulldelimiterspace} 2}\end{array}\]

Thus, the firstladder network is

Here, \[{v_1} = 12,{i_1} = 6,\] and \[{v_3} = 9,{i_3} = 4\].

03

Design the second ladder network

Consider the second ladder network as a shunt circuit with resistance \[{R_2}\], followed by a series circuit with resistance \[{R_1}\]. Therefore, the transfer matrix for this network is

\[\left[ {\begin{array}{*{20}{c}}1&{ - {R_1}}\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

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

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{1 + {{{R_1}} \mathord{\left/

{\vphantom {{{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&{ - {R_1}}\\{ - {1 \mathord{\left/

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

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{array}} \right]\].

To produce an output of 9 volts and 4 amps for an input of 12 volts and 6 amps, the transfer matrix must satisfy the equation:

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{1 + {{{R_1}} \mathord{\left/

{\vphantom {{{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&{ - {R_1}}\\{ - {1 \mathord{\left/

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

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{12}\\6\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\4\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{12 + {{12{R_1}} \mathord{\left/

{\vphantom {{12{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} - 6{R_1}}\\{ - {{12} \mathord{\left/

{\vphantom {{12} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} + 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\4\end{array}} \right]\end{array}\]

This implies that

\[\begin{array}{c} - {{12} \mathord{\left/

{\vphantom {{12} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} + 6 = 4\\{{12} \mathord{\left/

{\vphantom {{12} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} = 2\\{R_2} = 6\end{array}\]

And

\[\begin{array}{c}12 + {{12{R_1}} \mathord{\left/

{\vphantom {{12{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} - 6{R_1} = 9\\3 + {{12{R_1}} \mathord{\left/

{\vphantom {{12{R_1}} 6}} \right.

\kern-\nulldelimiterspace} 6} - 6{R_1} = 0\\3 + 2{R_1} - 6{R_1} = 0\\{R_1} = {3 \mathord{\left/

{\vphantom {3 4}} \right.

\kern-\nulldelimiterspace} 4}\end{array}\]

Thus, the second ladder network is

Here, \[{v_1} = 12,{i_1} = 6,\] and \[{v_3} = 9,{i_3} = 4\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

Suppose a linear transformation \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) has the property that \(T\left( {\mathop{\rm u}\nolimits} \right) = T\left( {\mathop{\rm v}\nolimits} \right)\) for some pair of distinct vectors u and v in \({\mathbb{R}^n}\). Can Tmap \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\)? Why or why not?

If a matrix \(A\) is \({\bf{5}} \times {\bf{3}}\) and the product \(AB\)is \({\bf{5}} \times {\bf{7}}\), what is the size of \(B\)?

The inverse of \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\C&I&{\bf{0}}\\A&B&I\end{array}} \right]\) is \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\Z&I&{\bf{0}}\\X&Y&I\end{array}} \right]\). Find X, Y, and Z.

Suppose \({A_{{\bf{11}}}}\) is invertible. Find \(X\) and \(Y\) such that

\[\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{\bf{0}}\\{\bf{0}}&S\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&Y\\{\bf{0}}&I\end{array}} \right]\]

Where \(S = {A_{{\bf{22}}}} - {A_{21}}A_{{\bf{11}}}^{ - {\bf{1}}}{A_{{\bf{12}}}}\). The matrix \(S\) is called the Schur complement of \({A_{{\bf{11}}}}\). Likewise, if \({A_{{\bf{22}}}}\) is invertible, the matrix \({A_{{\bf{11}}}} - {A_{{\bf{12}}}}A_{{\bf{22}}}^{ - {\bf{1}}}{A_{{\bf{21}}}}\) is called the Schur complement of \({A_{{\bf{22}}}}\). Such expressions occur frequently in the theory of systems engineering, and elsewhere.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free