Final%20Exam%20S08%20wSolutions

Final%20Exam%20S08%20wSolutions - Name : _ CWRU e-mail:_...

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Name : _________________________ CWRU e-mail:____________ Department of Electrical Engineering and Computer Science ENGR 210. Introduction to Circuits and Instrumentation (4) ENGR 210 - SPRING 2008 FINAL EXAMINATION–5/6/08 AVERAGE 1. 5.8 10 DC Circuit Analysis 2. 7.4 10 Thevenin/Norton equivalent circuits 3. 6.8 10 Basic transients 4. 7.1 10 Basic phasors 5. 6.5 10 Multiple op-amp circuits 6. 6.5 10 Measurement circuits 7. 6.4 10 Capacitors and inductors 8. 6.4 10 Sinusoidal analysis 9. 6.1 10 Sinusoidal frequency response 10. 5.9 10 Sequential switching SCORE 63.2 100 ACADEMIC DISHONESTY All forms of academic dishonesty including cheating, plagiarism, misrepresentation, and obstruction are violations of academic integrity standards. Cheating includes copying from another's work, falsifying problem solutions, or using unauthorized sources, notes or computer programs. Misrepresentation includes forgery of official academic documents, the presentation of altered or falsified documents or testimony to a university office or official, taking an exam for another student, or lying about personal circumstances to postpone tests or assignments. Obstruction occurs when a student engages in unreasonable conduct that interferes with another's ability to conduct scholarly activity.
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Name : _________________________ CWRU e-mail:____________ NOTES:
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Name : _________________________ CWRU e-mail:____________ Problem 1 DC Circuit Analysis (10 points) + - R 1 R 3 v S1 i S2 10v x v x + - i 1 i 2 i 3 + - + - v o R 2 Use R 1 =6k Ω R 2 =8k Ω R 3 =4k Ω v s1 =6 V i s2 =12mA (a) [1 point] How many nodes are there in the above circuit? __________ (b) [3 points] Use the mesh current method to solve for the mesh currents i 1 , i 2 and i 3 . Write each mesh- current equation in normal form. _____ i 1 + ______ i 2 + ______ i 3 = ________ _____ i 1 + ______ i 2 + ______ i 3 = ________ _____ i 1 + ______ i 2 + ______ i 3 = ________ SOLUTION: Writing KVL equations gives 6 + 6000 i 1 6000 i 2 + 4000 i 1 4000 i 3 = 0 (1) 10 v x + 8000 i 2 8000 i 3 + 6000 i 2 6000 i 1 = 0 (2) i 3 = .012 (3) Rewriting (1) and substituting (3) 10000 i 1 6000 i 2 = 6 + 4000 .012 ( ) = 6 48 = 42 Doing the same for (2) 40000 i 1 40000 i 3 + 8000 i 2 8000 i 3 + 6000 i 2 6000 i 1 = 0 Combining terms 34000 i 1 + 14000 i 2 = 48000 .012 ( ) = 576 (c) [3 points] What are the values of the mesh currents i 1 = _______ i 2 = _______
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Name : _________________________ CWRU e-mail:____________ i 3 = _______ SOLUTION: From (b) we have the equations 10000 i 1 6000 i 2 = 42 34000 i 1 + 14000 i 2 = 576 These can be solved on a calculator to give the above solutions. (d) [3 points] What is the value of v o ? _____ SOLUTION: v o = 4000 i 1 i 3 ( ) + 8000 i 2 i 3 ( ) v o = 4000 11.76 + 12 ( ) × 10 3 + 8000 12.59 + 12 ( ) × 10 3 V o = + 0.96 4.72 = 3.76
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Name : _________________________ CWRU e-mail:____________ Problem 2 Thevenin/Norton equivalent circuit (10 points) + - R 1 R 2 R 3 0.5i 1 v o + - a b v s i 1 Use R 1 =4 Ω R 2 =6 Ω R 3 =12 Ω v s =12V (a) [3 points] Determine the open-circuit voltage at terminals a-b. V ab,oc =____ SOLUTION + 12 v o 4 v o 0 6 + 1 2 i 1 v o 0 12 = 0 + 12 v o 4 v o 6 + 1 2 12 v o 4 v o 12 = 0 + 12 4 + 1 2 12 4 = v o 4 + v o 6 + v o 8 + v o 12 108 = 15 v o v o = + 7.2 volts (b) [3 points] Determine the short circuit current (from a to b) at terminals a-b.
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Final%20Exam%20S08%20wSolutions - Name : _ CWRU e-mail:_...

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