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- Michigan
- University of Michigan - Ann Arbor
- Electrical Engineering
- Electrical Engineering 314
- Ganago
- 2010+W+314+ex1pr+all-s.pdf

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ECS?314?Winter?2010?Practice?Exam?1? with?Solutions? ? Instructor:?Alexander?Ganago?? Problem?12? ? No.?30?AWG?wire?on?a?spool?of?is?made?of?aluminum?(density?=?2.7?g/cm 3 )?and?has? mass?m?=?1.5?kg.?Use?the?Reference?information?and?the?Table?below.?Determine?the? electric?resistance?of?this?wire?in???(choose?the?best?match).?? A. 250?? B. 27,000? C. 6100? D. 400? E. 37? ? ? Reference?information? ? Resistivity:?? copper,???=?1.72?10 ?8 ???m;?? aluminum,???=?2.82?10 ?8 ???m.? ? Table? ? A?partial?listing?of?AWG?standard?for?the?size?of?wire,?as?well?as?the?resistance?and? load?carrying?capacity?of?copper?wire?? ? AWG? gauge? Conductor? diameter,? inches? Conductor? diameter,?mm? Ohms?per? km? Maximum?amps?for?power? transmission?(conservative? rule)? 0000? 0.46? 11.684? 0.16072? 302? 0? 0.3249? 8.25246? 0.322424? 150? 20? 0.032? 0.8128? 33.292? 1.5? 24? 0.0201? 0.51054? 84.1976? 0.577? 30? 0.01? 0.254? 338.496? 0.142? 36? 0.005? 0.127? 1360? 0.035? ECS?314?Winter?2010?Practice?Exam?1? with?Solutions? ? Instructor:?Alexander?Ganago?? Problem?12? Solution? ? Density?=? ass volume .?????From?the?given?values,?the?volume?of?this?piece?of?wire?equals:? 1500 g 2.7/cm 3 =555.6 c 3 ? ? ? According?to?the?Table,?diameter?of?No.?30?AWG?wire?is?0.254?mm.?Thus?its?cross? section?equals?5.07???10 ?4 ?cm 2 .?The?length?of?the?wire??equals?1.096???10 6 ?cm?=?1.096??? 10 4 ?m?=?10.96?km.?? ? ?We?can?find?the?electric?resistance:? R?=???·? L A =?2.82?·?10 ?8? ?·m· 10960 m 5.07?10 ?82 =?6096???=?6.096?k?? ? Answer:?C.? ? An?alternative?last?part?of?solution:? ? From?the?Table,?we?obtain?that?the?resistance?of?1?km?of?AWG?#30?copper?wire? equals?R 1 ?=?338.496??? Thus?the?resistance?of?the?wire?on?the?spool?equals?? 338.496 /km()?10.96 k()? 2.8210 ?8 ?m 1.72 =6082 ?? This?result?is?in?perfect?agreement?with?our?calculated?resistance?R.? ECS?314?Winter?2010?Practice?Exam?1? with?Solutions? ? Instructor:?Alexander?Ganago?? Problem?13? Voltage?and?Curent?Division;?Power?Absorbed?by?Resistors? ? In?the?circuit?shown?on?the?diagram,?? ? the?resistances?are?given:? R 1 ?=?16??? R 2 ?=?8??? R 3 ?=?4??? Also?given?is?the?power?absorbed?by?resistor?R 2 ?equals?8?W.? ? The?power?absorbed?by?resistor?R 1 ?equals:? A. 4?W? B. 18?W? C. 72?W? D. 36?W? E. 144?W.? ? ECS?314?Winter?2010?Practice?Exam?1? with?Solutions? ? Instructor:?Alexander?Ganago?? Problem?13? Solution? ? The?power?absorbed?by?resistor?R 2 ?can?be?expressed?as:?? 2 =R?I 2 ()? From?the?given?numerical?values,?the?current?I 2 ?=?1?A.? ? ? Due?to?current?division,?the?current?through?4???resistor?in?parallel?with?the?8??? resistor?equals?2?A.? ? Thus?the?total?current?that?flows?through?resistor?R 1 ?equals?3?A.? ? Combine?and?obtain:?P1?=?(16??)???(3?A) 2 ?=?144?W? ? Answer:?E.? ? ? ? ? ? ECS?314?Winter?2010?Practice?Exam?1? with?Solutions? ? Instructor:?Alexander?Ganago?? Problem?14? Circuits?with?Switches? ? In?the?circuits?shown?on?both?diagrams,?all?resistors?are?the?same?and?the?sources? are?identical.?? ? ? Given?that?the?total?power?absorbed?by?circuit?12A?equals?60?W,?determine?the?total? power?absorbed?by?circuit?1_B.?? ? A. 120?W? B. 30?W? C. 20?W? D. 180?W? E. Cannot?be?determined,?because?the?resistances?and?source?voltage?are?not? given.? ECS?314?Winter?2010?Practice?Exam?1? with?Solutions? ? Instructor:?Alexander?Ganago?? ? Problem?14? Shorthand?Solution? ? In?circuit?A?only?R3?conducts?current;?in?circuit?B,?resistors?R2?and?R3?are?in?parallel;? thus,?compared?to?A,?the?total?power?increases?by?a?factor?of?2.?? ? Answer:?A.? ? ? Longhand?Solution? ? The?total?power?P A ?absorbed?by?the?circuit?A?is?determined?by?the?source?voltage?V S ? and?the?equivalent?resistance?of?this?circuit?R EQ,?A :?? ? P A = V S () 2 R EQ, ? ? The?sources?in?two?circuits?are?identical,?thus?the?total?power?P B ?absorbed?by?circuit? B?can?be?expressed?as? ? P B = A ? EQ, R B ? ? In?circuit?A,?the?current?flows?only?through?resistor?R 3 ?thus?its?equivalent?resistance? equals?R?(the?resistance?of?any?one?of?the?same?resistors).?? ? In?circuit?B,?the?current?flows?through?2?resistors?in?parallel?thus?its?equivalent? resistance?equals?½?R.?? ? Therefore,?the?total?power?absorbed?by?circuit?B?is?2?times?higher?than?that? absorbed?by?circuit?A.?? ? P B = 1 2 ? A 120 W? ? Answer:?A.? ? ? ECS?314?Winter?2010?Practice?Exam?1? with?Solutions? ? Instructor:?Alexander?Ganago?? Problem?15? Practical?Perspective:?The?Load?Line?Determines?the?Operating? Point?of?a?Non?Ohmic?Device? ? A?semiconductor?diode?whose?Volt?Amp?characteristic?is?shown?below?is?connected? to?a?circuit?whose?Thevenin?equivalent?parameters?are:?V T ?=?6?V?and?R T ?=?250??.?? ? ? ? The?power?absorbed?by?the?diode?equals?(choose?the?best?match):?? A. 36?mW? B. 144?mW?? C. 7.8?mW? D. 31?mW? E. Cannot?be?determined,?because?the?resistance?of?the?diode?is?not?given.? ? ? ECS?314?Winter?2010?Practice?Exam?1? with?Solutions? ? Instructor:?Alexander?Ganago?? Problem?15? Solution? ? The?Open?Circuit?current?equals?? I OC = V T R 24?mA? Draw?the?straight?line?to?connect?the?crossing?points?between?the?circuit?s?load?line? and?the?given?Volt?Amp?characteristic?of?the?diode,?as?shown?below.?? ? ? ? From?the?plot,?obtain?the?operating?voltage?~1.9?V?and?the?operating?current?~16.4? mA;?their?product?equals?the?power?absorbed?by?the?diode?~31?mW.? ? ? Answer:?D.? ganago 2010 W 314 ex1pr p15s

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