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EE-301 Spring AY 2016 6 Week Exam 22 FEB 2016 Time Limit: 50 Minutes Name (Print): Section This exam contains 9 pages (including this cover page) and 5 problems. Check to see if any pages are missing. Enter all requested information on the top of this page, and put your initials on the top of every page, in case the pages become separated. The course equation sheet is attached to the last page of the examination. The examination is closed-book/closed-notes. An approved calculator may be used on the exam. Notes, equations, formulas, example problems, etc., may not be stored in the memory of your your calculator. You may not use a laptop computer, cell phone or any other device with wireless connectivity in lieu of an approved calculator (TI NSPIRE for example). You are required to show your work on each problem. The following rules apply: • Organize your work, in a reasonably neat and coherent manner, in the space provided. Work scattered all over the page without a clear ordering is difficult to grade and as a result may receive very little credit. • Show your work. A correct answer, unsupported by calculations, explanation, or algebraic work will receive no credit; an incorrect answer supported by substantially correct calculations and explanations may receive partial credit. • Express all answers using proper Engineering prefixes and units. Express all numeric answers to at least 3 significant digits. • If you need more space, use the back of the pages; clearly indicate when you have done this. • Similar makeup exams will be taken by individuals at different times. No communication is permitted concerning the content of this exam with any individual who has not yet taken the examination. Do not write in the table to the right. Problem Points 1 25 2 30 3 20 4 20 5 5 Total: 100 Score EE-301 6 Week Exam - Page 2 of 9 PAGE INTENTIONALLY BLANK 22 FEB 2016 EE-301 6 Week Exam - Page 3 of 9 I0 20V 30Ω I1 −V2 + 60Ω 22 FEB 2016 + V1 − 10Ω Figure 1: Problem 1 1. (25 points) Consider the circuit in Fig. 1, (a) (5 points) Determine the total resistance, RT otal , seen by the source. (b) (5 points) Determine the source current, I0 . 60Ω EE-301 6 Week Exam - Page 4 of 9 22 FEB 2016 (c) (5 points) Determine the current I1 through the 30Ω resistor using the current divider rule. (d) (5 points) Determine the voltage, V1 , across the 60Ω resistor. (e) (5 points) Determine the power dissipated in 10Ω resistor. EE-301 6 Week Exam - Page 5 of 9 100Ω 22 FEB 2016 50Ω 100Ω 100Ω RLoad 200V 50Ω Figure 2: Problem 2. 2. (30 points) Determine the Thevénin equivalent circuit for the circuit displayed in Fig. 2. (a) (10 points) Determine the Thevénin equivalent voltage, ET h . (b) (10 points) Determine the Thevénin equivalent resistance, RT h . (c) (10 points) If RLoad = 300Ω, determine the load power, PLoad . EE-301 6 Week Exam - Page 6 of 9 10Ω 4Ω + 20V 22 FEB 2016 V − I 5Ω 6Ω 30V Figure 3: Problem 3. 3. (20 points) Consider the circuit in Fig.3 , use Nodal Analysis to determine: (a) (15 points) The voltage, V , across the 5Ω resistor. (b) (5 points) The current, I, through the 4Ω resistor. EE-301 6 Week Exam - Page 7 of 9 5Ω 22 FEB 2016 a 50V 3A 30Ω 10Ω 15Ω b Figure 4: Problem 4. 4. (20 points) For the circuit in Fig. 4, (a) (10 points) Perform a source conversion of the 50V battery and draw the circuit with the converted source. (b) (10 points) Determine the voltage across the 30Ω resistor. EE-301 6 Week Exam - Page 8 of 9 22 FEB 2016 PIN = 200W η1 = 0.9 POUT η2 = 0.8 η3 = 0.5 Figure 5: Problem 5. 5. (5 points) Given the system in Fig. 5, determine the output power, POU T , in Watts. Voltage, current, resistance Elementary charge q = 1.6 × 10−19 C V= Capacity (A. hr) Battery life (hr) = Discharge rate (A) W Q KVL � � closed loop ρ𝑙𝑙 A 2 P = VI = I2 R = V �R Max power transfer R L = R Th Pmax = 𝑖𝑖C (𝑡𝑡) = C W or E = 2 Transient analysis τ = RC −𝑡𝑡� τcharging 𝑣𝑣C (𝑡𝑡) = Vfinal �1 − e −𝑡𝑡� τdischarging � 𝑣𝑣C (𝑡𝑡) = Vfinal + (Vinitial − Vfinal ) e Electrical −𝑡𝑡� τ (general) Pmech loss=Tmech lossω Pd = Pin − Pelec loss = Pmech loss + Pout VDC − Ia R a = Ea R3 R1 R2 R1 +R2 +⋯� CDR −1 (parallel) (2 parallel) R2 � (2 parallel) R1 +R2 d𝑣𝑣C (𝑡𝑡) d𝑡𝑡 𝑣𝑣L (𝑡𝑡) = L 1 2 CV 2 W or E = Inductors τ = L�R −𝑡𝑡� τdischarging � d𝑖𝑖L (𝑡𝑡) d𝑡𝑡 1 2 LI 2 (charging) (discharging) −𝑡𝑡� τ (general) RPM ω = 2π � � rad/sec 60 Pin = Pelec loss + Pmech loss + Pout VDC 1 Linear Motor Mechanical Loss Ia + Inductors Pout=Tloadω 1 hp = 746 W Ra R2 RT = 𝑖𝑖L (𝑡𝑡) = Ifinal + (Iinitial − Ifinal ) e DC Motor 1 I1 = IEQ � 𝑖𝑖L (𝑡𝑡) = Iinitial × e (discharging) Pd=Tdω Pelec loss=Ia2R R1 −𝑡𝑡� τcharging Mechanical Electrical Loss + 𝑖𝑖L (𝑡𝑡) = Ifinal �1 − e (charging) Pd=EaIa=KvIaω Pin=VDCIa 1 (series) R EQ IX = IEQ � � RX Capacitors Pout Pin Capacitors RT = � RX VX = VEQ � � R EQ node VTh 4 × R Th R T = R1 + R 2 + R 3 + ⋯ VDR I=0 η= Resistances V=I×R � Iin = � Iout Power 𝑣𝑣C (𝑡𝑡) = V𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 × e R= Ohm’s law KCL V=0 � Erise = � Vdrop Q t I= DC Ea VB I R Fd = ILB (Newton) Eind = uBL (Volts) VB − Eind = IR Pin = VB I Pout = Eind I = FLoad u Eind