222 - EE 234: Electronics & Microcontrollers First Major Exam Wednesday, Feb 15, 2023, at 6:00 - 7:30 PM (H) Student Name: Student ID: Section Number: Teacher Name Some extra white pages are available at the end if needed Q 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans D E B A A D B D C B D D D E C Q1. The power, and energy transferred from t = 0 to t =10 s of the following element, respectively, are A) B) C) D) E) 200 − 20π‘ W, 1000 J −200 − 20π‘ W, 1000 J 200 − 20π‘ W, − 1000 J −200 + 20π‘ W, − 1000 J 200 + 20π‘ W, − 1000 J Q2. The voltages π£π , π£π , and π£π in the following circuit, respectively, are A) B) C) D) E) −5 V, −25 V, 35 V −5 V, 25 V, 35 V 5 V, −25 V, −35 V 5 V, 25 V, −35 V 5 V, −25 V, 35 V Q3. The power of the voltage source ππ in the following circuit is A) B) C) D) E) −48 W 48 W −64 W 64 W 36 W 2 Q4. The value of πΌπ₯ in the following circuit is A) B) C) D) E) 12 A 9A −12 A −9 A 6A Q5. The node-voltage equations of the following circuit in matrix form are 0.45 (A) [−0.25 0 −0.25 0.85 −0.2 π£1 0 −3.5 −0.2] [π£2 ] = [ 3.5 ] 0.3 π£3 2 0.45 (B) [ 0.6 0 0.45 (C) [−0.25 0 −0.25 0.75 −0.2 π£1 0 −3.5 −0.2] [π£2 ] = [ 3.5 ] 0.3 π£3 2 0.45 (D) [−0.25 0 0.45 (E) [−0.25 0 −0.25 0.85 −0.2 π£1 0 −3.5 π£ ] [ 2 ] = [ 3.5 ] −0.2 −0.1 π£3 2 −0.25 0.85 −0.2 −0.25 0.85 −0.2 π£1 0 −3.5 −0.2] [π£2 ] = [ 3.5 ] 0.3 π£3 2 0 π£1 −3.5 0.4] [π£2 ] = [ 3.5 ] 0.3 π£3 2 Q6. The value of the mesh current I2 in the following circuit is A) B) C) D) E) 2.75 A -4.75 A -2 A 4.75 A 2A 3 Q7. The value of the Thevenin voltage (Vt) in the following circuit (between points a, b) is A) B) C) D) E) 12 V -12 V -18 V -5 V 18 V Q8. The value of the Norton current in the following circuit (between points a, b) is A) B) C) D) E) -1 A 1A -3 A 3A 9/6 A Q9. Find RL for maximum power transfer from the circuit below. Also, find the maximum power of RL A) B) C) D) E) 2 β¦, -4.5 W 6 β¦, 4.5 W 2 β¦, 4.5 W 2 β¦, 9 W 9 β¦, 4.5 W 4 Q10. The node-voltage equation of V1 in the following circuit is 1 1 4 1 2 1 4 1 2 1 4 1 2 1 A) π1 ( + B) π1 ( + C) π1 ( + D) π1 ( + 4 2 1 6 1 6 1 6 1 6 1 6 1 6 1 + ) − π2 ( ) = 4 + ) + π2 ( ) = 4 + ) − π2 ( ) = 4 6 1 1 4 2 E) −π1 ( + 1 + ) − π2 ( ) = −4 3 1 1 6 6 + ) − π2 ( ) = 4 Q11. Find the mesh currents in the following circuit A) π1 = 2.364 A, π2 = 1.118 A B) π1 = −2.364 A, π2 = 1.818 A C) π1 = 3.364 A, π2 = 1.818 A D) π1 = 2.364 A, π2 = 1.818 A E) π1 = 2.98 A, π2 = 1.818 A Q12. Determine V in the following circuit (you may combine the resistance series/parallel) A) B) C) D) E) 1/5 V 6V 2V −2 V −1/3 V 5 Q13. For the following circuit, the value for π that makes the equivalent resistor π eq = 50 Ω is (A) (B) (C) (D) (E) π = 41 Ω π = 12 Ω π =8 Ω π = 16 Ω π =4Ω Q14. Find the power dissipated in the 5 Ω resistor in the following figure (A) (B) (C) (D) (E) π = 42 W π = 50 W π = 60 W π = 100 W π = 80 W Q15. Find π0 in the circuit shown below (you may use source transformation) (A) (B) (C) (D) (E) π0 = 409.3 mA π0 = −409.3 mA π0 = 636.4 mA π0 = −636.4 mA π0 = 119.7 mA 6 Name: ID: Sec: 7 Name: ID: Sec: 8