Problems_after_Quiz_3

4 pts isrc 4 8 j7 9 j174 v 12 j 7 vth 918 v il

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Unformatted text preview: er absorbed by the load. (4 pts) ISRC 4 Ω 8 − j7 = 9 − j1.74 V; 12 − j 7 |VTh| = 9.18 V; + IL = VTh /6 = 1.5 − j 0.29 A; 12∠0° V – V1 = (3 – j 0.58)IL = 4.68 +j 0 V rms 12 − 4.68 ISRC = = 1.83 +j 0 A 4 (9.18 )2 PSRC = V1ISRC = 12×1.83 ≅ 22 W; PL = ≅ 7 W. 4×3 VTh = 12 j5.78 Ω V1 IL 8Ω 3Ω -j 7 Ω -j 5.2 Ω c. For a purely resistive load, determine its value Rmax for maximum power transfer and find the power absorbed by the load. (3 pts) Rm = (3)2 + (5.2)2 ≅ 6 Ω; |I| = |VTh|/|Z| = 9.18 ( 3 + 6 ) 2 + ( 5 .2 ) 2 = 0.883 A P = |I|2×6 = 4.68 W. d. For a purely resistive load with RL = 2Rmax ( Rmax of part c), determine the power absorbed by the load. (3 pts) |I| = |VTh|/|Z| = 9.18 2 (3 + 12) + (5.2) 2 = 0.578 A; P = |I|2×12 ≅ 4 W 1/1 17. If a capacitor with impedance Z2 is connected in parallel to a load Z1 = 300 +j450 Ω. What should be Z2 in ohms so that the equivalent load is purely resistive? a) -928.6 j b) -1112.5 j c) -650 j d) -750 j e) None of the above 18. What is the power factor of the equivalent load of the previous question? a) 0.8 b) 0.6 c) 0 d) 1 e) None of the above 19. If Vs = -9V, R1=5KΩ, R2=2KΩ, R3 =2KΩ, then the voltage Vc across the capacitors at time t=0 is: a) Vc= -9V b) Vc= 9V c) Vc= 20V d) Vc= -20V e) None of the above 20. In the circuit shown above find in milliseconds the value of tau for the time interval 3≤ t<6. a) tau= 0.08 msec b) tau= 0.48 msec c) tau= 0.96 msec d) tau= 0.64 msec e) None of the above 23. Find the maximum power dissipated in R if: Is=2mA, R1=100kΩ, Vs=50V. a) P= 12.5 mW b) P= 1.25 mW c) P= 50 mW d) P= 56.25 mW e) None of the above 24. Solve for V2 if VS= 2V, R1= 4Ω, R2= 4Ω, R3= 4Ω. a) V2= 1.5 V b) V2= 1.8 V c) V2= 2.5 V d) V2= 3 V e) None of the above 100 ohms 30.76 V Problem 3 Consider the periodic signal Vs(t) shown below. Assume this signal is applied to the circuit in the figure below, find an expression for the voltage Vo(t) across the terminals a,b as shown. A) Vo (t ) = Vm Vm + 8 2π V V B) V0 (t ) = m + m 8 4π V V C) V4 (t ) = m + m 4 2π D) V4 (t ) = Vm Vm + 4 2π ∑ 2 n n =1 ∞ ∑ 4 n ∑ sin n =1 ∞ ∑ n =1 cos nω o t nπ sin n =1 ∞ nπ sin ∞ cos...
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This test prep was uploaded on 03/11/2014 for the course EECE 210 taught by Professor Riadchedid during the Spring '07 term at American University of Beirut.

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