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- A balanced positive-sequence wye-connected 60-Hz three-phase source has line-to-line voltages of VL = 440 V rms. This source is connected to a balanced wye-connected load. Each phase of the load consists of a 0.5-H inductance in series with a 50-? resistance. Assume that the phase of Van is zero. A) Find the line-to-neutral voltage phasor Van. Enter your answer using polar notation. Express argument in degrees. B) Find the line-to-neutral voltage phasor Vbn. Enter your answer using polar notation. Express argument in degrees. C) Find the line-to-neutral voltage phasor Vcn. Enter your answer using polar notation. Express argument in degrees. D) Find the line-to-line voltage phasor Vab. Enter your answer using polar notation. Express argument in degrees.2. Design a two-stage amplifier that produces the following output: V₁ = 3√₁ + 2V2 + 4V3 a. Draw the block diagram for this system b. Implement the block diagram for this circuit using op-amps. Ensure that all resistors drawn have values. c. If V₁=1 V, V2=2 V, V3 = 2 V, what should be the minimum Vcc of both amplifiers so that neither stage is saturated?8. An R-L series circuit has a reactive power of 1234 VARs and an apparent power of 4329 VA. By how many degrees are voltage and current out of phase with each other? 9. An R-L series circuit contains a resistor and an inductor. The resistor has a value of 6.5 22. The circuit is connected to 120 V and has a current flow of 12 A. What is the inductive reactance of this circuit? 10. What is the voltage drop across the resistor in the circuit in Question 9? 11. A phase angle meter connected in an R-L series circuit indicates that the current is 28 degrees out of phase with the voltage. What is the circuit power factor? 12. An R-L series circuit has a power factor of 73%. If the apparent power is 560 kVA, what is the true power in the circuit?
- A centrifugal pump having four stages in parallel delivers 12 kiloLiters per minute of liquid against a head of 25m. The diameter of the impeller being 24 cm has a speed of 1800 rpm. A pump is to be made up with a'number of stages in series. In similar construction to that of the first pump to run at 1250 rpm and to deliver 15 kiloLiters/min against a total head of 250 m, find the number of stages required in this case. Select the correct response. 2. 3 7. 15 WQUESTION 3 BAND PASS FILTER Find the transfer function, Vo(s)/Vi(s), for the op-amp system shown in the figure. C R R C Vo Vi V₁(s) V₁(s) ○ V (s) 0 V₁(s) ○ V (s) A V₁(s) WH RCs RCS +1 1 (RCs+1) 2 RCS+1 RCs V₁(s) RCs V₁(s) (RCs+1)²5. Sketch the following function in frequency domain plot: i. Y(t) = 3 sin (4лt) + 4 cos (4 лt) ii. Y(t) = 12 sin (10nt) + 5 sin (20лt) + sin (100лt)
- Design an analog Butterworth filter with the following properties. Find the transfer function? Ap=2dB (Passband attenuation Wp-20rad/sec pass bondi frequency As 10 dB (Stop band attenuation WS-30 rad/sec stop body frequencyIf f(x) = x² + cos(x) is a periodic function with period 2W, then a. It is an odd function which gives a value of a = 0 b. Its Fourier series is classified as a Fourier cosine series where a = 0 c. it is neither odd nor even function, thus no classification can be deduced. d. it is an even function which gives a value of b₁ = 02- for a DSB-SC modulator with calculations, assuming perfect carrier pression: usdal Modula ng Signal cerario: Carrier freque Modulating c) = 1 t), whe m = 1 kHz (20 marks)
- - Consider the positions servomechanism is shown in figure below. Assume the following numerical values for the system constant are. Kp = gain of potentiometer detector = 7.64 v/ rad, Kb = back e. m. f. constant -5.5*10-2 V/rad/sec K₁ = motor torque constant = 6*10-5 Nm/A C₁ = 4*10-2 Nm/ (rad /sec). Determine the natural frequency and damping ratio of the system. Amp. % Ra= armature winding resistant =20, Ka=Gain of amplifier =10 Jm= 1*10-5 Kg.m². J₁=4.4*10-³ Kg.m² n = gear ratio=1/10, Neglect Cm and la constant 0050 N₁ DIDamped natural frequency is not valid for an underdamped response. True False15. Find the value of R, that will produce the indicated closed-loop gain in each amplifier in Figure 12-66 Ry W Aj=50 ww 1.0k Figure 12-66 10 kn Ag=-300 Aj=8 12 k V₁₂₁M 2.2k (d) Ag=-75