Homework_4

The following data were obtained on a 50 kVA 2400/120 V transformer: Open circuit test, instruments on low voltage side: Waltmeter reading = 396 Watts Ammeter reading = 9.65 amperes Voltmeter reading = 120 Volts Short circuit test, instruments on high voltage side: Waltmeter reading = 810 Watts Ammeter reading = 20.8 amperes Voltmeter reading = 92 Volts Find the efficiency when rated kVA is delivered to a load having a power factor of 0.8 lagging.The following data were obtained on a 50 kVA 2400/120 V transformer : Open circuit test, instruments on low voltage side: Waltmeter reading = 396 Watts Ammeter reading = 9.65 amperes Voltmeter reading = 120 Volts Short circuit test, instruments on high voltage side: Waltmeter reading = 810 Watts Ammeter reading = 20.8 amperes Voltmeter reading = 92 Volts Find the efficiency when rated kVA is delivered to a load having a power factor of 0.8 lagging.

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The arc length voltage characteristic of a DC arc is given by the equation V=24+4L where 'V' is the arc voltage in volts and L' is the length in mm. The static volt-ampere characteristic of the power source is approximated by a straight line with no load voltage of 80V and the short circuit current of 600A. Then the optimum arc length is (mm).

4. A circuit contains a 20-2 resistor and an inductor with an inductance of 0.093 H. If the circuit has a frequency of 60 Hz, what is the total impedance of the circuit? 5. An R-L series circuit has a power factor of 86%. By how many degrees are the voltage and current out of phase with each other? 6. An R-L series circuit has an apparent power of 230 VA and a true power of 180 W. What is the reactive power? 7. The resistor in an R-L series circuit has a voltage drop of 53 V, and the inductor has a voltage drop of 28 V. What is the applied voltage of the circuit?

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.

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7. A step-down transformer has a turn ratio Np/Ns of 4:1. If the primary voltage Vp is 120 V(rms), the secondary voltage Vs is (а) 480 V (b) 120 V (c)30 V (d) It cannot be determined.

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Example 5: A 2400 V/400V single-phase transformer takes a no-load current of 0.5A and the core loss is 400 W. Determine the values of the angle p. and the magnetising and core loss components of the no-load current.

2. For the circuit shown in fig. 2. Find the Node Voltages using Nodal Analysis. W voor 20 200 10 I - 6A 20° 20 100 宣 Fig.2 E-30V 20°

Find out the error for the given program io = 1.7e-8; % dark saturation current in amperes tc = 27; % cell temperature in degree celsius isc = 250 ; % short circuit current density A/ square meter pin = 820; % solar radiation in w/square meter k =1.381e-23; % Boltzmann's gas constant e= 1.602e-19; % electronic charge t= 273+tc % cell temperature in Kelvin vmax=0.526; % mayi mum voltage in volts imax = (e*vmax)*(isc+io)/(k*t+e*vmax) pmax= vmax*imax efficiency = (pin/pmax)*100 O error in efficiency O no error O Error in pmax O Error in imax O no output

If the impedance Z1 = 10 +j10 and the impedance Z2 = 10 - j0 are connected in series, the equivalent impedance would be: O A. 20-j10 ohms O B. 20+ j20 ohms O C. None of the other choices are correct O D. 20 ohms O E. 20+ j 10 ohms

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In the network below, find the power associated to the current source, for V1 = 10 V. Vi + 2 A (T 14 V Select one: а. -14 b. -4 С. -8 d. -20 е. -18 2.