2. Given ID= -4mA at VGs=-2V, determine loss if Vp=-5V. a.11.11mA b.22.22mA c.33.33mA d.44.44mA
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- Xc 10 ? 100 2 R 20 2 E = 120 V ZGiven dF1/dP = 20 + 0.4P, dF2/dP = 30 + 0.4P2, andP;oss = 0.0004P? + 0.0006P? Assume the load = 1000 MW. Using the coordination equation method, and starting with PO = 500MW, P = 500MW Perform one iteration to get: Select one: a. None of these Ob. P = 775MW, P2 = 475MW, and) = 550$/MWh О с. Р. 3 675МW, P, 3 575MW,and^ — 450$/MWh O d. P1 = 875MW, P2 = 375MW, and = 650$/MWh Clear my choiceQUESTION 4 -) With the aid of suitable diagram, discuss and make comparison between an ideal AC and DC signals. For the circuit of Figure Q4a, illustrate the circuit in phasor domain equivalent circuit and hence find the input impedance, Zn 5cos2000t V 4/45° A m 10 mH ↑ Figure Q4a b) For the circuit of Figure Q4b, solve for the phasor current Is, and the real and reactive power supplied by the current source. ww k ww 50 50 2 50 μF Figure Q4b 2012 un 150 2 ww -1102 2502 150 Ω
- Solve for: a. Id (atleast 4 decimal place) b. Vgs c. Vd (Atleast 3 decimal value) d. Vs a. Id-Blank 1 mA b. Vgs-Blank 2V c. Vd-Blank 3V d. Vs-Blank 4 V c). 1MQ 1.5V. 12 V 1.2k2 Ing + Voso loss = 12 mA V₂ = -4V 1 ...Exercise-1 Vs = 240V 60Hz Is=10A/....... Power system Is ww R 1k0 Xc=-j/(2pi60x0.00016)=-1994 XL=ZL=j2pifl=j2pi60x0.142=j53.54 142mH Is=Ir+IL+Ic=.........=Vs/Ztotal=....... 160μFWhich of the following statement is true? a. Angle between load voltage and load current in a pure capacitive load is 45 degree. b. Angle between load voltage and load current in a pure capacitive load is is 180 degree. c. Angle between load voltage and load current in a pure capacitive load is is zero. d. Angle between load voltage and load current in a pure capacitive load is is 90 degree.
- Determine the branch currents Is, Igandl, in ampere. TVⒸ so www 14 Eng al. The current in ampere 11 [Select] b). The current in ampere 12 [Select] 60 www d. The current in ampere 13 [Select] 40A) find Ztot in rectangular form B) find Ztot in polar form C) find V1 or VL in polar for. ( use voltage divider rule or ohm law).Consider the following circuit used to provide power for an induative RL load. The input voltage is V-100V and the load has a 50 impedance value. The thyristor is working at a frequency fs = 2 kHz. The discharge current is to be limited to 40A and the required dv/dt is 40V/us. If the value of Ce is equal to 0.14uF, then the snubber losses are equal to: R. V. Select one: O a 5.4W O b. 7.4W 1.4W Od. 3.4W
- An AC circuit is shown in the diagram. The source of emf has a peak voltage of 9.00 V and a frequency of 60.0 Hz. Assume the wires contribute a negligible amount to the resistance and the AC source has negligible internal resistance. 524 2 493 2 1.56 k2 a) Find the equivalent resistance of the circuit. Redraw the circuit showing the equivalent resistor and use this simplified circuit for the rest of the problem. 315 2 Re = b) Find the rms current. 9.00 V, 60.0 Hz c) If the current is zero at time t= 0, find the first two times (after time t-0) that the current is equal to its rms value. and d) Find the average power dissipated by the circuit.A circuit consists of an AC voltage source, a 10-ohm resistor, and a capacitor, all connected in series. The magnitude of the impedance of the capacitor is also 10 ohms. Which of the following is true about the source voltage and the current through the circuit? a. Insufficient information is given. b. The source voltage will lag the current. c. The source voltage will lead the current. d. The source voltage will be in-phase with the current.C2 6. Figure shows a series resistance-capacitance bridge to R1 measure the capacitor, C3 and its leakage resistance of R3. The variable capacitor, C2 and resistor, R2 is adjusted alternately to null the bridge. The values of the resistors are RA given as R1 = 22 kN and R4 = 8 kN. The power supply frequency is 120 Hz. (i) Derive the equations for C3 and R3. (ii) Calculate the values of capacitor, C3 and resistor, R3 if the bridge is balanced when C2 = 4.7 µF and R2 = 150 2. [Module Test 2, March 2021] (ii) R3 = 54.55 , C3= 12.93 µF) R4 (Answer: (i) R3 = R2R4 C3 = R,C2 R1