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- Draw a seven-segment display, show the number highlighted in figure 1 (d3 -d0 are the input, a -g are the outputs). III. Simulation Wavetorm Editor - Cers/Ahmad Fainuz/Desanop/DLD Lab Practice/bedseg - beaeg - bcaseg 2018o03121936sim.vwwt Ee ep Ea w smutation search altera.com Haster Time Bar: ops Pointer: s29 us Start End iterval 529 us o ps 0 ps २ 2.56 us S12 us 75 us 10.24 us 12 s 15.36 us 172 us value at Name Ops do Figure 1For the circuit shown below, determine the voltage of the -j10 capacitive reactance. -jsa jl0n 0.5L-90 ILOA(1 -j102 jsn 52 102 AASAP
- The sum of the following two e.m.fs will be e1 = 10 sin wt and e2 = 10 cos wt O 10 O 20 sin wt O 14.14 cos wt O 14.14 sin (wt + TT/4)Please show your detailed solution Number 14 The current in a series inductive circuit is 7.5 A at 25 Hz. The circuit takes 425 W, and the power factor is 0.47. The resistance of the circuit is A. 7.32 ohms C. 7.98 ohms B. 7.86 ohms D. 7.56 ohmsA certain impedance is given by 1012° N. Does the current lead or lag the voltage, and by how many degrees? Fill in the blanks. The current the voltage by
- 8PFE-5 What is the current I, in the circuit ? 6ej2t 16 -j1Q I₁₁ 0(b) Figure Q2 shows a power system network. The generators at Bus 1 and Bus 2 are represented by their equivalent current sources with the reactance in per unit on a 100MVA base. The lines are represented by n model where the series and shunt reactances are expressed in per unit on a 100MVA base as well. However, the load located at Bus 3 and Bus 4 are expressed in MW and MVar. Reconstruct the network impedance shown in the figure into the network admittance by assuming the voltage magnitude at Bus 3 and Bus 4 are 1.0 per unit, respectively. (i) (11) Calculate the bus admittance matrix of the system.P 269 W, Q = 150 VAR (capacitive). The power in the complex form is: %3D Select one: a. 150 + j269 VÀ b. None of these c. 150- j269 VA d. 269-j150 VÀ
- "If lab is equal to 595.8252306 and lb is 344296, the phase sequence is:" O POSITIVE O NEGATIVEFor this question i get all the way to the point where i have a + bj [ 0.0166 + 0.1502j ] however, calculating the phase of this specific question just does not match up with the answer provided by my university. the formula that i am using is that the phase = 90-arctan(Imaginary/Real) which in this particular problem should add output a phase of -96,31 degrees. However it does only output 6,31 and the input voltage does not contain -90 so i dont get it.Ql: For the circuits shown in figures. If the temperature has changed from 25°C to 65°C, find Alc. le=1.5mA at 25°C, R =250k2, R=1k and Re-0.5k2. 25°C 65°C B55 70 Ico Rf Re InA 16nA VBE 0.6 0.5 Re