24. Find V a) 25√22-5° V b) 25√2285° V c) 25√22175° V d) 25√22-95°V e) none of these 50240° V 25 Ω ww еее 5:2 ell. 760 j100V
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- The dioces are ideal. write the Fransfer Characteidie equati ons ( &, as a funchiar g &) Plot &o agaiwet &, imdicaling all intrcepts, slopes and Volt ope lerels. Sketch bo f bq =40 sin wt. Indicate al voltage lezels - %3D 1OK lok lovin V1 R2 100k SINE(0 10 100) R1 100k D1 D V2 0 out D D2 V3 0 DC offset[V]: 0 Amplitude[V]: 10 Freq[Hz]: 100 T delay[s]: Theta[1/s]: Phi[deg]: Ncycles: anter 2019-2220 Figure 3-5 Clipper Circuit for LTSPICE Simulation C4. To specify the type of simulation, select "Simulate>>Edit Simulation Cmd" from the menu. Choose "Transient" and enter 30m for Stop time, 0 for Time to start saving data, and 1m for Maximum Timestep. Click OK. C5. Click Run to start the analysis. C6. Place two voltage probes (voltage probes appear when placed on a wire during simulation), one at the input side (at the top of V1) and another at the output side (at the top of D2) of the circuit. What difference do you observe between the input and the output waveforms?(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.
- 2) IL a cherecterio kc eq. of a conto/ oystem iJ ginen as s456+25+ 4K = 0,?u+ this eg. ia a form 1+ KGG) = 0 QS %3D draw roe t lous cw.r.t. k . cen nohDraw 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 1DC-AC Circuits Express R1 using Rx, Ry, and Rz when performing wye to delta conversion R3 Rx Rz R2 R1 Ry O A. (Rx+Ry)(Ry+Rz)(Rx+Rz) Rx В. RyRz Rx+Ry+Rz O C. Ry+Rz RXRYRZ O D. RXRYRZ Ry+Rz OE. RxRy+RyRz+RXRZ Rx
- 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 Regoogle.com/forms/d/e/1FAlpQLScF2Wzu92iVI-tEM2nxdhqXrGdR460G4QCTxaUZZeC678PqTg/formResponse مطلوب /06 *:The potential at point P due to Q1 and Q2 is Q, = 1x10-12C 0.5 m 0.5 m Qz = 1x10-12C %3D 50 cm 0.01797 V Ov O -0.01271 V O 0.01271 V O Insert PIC F10 F11 F12 F6 F7 FB Backspace 9 € P Enter HIf Is = 1x1016A and V= 26MV, Then Vout is: RL1 k2 3 V 750 mV Q1 Select one: a. None of these O b. 1.31 V O c. 1.65 V d. 2.66 V e. 1.99 V 11:20 PM EN 4/15/2021
- Solve for Rc of Figure 42. 12 V Ic Rc RB VCE B=80 IB = 40 μA Reference: Boylestad, R. & Nashehky, L. (2013). Bectronic devices and circuit theory |11th ed.]. Upper Saddle River, New Jersey: Pearson Education Figure 42 1.5 k ohms 1.875 kohms 3.531 kohms 2.75 kohms Vc=6VLet's say you measure t1/2 to be 0.037 seconds and the resistance in the circuit is 68 Ohms. Then what is the capacitance in micro Farads? Answer in micro Farads.Calculate the difference A B if A=10236.87 and B=10453.13°. Show your solution graphically.