Data: C1= 6uF; R1=12 Q; L1 = 10 mH; C2 = 10UF; VL1 = 380V; ƒ=1000 Hz %3D C1 R1 50 A C2 Figure 3 – Series and parallel AC network
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calculate the magnitude of apparent power
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- LVDT produces an rms output voltage of 5.0 V for displacement of 3.5 mm. Calculate the sensitivity of LVDT please explain meFacts A clipper is a device that removes either the positive half (top half) or negative half (bottom half), or both positive and negative halves of the input AC signal. In other words, a clipper is a device that limits the positive amplitude or negative amplitude or both positive and negative amplitudes of the input AC signal. In some cases, a clipper removes a small portion of the positive half cycle or negative half cycle or both positive and negative half cycles. In the below circuit diagram, the positive half cycles are removed by using the series positive clipper. Question: In your own opinion. Why do we need to clip a certain amount of voltage in positive or negative or on both sides? What is the benefit of that in our devices or circuits in doing such thing?The waveforms show a voltage (v) and a current (i) for an AC circuit. The current signal has a value of 0A at t = 25ms, 75ms, and 125ms. The voltage signal reaches a maximum value of 5 volts at t = 83.33ms and 183.33ms. 6 5 4 3 i 1 1. 0.05 0.1 1 0.15 0.2 0.25 0.3 -1 1. -2 -3 -4 -5 -6 a) What is the frequency of the current signal, in Hz? b) What is the angular frequency of the voltage signal, in rad/s? c)What is the phase difference between the current and voltage [5 pts] signals (in degrees)? d)Does the voltage signal LEAD or LAG the current signal?
- Electrical engineering problemA single-phase AC voltage controller has a resistive load R = 10 Ohm, and the RMS input voltage is Vs = 230 V, 60 Hz. The delay angle is 120 degree. Then the RMS output voltage is: Select one: O a. 101 V O b. 39 V O c. None of the above O d. 143 V O e. 215 VFind I using TEC
- Damping ratio quettion.please write introduction about AC Circuits and Parallel CircuitsA single-phase AC voltage controller has a resistive load R = 10 Ohm, and the RMS input voltage is Vs = 230 V, 60 Hz. The delay angle is 120 degree. Then power factor of the circuit is: Select one: O a. 1 O b. 0.75 O c. 0.44 O d. None of the above O e. 0.9
- in 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?1. Explain the difference between the AC and DC? Give examples 2. Explain how RC and RL works in a circuit?التاریخ | موضوع الدرس Vincoms) -220v AL=10k C = 100 ME e The FwRcct. %3D Find: 1. Dcvaltage and Current AS/p=1/5 ii the ripple Ualtage Tit ripple factor