A. Explain the mode of operation for complementry commutation circuit. Find the circuit turn off time if the load resistances R1-R2-5 capacitance C-7.5 μF. V = 100 volts.
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off time if the load resistances R1-R2-5 capacitance C-7.5 μF. V = 100 volts."
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- 3 A peak rectifier (peak tollower) Cir cuit is gian below. Vs is a GOH3 Sinuscidal Vollge with peak value Upz50V. 50V. The load resistance R= Sk . Find the Vallke of the capacitance C Such that the peak- to-peak ripple 2 volts. (idenl diede) Voltage Vr VsPlease show steps and work1) AC power in a load can be controlled by using a. Two SCR's in parallel opposition b. Two SCR's in series c. Three SCR's in series d. Four SCR's in series 2) The advantage of using free - wheeling diode in half controlled bridge converter is that a. There is always a path for the ac current independent of the ac line b. There is always a path for the dc current independent of the ac line c. There is always a path for the dc current dependent of the ac line d. There is always a path for the ac current dependent of the ac line 3) Silicon controlled rectifier can be turned on a. By applying a gate pulse and turned off only when current becomes zero b. And turned off by applying gate pulse c. By applying a gate pulse and turned off by removing the gate pulse d. By making current non zero and turned off by making current zero
- In the full-wave rectifier circuit of figure , the transformer has a turns ratio of 1:2. The transformer primary winding is connected across an AC source of 230V (rms), 50 Hz. The load resistor is 50ohms. For this circuit, determine the peak-to-peak ripple in the output voltageIn the full-wave rectifier circuit of figure, the transformer has a turns ratio of 1:2. The transformer primary winding is connected across an AC source of 230V (rms), 50 Hz. The load resistor is 50ohms. For this circuit, determine the peak-to- peak ripple in the output voltage * Di i v2 = VmSin ot RL Vs(N V1 Vo + V3 = VmSin ot D2 Full-wave rectifier- Circuit operation during positive half cycle 460 V 207 V 325.3 V None of the above ell ellIn the full-wave rectifier circuit of figure, the transformer has a turns ratio of 1:2. The transformer primary winding is connected across an AC source of 230v (rms), 50 Hz. The load resistor is 50ohms. For this circuit, determine the peak-to- peak ripple in the output voltage * Di v: = VmSin ot Vs V. + Vy = VSin ot D2 Full-wave rectifier. Circuit operation during positive half cycle ell + ミ」
- An unipolar PWM full bridge circuit has a Vr.rms of 125V at a duty cycle D1 of 0.7. What is Vd? What is Vo.rms?In the full-wave rectifier circuit of figure , the transformer has a turns ratio of 1:2. The transformer primary winding is connected across an AC source of 23oV (rms), 50 Hz. The load resistor is 50ohms. For this circuit, determine the peak-to- peak ripple in the output voltage * Di V2 = VmSin cot Vs (N Vo + V3 = VmSin ot D2 Full-wave rectifier- Circuit operation during positive half cycle 460 V 207 V 325.3 V O None of the above ell ellIn the circuit of the following figure, the input voltage Vs is 15 volts rms with a frequency of 60 Hz, R equals 150 Ohms and C equals 100,000 Pico Farads. The diodes are Germanium (Vd = 0.2 volts) and the Zener diode is 12 volts. a) The magnitude of the ripple voltage at Cb) The Magnitude of the Peak Inverse Voltage (PIV) for D1 and D2.
- A three phase full wave rectifier is shown below along with peak phase voltage Vm-169.7V. The load is purely resistive. The rectifier delivers Ipc = 100 A and the source frequency is 60 Hz. The DC output voltage is %3D VDc=280.7V and the output RMS voltage is equal to VRMS=280.93V. The efficiency, FF and RF are respectively equal to: Secondary D D D, R AD. Z D. D, Select one: O a. 99.83%, 100.08%, 4% O b. 99.83%, 55.08%, 4% O C. 99.83%, 100.08%, 16% O d. 87.83%, 100.08%, 4%A buck-boost converter operates with an input battery. It converts +12 V to –12 V at apower level of about 75 W. the switching frequency is 120 kHz. The switches have200 ns switching time. The battery has an internal series resistance of 0.2 and seriesinductance of 200 nH. (a) What is the operating value of the duty ratio? What power is lost in the battery resistance? (b) Propose an interface structure to improve operation and decrease losses. What are the duty ratio and battery resistance loss with your interface in place?Please help with the solution.
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