Q2// The supply voltage for a single-phase full-wave uncontrolled rectifier is vs-240v2 sin3141. The rectifier is loaded with resistive load R= 100 ohms. Find the DC power absorbed by the load.
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- C4 R1 Vout For the precision rectifier circuit shown in Figure C4 what is the correct operation for the circuit if a Sine wave is 10k D1 R2 LM324 VEE applied to the input signal VIN? Vin D2 10k OUT V3 U1A Figure C4 A. When VIn is negative the circuit operates as unity voltage follower providing an in phase sine wave at VouT. When VIN is positive the circuit conducts but only to one diode drop, B. When VIn is negative the circuit operates as an inverting amplifier providing an inverted sine wave at VouT. When VIN is positive the circuit is driven to very close to zero due to amplifier gain reducing the diode drop voltage, c. When VIn is negative the circuit operates as an inverting amplifier providing an in-phase sine wave at VOUT. When VIn is positive the circuit is driven to very close to zero due to amplifier gain reducing the diode drop voltage, D. When VIn is negative the circuit operates as unity voltage follower providing an inverted sine wave at Vour. When VIn is positive the…A full-wave rectifier uses 2 diodes. The internal resistance of each diode is 20 02. The transformer RMS secondary voltage from centre tap to each end of the secondary is 50 V and the load resistance is 980 2. Mean load current will beA 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%
- Tiguie Question 3 Figure 2 shows a center-tapped full wave rectifier circuit when a 100 V peak sine wave is applied to the primary winding. If the peak input in Figure 2 is changed to 220V draw the voltage waveforms across each half of the secondary winding and across Ri for TWO (2) complete input cycle voltage. Also determine the Vout, current through R. and PIV rating for each diode. Assume Dị and D2 are practical diode model. D 2:1 +100 V -- IN4001 V OV RL -100 V 10 k2 IN4001 Figure 2Zener Diode conducts current in a. reverse direction O b. both reverse and forward O c forward direction Od. Depend of the voltagePower supply circuit is delivering 0.5 A and an average voltage 20 V to the load as shown in the circuit below. The ripple voltage of the half wave rectifier is 0.5 V and the diode is represented using constant voltage model. The smoothing capacitor value is equal to 220V omsh RL VI-DC20V 0.01 F 0.02 F 0.0167 F None of the above
- Power supply circuit is delivering 0.5 A and an average voltage 20 V to the load as shown in the circuit below. The ripple voltage of the half wave rectifier is 0.5 V and the diode is represented using constant voltage model. The smoothing capacitor value is equal to 220V ams 5OHZ İL-DC =05A RL VL-DC =20VPower supply circuit is delivering 0.5 A and an average voltage 20 V to the load as shown in the circuit below. The ripple voltage of the half wave rectifier is 0.5 V and the diode is represented using constant voltage model. The smoothing capacitor value is equal to IL-DC =0:5A RL VL-DC =20V 220V omsb O 001 F O 0.02 F O 0.0167F O None of the above ActivateDraw the input waveform and output waveform for the circuit given below with proper values marked in the figure. Assume D1 as germanium and D2 as silicon diodes. Input Vpp=20V, V1=5 v and V2=8 V. R DZ D1 Vin Vout V2 Maximum voltage of output waveform Minimum voltage of output waveform Windows hui