Calculate the value of the capacitor in the ful wave rectifier filter as shown below. Knowing that the ripple percent is 5% 10.1 120 V Ourpon 750 uF O 1000 uf O 660 uF O 2200 uF Other value O
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- What is CEMF?A 60 Hz sinusoidal voltage is applied to an SCR full-wave rectifier. If the DC output voltage is 58 V and the firing angle of the SCR is 135 degrees, the RMS voltage at the output would be: O A. 98 V O B. None of the other choices are correct O C. 266 V O D. 399 V O E. 133 VA 60 Hz sinusoidal voltage is applied to an SCR full-wave rectifier. If the DC output voltage is 74 V and the firing angle of the SCR is 60 degrees, the RMS voltage at the input would be: O A. 220 V O B.99 V O C. None of the other choices are correct OD. 110 V O E.256 V
- CIRCUIT DIAGRAM : D1 IÑ4001 AC Source RL 1K Q VOUT PROCEDURES : 1. Construct the half-wave rectifier circuit as shown. 2. Monitor the input voltage using the oscilloscope. Adjust the oscilloscope to show at least two complete cycles of the input signal. Draw the input waveform in Graph 1 and measure the peak-to-peak value. Record this in Table 1. Convert this value into RMS value using the formula: 3. 4. Vp-p Vp-p P-P VRMS 2.828The power output of the half-wave rectifier is rated at 102.4 W with a load resistor of 40 ohms. The RMS voltage at the output is: O A. None of the other choices are correct O B. 12.5 V O C. 102 V O D. 25.5 V O E. 51 VPower 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 Activate
- A 1 - ∅ full-wave rectifier is made by using thyristors. If the peak value of the sinusoidal input voltage is Vm and the value of the delay angle is 45 degree, find the average value of output voltage Select one: a. 0.65 VM b. 0.85 VM c. 0.45 VM d. 0.25 Vm/1/A single-phase, half-wave, controlled has a source 220V at 50 HZ and is eding a load R=2002 and L=25 mH, the firing angle a=60° and the current tinction angle ß=225° Determine the average load voltage and current. O Determine the rms load voltage and current. led rectifier the value of the outputQuestion 3 Consider the following cirucit: in + V1 R1 5 PULSE (0 2 1 1 1 1) (Note again the unrealistic diode characteristics.) The input is a ramp: 2.0V- 1.6V- 1.2V- 0.8V- 0.4V- 0.0V- 0.0s 0.2s 0.4s 0.6s 0.8s out .tran 2 .model diode D(Ron=5 Vfwd=1 Roff=1G) V(in) D1 diode 1.0s 1.2s 1.4s 1.6s 1.8s 2.0s The output voltage is similar to the input voltage, except that it breaks at a certain voltage and reaches a different final value. Find the break voltage and the final value of the output.
- A 60 Hz sinusoidal voltage with an RMS value of 120 V is applied to a single-phase full-wave SCR rectifier. The DC output voltage of the rectifier is 101 V, the firing angle (in degrees) of SCR is would be: O A. 36 O B. 45 O C. None of the other choices are correct O D. 30 O E. 60Power 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 c 20:5A L VI-DC20V 220V omsh T 0.01 F 0.02 F 0.0167 F None of the above euerDuring the operation of a half-wave rectifier with a capacitor- filter, the ripple factor can be lowered by the value of the filter capacitor or. the load resistors. O increasing, increasing O decreasing, decreasing O increasing, decreasing O decreasing, increasing