For the circuit shown below what is the current flowing through the 1.8 kohm resistor? (Determine the direction of the current flow first) 8.08 mA 11.70 mA 6.50 mA 6.28 mA 12V + Ge Si Si w 1.8 kohm 本 Si
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- What is CEMF?A three phase full wave rectifier is shown below. The input phase voltage peak is Vm=311.128 V. The load is purely resistive of value 52. The DC and RMS output voltages are respectively equal to: Secondary 太D太D,太D, 太D。太De太D, b. Select one: O a. 180V, 240.2V O b. 514.6V, 515.04V O c. 180V, 197.2V O d. 220V, 249.2VThe input to a full-wave rectifier is 20 V AC (RMS). The load resistor is 20 ohms and rectifier losses are 4W. The ripple factor of the rectifier is: O A. 0.96 OB. 0.33 O C. 0.48 O D. None of the other choices are correct O E. 0.75
- FAIRCHILD Discrete POWER & Signal Technologies SEMICONDUCTOR ru 1N4001 - 1N4007 Features • Low torward voltage drop. 10 a14 * High aurge eurrent cepablity. 0.160 4.06) DO 41 COLOR BAND DGNOTEs CAT-Cos 1.0 Ampere General Purpose Rectifiers Absolute Maximum Ratings T-26*Cuness atnerwioe rated Symbol Parameter Value Units Average Recttied Current 1.0 375" lead length a TA - 75°C Tsargei Peak Forward Surge Current 8.3 ms single halr-sine-wave Superimposed on rated load JEDEC method) 30 A Pa Total Device Dissipetion 2.5 20 Derste above 25°C Ra Tag Thermal Resistence, Junction to Amblent 5D Storage Temperature Range 55 to +175 -55 to +150 Operating Junetion Temperature PC "These rarings are imithg valuee above whien the serviceatity or any semiconductor device may te impaired. Electrical Characteristics T-20'Cunieas ofherwise roted Parameter Device Units 4001 4002 4003 4004 4005 4006 4007 Peak Repetitive Reverse Vellage Maximum RME votage DC Reverse Voltage Maximum Reverse Current @ rated VR…The input to a full-wave rectifier is 30 V AC (RMS). The load resistor is 10 ohms. The DC power at the output is: O A. 73 W O B. 18 W C. None of the other choices are correct D. 36 W O E. 146 WI need the answer as soon as possible
- In 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.Quèstion 15 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 IL-DE 205A iL-DC 220V ams 5OH23 VL-DC =20V 0.01 F 0.02 F 0.0167 F None of the above SHOT ON RED MAGIC 5S POWERED BY NUBIAThe 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 V
- What will be the output of the following circuit? (Assume 0.3V drop across the diode) Marko 18H62 V1 12V· A single-phase controlled rectifier bridge consists of one SCR and three diodes as shown in figure. The average output voltage and power delivered to battery for a firing angle of 45° are R = 52 D3 230V L= 8mH %3D 50 Hz D D2 E= 100V (a) 88.37 V, 1.837 kW (b) 191.88 V, 1.876 kW (c) 100 V, 3.525 kW (d) 88.37 V, 3.525 kWQuèstion 14 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 L-Dc こosA 22V rmsら 50H2 RL VL-DC =20V 7. 0.01 F 0.02 F 0.0167 F None of the above