8) The resusltant value of two currents that are out of phaso writh each other by 60 oloc đeg is 70 mp. If one of them is 50 cp, What is the other? Ans. 30 anp.
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- ide cal diode effechive value Frequency equal to 60 He. the voltage ripple at the terminals of R is equeal the source is sinu soidal, with equal to120V and a to 6.06 % of the arerage value of he voltage of R. at the terminals value of Hhe vollage the cueraye value of the volkage R is ĕ guual to: across O lo5 V l10 v (3 165 V 160 VExe 1. A half-wave rectifier has a source of 120 V rms at 60 Hz and an R-L load with R=10 Ω and L= 10 mH.Determine, an expression for the load current, the average current, the power absorbed by the resistor, and the power factor. Hint: Beta=3.5 Rad7) sine wave. Sketch the waveform (1 period is enough) for the resulting Vout- What are its positive and negative peak values? (AVforward,ideal = 0) For each of the ideal-diode circuits below, the input Vin is a 1kHz 5-V peak a) b) Vin Vaut Vin Vaut 1kN 1kN c) c) Vin Vout Vin Veut 1kN 1kN e) f) Vin 1kN Vaut Vin 1kN Vout 太太
- In a rectifier circuit , DC output voltage and RMS output voltage is 0.3 V & 0.55 V respectively. The form factor of this circuit is a. 3.67 b. 0.55 c. 1.35 d. 1.83Z01 @ C Asiacell l. positive clipper negative cli 03_diode_clipper_clamper_multiplie... pi 5/12 7. Questions: A. The positive peak voltage of a positive clipper is: 1- 0 V 2- 0.6 V 3- Equal to the input peak voltage 4. 1.2 V B. Why is the positive peak voltage in the negative clipper not cut? 1- The diode is forward biased 2- The diode is reversed biased C. In a positive polarized clipper we found the voltage source in series to the diode equal to be +5V. Which is the cut level of the positive voltage? 1- 0.6 2- Equal to the input peak voltage 3- 5 V 4- 5.6 V 6/121. For the circuit shown to the right vin is a 100 volt peak amplitude sine wave and D1 is an ideal diode. Determine the average power and the apparent delivered by this source. Also determine the Power Factor of the source. Vin D1 Dideal RL 10
- Find: The Current across Series Resistance in Ampere, The Power Dissipated across Zener Diode, The Vout, The Current across Zener Diode in mA, The Current across Load Resistance in mAFor a certain 12 V zener diode, a 10 mA change in zener current produces a 0.05 V change in zener voltage. The zener impedance for this current range is O a. 50 b. 10 c.0.10 Od. 100Quèstion 3 NPN BJT is considered as two diodes their anodes are connected together. O True False A Moving to another question will save this response. Chp
- The impedance of Zener diode in the Circuit when the reverse current in the diode increases from 20 mA to 30 mA as the zener voltage changes from 4.6 V to 4.71 V is O a. 90 ohms O b. 11 ohms O c. 35 ohms O d. None of the AbovePower 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 euerFor a certain 12 V zener diode, a 10 mA change in zener current produces a 0.05 V change in zener voltage. The zener impedance for this current range is Ο ° a. 50 b. 1Ω C.0.1Ω d. 10 Q