2. Using figure below, change RL to 2 kand calculate the new ideal dc load voltage and ripple. 1N4001 5:1 120 V 60 Hz RL ► 5 kQ 100 µF
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Solve question no. 02 and show a clear and organized solution. Thanks!!!
Note: Since ideal is asked, use First Approximation.
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- Q ) For the full- wave uncontrolled rectifier circuit supplying a series resistive Thesupply voltage is v =300 sinωt , the supply frequency is 50 Hz and the loadparameter values are: R =05 Ω,a) Sketch the diagram of electrical circuitb) Sketch the load voltage and current waveforms .c) Calculate the load currentd) Calculate the average DC load voltage Vdc .e) Calculate the supply power to the loaQ ) For the full- wave uncontrolled rectifier circuit supplying a series resistive The supply voltage is v =300 sinωt , the supply frequency is 50 Hz and the load parameter values are: R =05 Ω, a) Sketch the diagram of electrical circuit b) Sketch the load voltage and current waveforms . c) Calculate the load current d) Calculate the average DC load voltage Vdc . e) Calculate the supply power to the load.In the Boost DA-DA converter circuit, given below; 1) Source voltage (E) value: In the voltage range [ AB ] (Volts) 2) Switching frequency (f) value: M kHz 3) Voltage between load terminals (Vload) F (Volt) 4) Capacitance value 470 (uF) 5) Pload 2 H (Watts) consumed per load The lowest L inductance value that can keep the DC-DC converter in continuous current condition under Which option do you see below as pH ? A = 10 Volts B = 19 Volts M = 23kHz F = 26 Volts H = 8 Watts
- Three-phase half wave rectifier supply resistive load with DC power of 4 kW; the ammeter connected with D1 indicates 8 A r.m.s current, and the source frequency is 50 Hz Due to system inductance the average voltage drops to 96.2% of idealized value ( when the inductance is neglected). How much will the average and r.m.s voltage s be? The r.m.s phase current and average load current Is=? & Idc=? The secondary r.m.s voltage and transformer capacity Vs & KVA? The inductances causes the voltage drop Lc=? The diode forward current and maximum inverse voltage? Now, if the phase B is failed ( disconnected), how much with new dc voltage be in %..Vdc=….% ??Design a DC power supply using FWR to perform the following requirements. Find V_dc, V_rms, V-ripple, PIV of the diode. draw the waveform of (V_in, V_out, L_in, I_load, V_diode and I_diode). Input voltage = 25 Transformer turn ratio 0.1 load resistance = 1k ohm Ripple factor = 0.025 Diode type =SiIn the figure given we have u(t)=10· coswt [V]. We assume the diodesand the A-meter (A) to be ideal.a) Plot the waveform of the current flowing through the A-m in scale.b) What is the reading of the A-m, if it is moving-coil type?c) What is the reading of the A-m, if it is moving-iron type?d) Calculate the power factor of the WHOLE structure.
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- I need a diagram of a circuit by the transform and rectification block, using the bridge rectifier circuit (KBJ1008). It can be a source of any value. You don't need to do any calculations, just a drawing of how this circuit will look, please.Make the control circuit made up of a TRIAC and an optocoupler, for one: 2000W lamp, 120vac You must choose from the following elements: optocoupler: OPTO 1: MOC3021 OPTO 2: MOC3051 Transistor: TRIAC 1:Q20115L5 TRIAC2:Q2025R5For a Full wave rectifier circuit, the AC voltage input to transformer primary is 115V. Transformer secondary voltage is 50V. Load resistor is 25 ohms. Determine Peak and Average DC Component of load voltage. (Assume ideal diodes) a.VDC (peak) = 25.4 V; VDC (average) = 18.5V b.VDC (peak) = 35.4 V; VDC (average) = 22.5V c.VDC (peak) = 38.4 V; VDC (average) = 23.5V d.VDC (peak) = 50 V; VDC (average) = 34.5V