A rectifier ammeter has the following components: PMMC instrument with FSD = 200 !LA and Rm =900 n; current transformer with Ns == 600 and Np == 5; diodes with VF = 0.3 V; meter series resistance Rs =270 kn; transformer shunt secondary resistance RL =98.7 kn. Calculate the level of transformer primary current for instrumentFSD.

the three phase half wave uncontrolled rectifier is operated from 500v,50Hz supply at secondary side and the load resistance is 25 ohms. if the source inductance is negligible, determine a) rectification efficiency b) form factor  c) ripple factor d) transformer utilization factor e) peak inverse voltage of each diode

A load is connected in the Full Bridge rectifier circuit (Four Diodes). Usedtransformer primary voltage value is 220 V (rms). The transformer winding ratio is 1:10.Using the 0.7 V model of the diode,Load Rating: 7 K Ohmsa) How many % of conduction does each diode branch stay on?b) Find the average values ​​of the load voltage.c) Find the average values ​​of the load current.d) Find the effective value of the load voltage.e) Find the effective value of the load current.

A step-down transformer supplies 25Vrms to a simple half wave rectifier power supply which is connected to a load resistance of 912ohm. The diode breakdown voltage, Vf is 0.7 V. Calculate the peak signal voltage received by the diode

A R-load half-wave rectifier used sinusoidal voltage source with a peak value (35.353kV). The firing angle is (30). The total resistance is 50k2 and frequency is 50HZ. Calculate (in details): load direct voltage, maximum direct voltage, normalized voltage and effective voltage, power delivered to the load, power factor, form factor, ripple factor?

A three phase full wave rectifier is shown below along with the peak input 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 Vpc=280.7V and. The RMS transformer secondary current is equal to 81.72A, the output RMS voltage is equal to VRMS=280.93V and the output resistance is equal to 2.8070. The TUF would be equal to: Secondary a平a字a辛 R 太D。太D。太D. Select one: 0.875 Ob. 0.955 O C. 0.551 O d. 0.432 Compared with power transistors, thyristors have the following characteristic

The main advantage of a bridge rectifier using tha same input transformer as a full-ave rectifier is that _______________. a) İt will double the output voltage. b) less current is required fort he diodes to conduct. c) the ripple frequency is twice that of the full wave rectifier. d) the ripple frequency is on-half that of the full wave rectifier.

For a full wave controlled rectifier with pure resistive load (R=100),the supply voltage is (V-50Sin314t). When the firing angle (a =0%) determine the followings: 1) The mean (dc) value of the load current. 2) The RMS value of the load voltage.

The peak output voltage for the bridge rectifier in the figure is_______Volts. Assuming ideal diodes, the PIV rating required for the diodes is ______Volts. The transformer is specified to have 12 Vrms secondary voltage for the standard 110 V across the primary.

A single-phase center-tapped rectifier has an R-L of R= 20 0, L = 50mH, and a 240 V ac 50 Hz source voltage. Determine: i. The average and rms currents of the diodes and the load. ii. The rms and average 50 Hz source currents. iii. The power absorbed by the load. iv. The supply power factor.