Question Solved1 Answer Name: Country: A diode beilge rectifier with filtering capacitor feeds a lead. Diode characteristics is approximatel by a rectangulat line approximated diede chacteristics . lue- c tusf本.. Given preters: Input voltage RMS - 230 V line frequency 1-50 Hz lead active power P-1800 W Calculate for maximum voltage drop on the capacitor ay = 50 V and for given parameters and approximated diode characteristics average capacitor voltage U average load current 1, average diode curentia maximun diede voltage U power dissipation on diedes AP value of filtering capacitor C D

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Transcribed Image Text: Name: Country: A diode beilge rectifier with filtering capacitor feeds a lead. Diode characteristics is approximatel by a rectangulat line approximated diede chacteristics . lue- c tusf本.. Given preters: Input voltage RMS - 230 V line frequency 1-50 Hz lead active power P-1800 W Calculate for maximum voltage drop on the capacitor ay = 50 V and for given parameters and approximated diode characteristics average capacitor voltage U average load current 1, average diode curentia maximun diede voltage U power dissipation on diedes AP value of filtering capacitor C D
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Transcribed Image Text: Name: Country: A diode beilge rectifier with filtering capacitor feeds a lead. Diode characteristics is approximatel by a rectangulat line approximated diede chacteristics . lue- c tusf本.. Given preters: Input voltage RMS - 230 V line frequency 1-50 Hz lead active power P-1800 W Calculate for maximum voltage drop on the capacitor ay = 50 V and for given parameters and approximated diode characteristics average capacitor voltage U average load current 1, average diode curentia maximun diede voltage U power dissipation on diedes AP value of filtering capacitor C D
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1) The average load voltage and capacitor voltage are the same, as they are connected in parallel. * So, the average capacitor voltage ( UC ) is   UC = 2×Vm / 8 * Where, * Vm = Vp - ( 2 × Up ) * At a time, two diodes will conduct in bridge rectifier. * So, the voltage is double the forward voltage drop of individual diode. * Vp = Uin × √2          = 230 × √2 * Vp = 325.269 V * So,   Vm = 325.269 - 1.8  * Vm = 323.469 V and * The average capacitor voltage ( UC ) is     UC = 2 × 323.469 / 8 * The average capacitor voltage ( UC ) = 205.9 V. 2) The average load current ( IL ) is      IL = P_active / Vavg            = 1800 / 205.9 * So, the average load current ( IL ) = 8.74 amps 3) The load current is the sum of two path current and * The each path carries 1/2 times the load current via dio ... See the full answer