Question

1. A full wave rectifier, used as a battery charger, is shown in the figure below. D4 Assume battery voltage, E 12 V, battery capacity 50 watt-hours, average charging current, Io-4 A, primary input voltage is 120 V rms, 60 Hz, and the transformers (primary-secondary) turns ratio 4:1. Calculate (a) the conduction angle of the diode, (b) the value of the current limiting resistance R, (c) the power rating of R, (d) the charging time in hours.

Answer 1

Given Full wave rectifier as a battery charges since it is given with a Transformer, So output of Transformer is v_1/v_2 = N_1/N_2 = > 120/v_2 = 4/1 = > v_2 = 30v Now we can draw the circuit as We can also draw waveform as Here we can find theta = sin^-1 (E/V_m) = > theta = sin^-1 (12/30 squareroot 2) theta = 16.43 degree As we see in the waveform the conduction angle of diode is gamma = (pi - theta) - theta gamma = pi - 2 theta condition angle of diode = pi - 2 theta = > gamma_Diode = 180 degree - 2 (16.43 degree) = > gamma_Diode = 147.14 degree \r\n We know For RE Load the value of Avg current fro discontinuous condition is I_o = 1/pi R [V_m (2 cos theta) - E (pi - 2 theta)] = > But I_o given = 4 Amp Therefore 4 = 1/pi R [30 squareroot 2 [1.92] - 12 [pi - 0.574]] = > R = 1/4 pi [81.458 - 30.81]] = > R = 4.03 Omega Current Limiting Resistance Power rating = I^2 R = > power rating = (4)^2 4.03