Question

In a center tapped full wave rectifier, the rms half secondary voltage is 9V. Assuming ideal diodes and load resistance RL = 1K. solve
(i). Peak current
(ii).D.C load voltage
(iii). RMS current.
(iv). Ripple factor
(v). Efficiency.​

Answer 1

Answer:

Explanation:

We have to load voltage

The voltage across half the secondary windings is

V=110sin100πt

(i) Here peak value of voltage, V0=110V

Peak value of current, I0=V0Rd+RL

11020+1000=0.108A=108mA

(ii) d.c. value of current, Ia.c.=2I0π=2×1083.142=68.7mA

(iii) rms value of current, Irms=I02–√=1081.414=76.