Question

In a centre tapped full wave rectifier, RL=1KΩ and for diode Rf=10Ω. The primary voltage is 800sinωt with transformer turns ratio=2. The ripple factor will be _________

(a) 54%

(b) 48%

(c) 26%

(d) 81%

This question was posed to me during an online interview.

My question is taken from Full-wave Rectifier topic in division Application of Diodes of Electronic Devices & Circuits

Answer 1

Correct answer is (b) 48%

To elaborate: The ripple factor ϒ= [(IRMS/IAVG)^2 – 1]^1/2. IRMS =Im /√2=Vm/(Rf+RL)√2=200/1.01=198.

(Secondary line to line voltage is 800/2=400. Due to centre tap Vm=400/2=200)

IRMS=198/√2=140mA, IAVG=2*198/π=126mA. ϒ=[(140/126)^2-1]^1/2=0.48. So, ϒ=48%.