Question

A single phase half bridge inverter has RLC load. The dc input voltage (Vs/2) = 115 V and the output frequency is 50 Hz. The expression for the load voltage up to the fifth harmonic will be given by

(a) 146 sin 314t + 48.81 sin 314t + 58.57 sin 318t + 28.31 sin 318t + 3.686 sin 318t

(b) 146 sin 314t + 48.81 sin (3 x 314t) + 29.28 sin (5 x 318t)

(c) 146 sin 314t + 48.81 sin (2 x 314t) + 58.57 sin (3 x 318t)

(d) none of the mentioned

I got this question by my school principal while I was bunking the class.

My doubt stems from Single Phase VSI-5 topic in division Inverters of Power Electronics

Answer 1

The correct choice is (b) 146 sin 314t + 48.81 sin (3 x 314t) + 29.28 sin (5 x 318t)

For explanation I would say: In a single phase HALF bridge inverter only odd harmonics are present. i.e. 1,3,5 etc.

Vo = (2Vs/π) sin ωt + (2Vs/3π) sin 3ωt + (2Vs/5π) sin 5ωt . . .

(2Vs/π) = 146 V

ωt = 2 x f x π x t = 2 x 3.14 x 50 x t = 314t.