Question

A single phase full bridge inverter has RLC load with R = 4 Ω, L = 35 mH and C = 155 μF. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the angle by which the third harmonic current will lead/lag the third harmonic output voltage.

(a) 67.25°

(b) 96.4°

(c) 49.87°

(d) 81.3°

I had been asked this question during an interview.

The question is from Single Phase VSI-5 topic in portion Inverters of Power Electronics

Answer 1

Correct option is (d) 81.3°

The best I can explain: XL = 2 x 3.14 x 50 x 0.035 = 10.99 Ω

XC = 1/(2 x 3.14 x 50 x 155 x 10^-6) = 20.54 Ω

For the third harmonic component

XL(3rd harmonic) = 10.99 x 3 = 33 Ω (approx.)

XC(3rd harmonic) = 20.54/3 = 6.846 Ω

R = 4 Ω

P = tan^-1 (XL – XC)/R = 81.3°.