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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.

Correct Answer: 81.3°
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°.

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