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jangyasinni
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Asked: 2024-08-18T23:01:18+05:30 In:

A single phase full bridge inverter has RLC load. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the expression for the load voltage up to the fifth harmonic

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Poll Results

0%a) 292 sin 314t + 97.62 sin 314t + 58.57 sin 318t + 28.31 sin 318t + 3.686 sin 318t ( 0 voters )
100%b) 292 sin 314t + 97.62 sin (3 x 314t) + 58.57 sin (5 x 318t) ( 1 voter )
0%c) 292 sin 314t + 97.62 sin (2 x 314t) + 58.57 sin (3 x 318t) + 28.31 sin (4 x 318t) + 3.686 sin (5 x 318t) ( 0 voters )
0%d) 292 sin 512t + 25.62 sin 249t + 6.74 sin 508t ( 0 voters )
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    2 Answers

    1. jangyasinni
      Teacher

      b

      Explanation: In a single phase full bridge inverter only odd harmonics are present. i.e.

      1,3,5 etc.

      Vo = (4Vs/π) sin ωt + (4Vs/3π) sin 3ωt + (4Vs/5π) sin 5ωt

      (4Vs/π) = 292 V

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

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    2. Quearn
      Best Answer
      Quearnist

      To find the expression for the load voltage of a single-phase full bridge inverter with an RLC load, we can utilize Fourier series expansion to account for the fundamental frequency and the higher harmonics up to the fifth harmonic.

      1. DC Input Voltage (V_dc): 230 V
      2. Output Frequency (f): 50 Hz
      3. Angular Frequency (ω): ( omega = 2pi f = 2pi times 50 = 100pi ) rad/s
      4. Harmonics: The harmonics for a full bridge inverter can be expressed as ( n times f ) where ( n ) is an integer.

      The fundamental (first) harmonic and its higher harmonics up to the fifth can be expressed as:

      – Voltage waveforms:

      – Fundamental (1st harmonic): ( V_1(t) = frac{2V_{dc}}{pi} sin(100pi t) )

      – 3rd harmonic: ( V_3(t) = frac{2V_{dc}}{3pi} sin(300pi t) )

      – 5th harmonic: ( V_5(t) = frac{2V_{dc}}{5pi} sin(500pi t) )

      Overall Load Voltage Expression:

      The overall load voltage ( V_{load}(t)

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