Question

A 1-\(\phi\) inverter is fed with an RLC series load of \( R = 1\,\Omega, X_L = 6\,\Omega \) and \( X_C = 7\,\Omega \). Then the phase angle between fundamental current and third harmonic voltage is

• A. \( -\tan^{-1}(18) \)
• B. \( -\tan^{-1}\left( \frac{47}{3} \right) \)
• C. \( +\tan^{-1}(18) \)
• D. \( +\tan^{-1}\left( \frac{47}{3} \right) \)

Hint:

To find phase difference between current and harmonic voltage: calculate angle of respective harmonic impedance.

Answer 1

The Correct Option is B

Impedance of RLC load for fundamental: \[ Z_1 = R + j(X_L - X_C) = 1 + j(6 - 7) = 1 - j \Rightarrow \theta_1 = -\tan^{-1}(1) \] Impedance at 3rd harmonic: \[ X_{L3} = 3X_L = 18,\quad X_{C3} = \frac{X_C}{3} = \frac{7}{3} \Rightarrow Z_3 = 1 + j(18 - \frac{7}{3}) = 1 + j\left( \frac{47}{3} \right) \Rightarrow \theta_3 = \tan^{-1}\left( \frac{47}{3} \right) \] Phase difference = \( \theta_3 - \theta_1 = \tan^{-1}\left( \frac{47}{3} \right) - (-\tan^{-1}(1)) \)