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Question: A resistance of 300 Ω and an inductance of \(\frac{1}{\pi}\) henry are connected in series to a ac v...

A resistance of 300 Ω and an inductance of 1π\frac{1}{\pi} henry are connected in series to a ac voltage of 20 volts and 200 Hz frequency. The phase angle between the voltage and current is

A

tan⁡−143\tan^{- 1}\frac{4}{3}

B

tan⁡−134\tan^{- 1}\frac{3}{4}

C

tan⁡−132\tan^{- 1}\frac{3}{2}

D

tan⁡−125\tan^{- 1}\frac{2}{5}

Answer

tan⁡−143\tan^{- 1}\frac{4}{3}

Explanation

Solution

Phase angle tan⁡φ=ωLR=2π×200300×1π=43\tan\varphi = \frac{\omega L}{R} = \frac{2\pi \times 200}{300} \times \frac{1}{\pi} = \frac{4}{3}

∴φ=tan⁡−143\therefore\varphi = \tan^{- 1}\frac{4}{3}