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Question: A 12 ohm resistor and a 0.21 henry inductor are connected in series to an ac source operating at 20 ...

A 12 ohm resistor and a 0.21 henry inductor are connected in series to an ac source operating at 20 volts, 50 cycle/second. The phase angle between the current and the source voltage is

A

30°

B

40°

C

80°

D

90°

Answer

80°

Explanation

Solution

tan⁡φ=ωLR=2π×50×0.2112=5.5⇒φ=80o\tan\varphi = \frac{\omega L}{R} = \frac{2\pi \times 50 \times 0.21}{12} = 5.5 \Rightarrow \varphi = 80^{o}