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Q.

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 (tan^-15.5 = 80⁰ approx)

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a

40°

b

90°

c

30°

d

80°

answer is C.

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Detailed Solution

$tanφ = \frac{ωL}{R} = \frac{2 π \times 50 \times 0 .21}{12} = 5 .5 \Rightarrow φ = 80^{o}$

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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 (tan-15.5 = 800 approx)