A 0.07 H inductor and a 12Ω resistor are connected in series to a 220 V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take π as 22/7]

Option 1 -

88 A and tan⁻¹(11/6)

Option 2 -

8.8 A and tan⁻¹(11/6)

Option 3 -

0.88 A and tan⁻¹(11/6)

Option 4 -

8.8 A and tan⁻¹(6/11)

4 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

L = 0.07H and R = 12Ω
Vs = 220V, 50Hz
XL = 2πfL = 2 (22/7) (50) (0.07) = 22Ω
Z = √ (R² + XL²) = √ (12² + 22²) = √ (144+484) = √628 ≈ 25Ω
i = V/Z = 220/25 = 8.8A
tan φ = XL/R = 22/12 = 11/6
φ = tan? ¹ (11/6)

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