# Simple alternating current circuits

### Explanation of question 7

7 A 480mH coil of negligible resistance is connected to a 24V, 50Hz supply. The current taken by the coil is approximately

1. 50mA, lagging the supply voltage by 90°
2. 50mA, leading the supply voltage by 90°
3. 160mA, lagging the supply voltage by 90°
4. 160mA, leading the supply voltage by 90°

The answer is :- 3 XL = 2 x PI x F x L = 2 x 3.14 x 50 x 0.48 = 150 (approximately)

I = V / XL = 24 / 150 = approximately 160mA

For a pure inductor, the current will lag the voltage by 90°

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