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Question
A 220V, 50Hz ac source is connected to a coil having coefficient of self-induction of 1H and a resistance of 400 Ω. Calculate:
- the reactance of the coil.
- the impedance of the coil.
- the current flowing through the coil.
Solution
1. Reactance of coil, XL= `2pi f L`
= 2 × 3.14 × 50 × 1 Ω
= 314 Ω
2. Impedance of coil,
Z = `sqrt(R^2 + X_L^2)`
= `sqrt(400^2 + 314^2)`
= `sqrt(160000 + 98596)`
= `sqrt(258596)`
= 508.5 Ω
3. Current flowing in coil, `I =V/Z`
= `220/508.5 A`
= 0.43 A
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