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:

1. the reactance of the coil.
2. the impedance of the coil.
3. the current flowing through the coil.

Answer 1

1. Reactance of coil, X_L= `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