Question

A coil of self inductance 20mH is connected to a a.c source of 220v and of frequrncy 50Hz .What is the inductive reactance and r.m.s current in the circuit?​

Answer 1

The inductive reactance XL = 6.25 ohm and current Iv =  35.2 A

Explanation:

Given data:

• Inductance "L" = 20.0 m.H = 25 × 10^−3 H
• Ev = 220 V
• Frequency "V" = 50 Hz

XL=?

Iv=?

Solution:

XL = ωL = 2πvL

XL = 2 × 22 / 7 × 50 × 20×10^−3

XL = 312.5 x 20 x 10^-3

XL = 6250 x 10^-3 = 6.25 ohm

Iv = Ev / XL = 220 / 6.25 = 35.2 A

Thus the inductive reactance XL = 6.25 ohm and current Iv =  35.2 A

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Answer 2

Answer:

The inductive reactance XL

6.25 ohm and current lv =

35.2 A