Question

Find out the inductive reactance for 100 mH inductance connected through alternating current of 1 kHz frequency. If voltage of the source is 6.28 V, then find out current in the inductance.

Answer 1

Frequency of a.c. current f = 1 kHz
= 1 × 10³ Hz
Coefficient of self induction (L) = 100 mH
= 100 × 10^-3H
Potential of source (V) = 6.28 V
Inductive reactance (X_L) = 2πfL
= 2 × 3.14 × 1 × 10³ × 100 × 10^-3
= 6.28 × 100
= 628 Ω
Current in inductor (I) = \(\frac{V}{X_{L}}=\frac{6.28}{628}\)
= \(\frac{1}{100}\) = 0.01A