Question

An alternating voltage of 220 V, 60 Hz is applied
inductor of inductance 0.2 mH. The
reactance of the inductor is
(1) 7.54 x 10-22
(2) 9.54 x 100
(3) 6,54 10%
(4) 3,54 x 10 % 2​

Answer 1

$\color{darkblue}\underline{\underline{\sf Given-}}$

• Frequency (f) = 60Hz
• Inductance (L) = ${\sf 0.2×10^{-3}H}$

$\color{darkblue}\underline{\underline{\sf To\:Find-}}$

• Inductive Reactance ${\sf (X_L)}$

❏$\Large\color{violet}\underline{\boxed{\sf X_L=\omega L}}$

❏$\color{violet}\Large\underline{\boxed{\sf \omega =2πf}}$

$\implies{\sf X_L=2πfL}$

$\implies{\sf X_L=2×3.14×60×0.2×10^{-3}}$

$\implies{\sf X_L=75.6×10^{-3}}$

$\color{red}\implies{\sf X_L=7.5×10^{-2}\:ohms}$

$\color{darkblue}\underline{\underline{\sf Answer-}}$

The reactance of the inductor is $\color{red}{\sf 7.5×10^{-2}\:ohms}$.

Answer 2

Answer:

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