Induction of inductor, L = 0.5 Hz

Resistance of the resistor R = 100 Ω

Voltage V = 240 V

Frequency v  = 10 KHz = 10⁴ Hz

Angular frequency  ω = 2πv = 2 x π x 10⁴ rad/s

1. a) Peak voltage,

Maximum current  I₀ = V₀/√ R² + ω²L²

                                       = 240√2/√ (100)² x (2 π x 10⁴)² X (0.50)²

                                       = 1.1 X 10^-2 A

(b) For phase difference ɸ

tan ɸ = ωL/R =  2 x π x 10⁴ X 0.5/100 = 100 π

ɸ = 89.82° = 89.82 x π/180 rad

so, ωt = 89.82 x π/180

t = 89.82 x π/180 x 2 π x 10⁴ = 25 μs

It can be observed that I₀ is very small. Hence at high frequencies, the inductore amounts to an open circuit.

In DC circuit, after a steady state is achieved ω = 0.

Hence inductor L behaves like a pure conducting object