Question

Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?

Answer 1

Given :

Inductance of Inductor =L= 0.5Hz

Resistance of resistor= R=100 ohms

Potential supply of voltage =v= 240V

Frequency of supply= 10kHz=10^ 4 Hz

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

a) Peak voltage,

Maximum current  I0 = V0/(√ R2 + ω²L²)

                                      = 240√2/√ (100)² x (2 π x 10⁴)² X (0.50)2²

                                      = 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

I0 is very small. Hence at high frequencies, the inductor amounts to an open circuit.

after a steady state is achieved ω = 0.

[ Applicable to Dc circuit]

Hence inductor L behaves like a pure conducting object