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

(a)

It is given that the inductance of the inductor, L=0⋅50H, the resistance of the

Resistor, R=100Ω, the supply voltage, V=240V and the frequency of supply,

ν=10kHz.

The formula of angular frequency is,

ω=2πν

Substitute the value of ν.

ω=2π× 10 4  rad/s

The formula of maximum current in the circuit is,

I 0 = 2 V R 2 + ω 2 L 2

Substitute the values.

I 0 = 2 ×240 ( 100 ) 2 + ( 2π× 10 4 ) 2 ( 0⋅50 ) 2 I 0 =1⋅1× 10 −2 A

Thus, the value of maximum current in the circuit is 1⋅1× 10 −2 A.

(b)

The equation of voltage is,

V= V 0 cos( ωt )

The equation of current is,

I= I 0 cos( ωt−ϕ )

The equation at t=0 is,

ωt−ϕ=0 t= ϕ ω (1)

From the formula of phase angle,

tanϕ= ωL R

Substitute the values.

tanϕ= 2π× 10 4 ×0⋅5 100 =100π ϕ= 89⋅82π 180  rad

Substitute the value of ϕ in equation (1).

t= 89⋅82π 180×2π× 10 4 t=25× 10 −6  s =25 μs

Thus, from the above calculation we can conclude that maximum current is very small at high frequencies so the inductor behaves as an open circuit. In a dc circuit after steady state is achieved, ω=0, the inductor behaves like a pure conducting element.