Obtain the answers (a) to (b) in Exercise 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?
Given: Inductor = 0.5 Hz
Resistor R = 100 Ω
Voltage V = 240 V
Frequency v = 10 kHz
In Hz, the frequency v = 104 Hz
(a) Angular frequency is given by the following relation
ω = 2πv
ω = 2π × 104 rad/s
Peak voltage is given by the relation: V = √2 × V
On substituting the values
V = 240√2V
The maximum current is given by the relation:
Substituting the values we get
⇒ 
On calculating, we get
⇒ I0 = 1.1 × 10-2A
(b) tan Ф = ωL/R
tan Ф = 
on calculating, we get
or Ф = = 89.820
in rad, Ф = 89.82 π/180 rad
ωt = 89.82π/180
or 
On calculating, we get
t = 25 μs
Since I0 is very small, therefore at higher frequency the inductor is at open circuit.
In case of a dc circuit, when steady state is obtained i.e. ω = 0. Therefore, the inductor behaves like a conducting object.