Q. 144.6( 8 Votes )

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?

Answer :

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.


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