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 = 10^{4} Hz

(a) Angular frequency is given by the following relation

ω = 2πv

ω = 2π × 10^{4} 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

⇒ I_{0} = 1.1 × 10^{-2}A

(b) tan Ф = ωL/R

tan Ф =

on calculating, we get

or Ф = = 89.82^{0}

in rad, Ф = 89.82 π/180 rad

ωt = 89.82π/180

or

On calculating, we get

t = 25 μs

Since I_{0} 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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Consider the LCR circuit shown in Fig 7.6. Find the net current i and the phase of i. Show that . Find the impedance Z for this circuit.

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