Name: Purely Inductive Circuit
Uploaded: 2019-07-19
Duration: 5 min 40 s
Description: How does an inductor behave in a circuit? This video will explain every parameter involved in the derivation of a pure inductor working with AC supply.

Question

What does meant by half powerpoint frequencies.

Obtain an expression for bandwidth in a LCR series circuit.

Show half power point frequency in curve between alternating current and frequency.

Accepted Answer

Half power point frequency: The input frequency at which the power dissipation in LCR circuit becomes half of maximum power.
Resonance condition:
Here,

$ω_{1} = ω_{0} - ∆_{ω}$

$ω_{2} = ω_{0} + ∆_{ω}$

Band width

$ω_{2} - ω_{1} = (ω_{0} + ∆_{ω}) - (ω_{0} - ∆_{ω})$

$ω_{2} - ω_{1} = 2 ∆ ω$

The $ω_{1}$ and $ω_{2}$ has frequency at which maximum current becomes $\frac{1}{\sqrt{2}}$ times of it's value

$V_{0} = I_{0} Z \frac{V_{0}}{Z} = \frac{I_{0}}{\sqrt{2}} \frac{I_{0} R}{Z} = \frac{I_{0}}{\sqrt{2}} [V_{0} = I_{0} R] \sqrt{2} R = Z$

$\sqrt{2} R = (\sqrt{R^{2} + {(X_{C} - X_{L})}^{2}}) 2 R^{2} = R^{2} + (\frac{1}{ω C} - ω L) R^{2} = {(\frac{1}{ω_{1} C} - ω_{1} L)}^{2}$

$R = \frac{1}{ω_{0} C (1 + \frac{∆ ω}{ω_{0}}) - ω_{0} L (1 + \frac{∆ ω}{ω_{0}})} R = ω_{0} L {(1 + \frac{∆ ω}{ω_{0}})}^{- 1} + ω_{0} L (1 + \frac{∆ ω}{ω_{0}}) R = ω_{0} L (1 - \frac{∆ ω}{ω_{0}}) - ω_{0} L (1 + \frac{∆ ω}{ω_{0}})$

$R = ω_{0} L [1 - \frac{∆ ω}{ω_{0}} - 1 - \frac{∆ ω}{ω_{0}}] R = ω_{0} L [\frac{2 ∆ ω}{ω_{0}}]$

$R = L 2 ∆ ω ∆ ω = \frac{R}{2 L} B a n d w i d t h$

Embibe Experts Solutions for Chapter: Alternating Current, Exercise 1: Rajasthan Board-2018

Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Chapter: Alternating Current, Exercise 1: Rajasthan Board-2018

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