Name: Energy Losses in a Transformer
Uploaded: 2019-07-19
Duration: 8 min 12 s
Description: There are different types of energy losses in a transformer. This video explains those losses in detail and also talks about the efficiency of ideal and prac...

Question

The amplitude of the current in series LCR circuit connected to an AC frequency $" ω "$ is given by,

$i_{m} = \frac{v_{m}}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}$ .

Where $X_{L} and X_{C}$ are inductive and capacitive reactants respectively and $" v_{m} "$ is the amplitude of the voltage. Starting from this equation show that the sharpness of resonance the circuit is equal to the quality factorof the circuit.

Accepted Answer

The depletion of an oscillating wave with respect to time is called the sharpness of the resonance.

The Q-factor of resonance is defined as the ratio of the resonance's centre frequency to its half-power bandwidth. It tells us how sharp, or steep, the resonance is. In LCR Circuit, the ratio of resonance frequency to the difference of its neighbouring frequencies so that their corresponding current is 1/2 times of the peak value is called Q-factor of the circuit.

Given that $i_{m} = \frac{v_{m}}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}} o r \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}} = {(\frac{i_{m}^{m a x}}{I_{m}})}^{} R$

Squaring both sides, we have, $R^{2} + {(ω L - \frac{1}{ω C})}^{2} = {(\frac{i_{m}^{m a x}}{I_{m}})}^{2} R^{2}$

We suppose that at an angular frequency $ω_{1}$ , $i_{m} = \frac{1}{\sqrt{2}} i_{m}^{\max}$ so we have,

$R^{2} + {(ωL - \frac{1}{ωC})}^{2} = {(\sqrt{2})}^{2} R^{2} = 2 R^{2} or {(ωL - \frac{1}{ωC})}^{2} = R^{2}$

As $ω_{1} = ω_{r} + △ ω$ , we get $(ω_{r} + △ ω) L = \frac{1}{(ω_{r} + △ ω) C} = R$

Or $ω_{r} L (1 + \frac{△ ω}{ω_{r}}) - \frac{1}{ω_{r} C (1 + \frac{△ ω}{ω_{r}})} = R$

Since the resonance frequency $ω_{r} = \frac{1}{\sqrt{LC}}$ , therefore in terms of $L$ , we get,

$ω_{r} L (1 + \frac{△ ω}{ω_{r}}) - \frac{ω_{r} L}{(1 + \frac{△ ω}{ω_{r}})} = R$

Or $ω_{r} L [(1 + \frac{△ ω}{ω_{r}}) - {(1 + \frac{△ ω}{ω_{r}})}^{- 1}] = R$ as $\frac{△ ω}{ω_{r}} < < 1$ , we get,

$ω_{r} L [(1 + \frac{△ ω}{ω_{r}}) - {(1 + \frac{△ ω}{ω_{r}})}^{}] = R or △ ω = \frac{R}{2 L}$ .

The sharpness of resonance is $\frac{ω_{r}}{bandwidth} = \frac{ω_{r}}{2 △ ω} = \frac{ω_{r} L}{R}$ .

Since $Q$ factor is given by $Q = \frac{ω_{r} L}{R} = \frac{1}{ω_{r} CR}$ , so we get,

$\frac{ω_{r}}{2 △ ω} = Q$ . This proves that sharpness is equal to $Q$ factor.

Important Questions on Alternating Current

Learn and Prepare for any exam you want !