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

Draw a graph between frequency and current for a series L-C-R ac circuit. Derive expression for band width, showing half power frequencies on the graph.

Accepted Answer

We know that the net impedance for a series LCR circuit can be given as:

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$ and thus the current through the circuit will be: $I_{0} = \frac{E_{m}}{Z} = [\frac{E_{m}}{\sqrt{R^{2} + {(ω L - \frac{1}{ω c})}^{2}}}]$ ---- (1)

The graph between frequency and current for different values of resistance can be plotted as :

The current amplitude has a peak value at the resonant frequency that is when

$ω = ω_{0} = \frac{1}{\sqrt{L C}}$ for any other value of $ω_{0}$ , the value of current is less than the maximum value.

As is clear from the figure that as the resistance in the circuit is reduced, the resonance curve becomes sharper. The faster the current amplitude ( $I_{m}$ ) falls for changes of frequency on both sides of resonant frequency ( $ω_{0}$ ), the sharper is said to be resonance. The sharpness of resonance of a circuit is described by its quality factor $Q$ , also known as Q-factor. It is defined as the ratio of the series resonant frequency to the difference taken between the two frequencies on both sides of resonant frequency such that at each of these frequencies, the current amplitude is equal to half its maximum value. Mathematically it is given by:

$Q = \frac{ω_{0}}{ω_{1} - ω_{2}}$

Since it's a ratio of frequencies, Q is a dimensionless quantity.

The smaller the value of $Δ ω$ , the sharper is the resonance.

Here, $ω_{1} = ω_{0} + Δ ω$ and $ω_{2} = ω_{0} - Δ ω$

Thus, adding the two equations,

$ω_{1} - ω_{2} = 2 Δ ω$ .

The bandwidth of the circuit is equal to the difference between the two resonant frequencies .

The current through the circuit : $\frac{I_{m}}{\sqrt{2}} = \frac{E_{m}}{\sqrt{2} R}$

${(\frac{I_{m}}{\sqrt{2}})}^{2} = \frac{1}{2} (\frac{{E_{m}}^{2}}{R^{2}}) = {(\frac{E_{m}}{Z})}^{2} \Rightarrow Z^{2} = 2 R^{2}$

Substituting the value of impedance, From the equation (1), we get :

$R^{2} + ({X_{L}}^{2} - X_{C})^{2} = 2 R^{2}$

$X_{L} - X_{C} = \pm R$

$ω_{1} L - \frac{1}{ω_{1} C} = - R . . . . (2)$ and $ω_{1} L - \frac{1}{ω_{2} C} = + R . . . . . . (3)$

Adding the two equations i.e : eq $(2) +$ eq $(3)$

$L (ω_{1} + ω_{2}) = \frac{1}{C} (\frac{1}{ω_{1}} - \frac{1}{ω_{2}}) = \frac{1}{C} (\frac{ω_{1} + ω_{2}}{ω_{1} ω_{2}})$

$\Rightarrow ω_{1} ω_{2} = \frac{1}{L C} = {ω_{0}}^{2}$

$∴ \sqrt{ω_{1} ω_{2}} = \frac{1}{\sqrt{L C}} = ω_{0}$

Subtracting the above two equations i.e : eq $(3) -$ eq $(2)$

$(ω_{1} - ω_{2}) [L + \frac{1}{C} (L C)] = 2 R$

$\Rightarrow ∆ ω = ω_{2} - ω_{1} = \frac{R}{L}$

The difference between the frequencies of the half power points is called the bandwidth.

$∴ ∆ f = f_{2} - f_{1} = \frac{∆ ω}{2 π} = \frac{R}{2 π L}$

This is the Bandwidth of the LCR circuit

Draw a graph between frequency and current for a series L-C-R ac circuit. Derive expression for band width, showing half power frequencies on the graph.

Important Questions on Alternating Current

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