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प्रश्न
A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation.
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उत्तर
Let an alternating Emf E = E0 sinωt is applied to a series combination of inductor L, capacitor C and resistance R. Since all three of them are connected in series the current through them is same. But the voltage across each element has a different phase relation with current.
The potential difference VL, VC and VR across L, C and R at any instant is given by
VL = IXL, VC = IXC and VR = IR
Where I is the current at that instant.
XL is inductive reactance and
XC is capacitive reactance.
VR is in phase with I. VL leads I by 90° and VC lags behind I by 90°
In the phases diagram,
VL and VC are opposite to each other. If VL > VC then resultant (VL − VC) is represent by OD. OR represent the resultant of VR and (VL − VC). It is equal to the applied Emf E.
`E^2 = V_R^2 + (V_L -V_C)^2`
`E^2 =I^2 +[R^2+(X_L -X_c)^2]`
`or I =E/sqrt (R^2 + (X_2 -X_c)^2)`
The term `sqrt(R^2 +(X_2 - X_c))` is called impedance Z of the LCR circuit.
`Z = sqrt(R^2 +(X_2 -X_c)^2) =sqrt(R^2 +(L omega-1/(comega))^2)`
Emf leads current by a phase angle Φ
`tan phi = (V_L -V_C)/R = (X_L - X_c)/R =(Lomega -1/(comega))/R`
When resonance takes place
`omegaL= 1/(omegac)`
Impedance of circuit becomes equal to R. Current becomes maximum and is equal to `E/R`
`omega_0 = 1/sqrt(LC)`
`f_0 = omega_0/(2pi) = 1/(2pisqrt(LC))`
This is the condition for resonance.
When at resonance f = f0 the current in the circuit is maximum and hence impedance of the circuit is maximum for values of f less than or greater than f0 comparatively small current flames in the circuit.
संबंधित प्रश्न
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