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Derive the Formula for Resonant Frequency of the Circuit with a Pure Capacitor in Parallel with a Coil Having Resistance and Inductance. - Basic Electrical and Electronics Engineering

Sum

Derive the formula for resonant frequency of the circuit with a pure capacitor in parallel with a coil having resistance and inductance. Find the expression for dynamic resistance of this parallel resonant circuit.

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Solution

Consider a parallel circuit consisting of a coil and a capacitor as shown below. The impedances of two branches are:-

`overline(Z_1)=R+jX_L`       `overlineZ_2=-jX_c`

`overlineY_1=1/Z_1=1/(R+jX_L)=(R-jX_L)/(R^2+X_L^2)`      `overline(Y_2)=1/overline(Z_2)=1/(-jX_C)=j/X_C`

Admittance of the circuit `overline(Y)=overline(Y_1)+overline(Y_2)`

`overline(Y)=(R-jX_L)/(R^2+X_L^2)+j/X_C   =  R/(R^2+X_L^2)-j(X_L/(R^2+X_L^2)-1/X_C)`

At resonance the circuit is purely resistive. Therefore, the condition for resonance is.

`X_L/(R^2+X_L^2)-1/X_C=0`

`X_L/(R^2+X_L^2)=1/X_C`

`X_LX_C=R^2+X_L^2`

`omega_oL 1/(omega_0C)=R^2+omega_0^2L^2`

`omega_0^2L^2=L/C-R^2`

`omega_0=sqrt(1/(LC)-R^2/L^2)`

`f_0=1/(2pi)sqrt(1/(LC)-R^2/L^2)`

Where 𝑓0 is called as the resonant frequency of the circuit.
If R is very small as compared to L then

`omega_0=sqrt1/(LC)`

`f_0=1/(2pisqrt(LC))`

DYNAMIC IMPEDANCE OF A PARALLEL CIRCUIT.

At resonance the circuit is purely resistive the real part of admittance is `R/(R^2+X_L^2)`.Hence the dynamic impedance at resonance is given by,

`Z_D=(R^2+X_L^2)/R`\

At resonance ,

`R^2+X_L^2=X_LX_C=L/C`

`Z_D=L/(CR)`

Concept: Series and Parallel Resonance
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