मराठी

A Series Lcr Circuit is Connected to a Source Having Voltage V = Vm Sin ωT. Derive the Expression for the Instantaneous Current I and Its Phase Relationship to the Applied Voltage.

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प्रश्न

A series LCR circuit is connected to a source having voltage v = vm sin ωt. Derive the expression for the instantaneous current I and its phase relationship to the applied voltage.

Obtain the condition for resonance to occur. Define ‘power factor’. State the conditions under which it is (i) maximum and (ii) minimum.

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उत्तर

v = vm sin ωt

Let the current in the circuit be led the applied voltage by an angleΦ.

`i= i_m sin(omegat +phi)`

The Kirchhoff’s voltage law gives`L ((di)/dt +Ri +q/C = v)`.

It is given that v = vm sin ωt (applied voltage)

`L(d^2q)/(dt^2) +R(dq)/(dt) +q/C = v_m sinomegat    ...... (1)`

On solving the equation, we obtain

`q = q_m sin(omegat + theta)`

`(dp)/(dt) = q_momega cos(omegat +theta)`

`((d^2)q)/(dt^2)  = -q_momega^2 sin(omegat +theta)`

On substituting these values in equation (1), we obtain

`q_momega[R cos(omegat +theta)+ (X_c -X_L)sin(omegat +theta)] = v_msinomegat`

`X_c = 1/(omegaC)  X_c = omegaL`

`Z = sqrt(R^2 +(X_c - X_L)^2`

`q_momegaZ[R/Z cos(omegat+theta)+((X_c -X_L))/Z sin (omegat+theta)] = v_m sin omegat         ........... (2)`

Let `cos phi = R/2` and `(X_c -X_L)/Z = sinphi`

This gives

`tan phi = (X_c - X_L)/R`

On substituting this in equation (2), we obtain

`q_momegaZcos (omegat +theta -phi) = v_msinomegat`

On comparing the two sides, we obtain

`V_m = q_momegaZ = i+mZ`

`i_m = q_momega`

and `(theta-phi) = -pi/2`

`I = (dp)/(dt ) =q_momega cos (omegat+theta)`

              `=i_m cos(omegat+theta)`

Or

`i = i_m sin(omegat +theta)`

Where,`i_m = (v_m)/Z = (v_m)/(sqrt(R^2 +(X_c - X_L)^2)`

And

`phi = tan^-1((X_c -X_L)/R)`

The condition for resonance to occur

`i_m = v_m/sqrt(R^2 +(X_C - X_L)^2)`

For resonance to occur, the value of im has to be the maximum.

The value of im will be the maximum when

`X_c = X_L`

`1/(omega C) = omegaL`

`omega^2 = 1/(LC)`

`omega = 1/(sqrtLC)`

`2pif = 1/sqrt(LC)`

`f = 1/(02pisqrt(LC)`

Power factor = cos Φ

Where,`cosphi = R/Z = R/(sqrt(R^2 +(X_c- X_L)^2)`

(i) Conditions for maximum power factor (i.e., cos Φ = 1)

  • XC = XL

Or

  • R = 0

(ii) Conditions for minimum power factor

  • When the circuit is purely inductive

  • When the circuit is purely capacitive

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2009-2010 (March) Delhi set 3

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