PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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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. - PUC Karnataka Science Class 12 - Physics

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Question

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.

Solution

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

  Is there an error in this question or solution?

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Solution 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. Concept: Ac Voltage Applied to a Series Lcr Circuit.
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