CBSE (Science) Class 12CBSE
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

A Voltage V = V0 Sin ωt is Applied to a Series LCR Circuit. Derive the Expression for the Average Power Dissipated Over a Cycle. - CBSE (Science) Class 12 - Physics

Login
Create free account


      Forgot password?

Question

A voltage V = V0 sin ωt is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition (i) no power is dissipated even though the current flows through the circuit, (ii) maximum power is dissipated in the circuit?

Solution

Voltage V=V0sinωt is applied to an series LCR circuit.

Current is  π2" role="presentation" style="position: relative;">π2

`I_0=V_0/Z`

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

Instantaneous power supplied by the source is

P=VI

=(V0sinωt)×(I0sin(ωt+ϕ)

`=(V_0I_0)/2[cosphi-cos(2omegat+phi)]`

The average power over a cycle is average of the two terms on the R.H.S of the above equation. The second term is time dependent; so, its average is zero.

 `P=(V_0I_0)/2cosphi`

`=(V_0I_0)/(sqrt2sqr2)cosphi`

 =VIcosϕ

P=I2Zcosϕ

cosϕ  is called the power factor.

Case I

For pure inductive circuit or pure capacitive circuit, the phase difference between current and voltage is `pi/2`

`:.phi=pi/2,cosphi=0`

Therefore, no power is dissipated. This current is sometimes referred to as wattless current.

Case II

For power dissipated at resonance in an LCR circuit,

`X_C-X_L=0, phi=0`

∴ cos ϕ = 1

So, maximum power is dissipated.

  Is there an error in this question or solution?
Solution A Voltage V = V0 Sin ωt is Applied to a Series LCR Circuit. Derive the Expression for the Average Power Dissipated Over a Cycle. Concept: Ac Voltage Applied to a Series Lcr Circuit.
S
View in app×