Why Is Power Factor Not Included in Series LCR Circuit Calculations?

[Prashasti]

Hello!

Homework Statement

A sinusoidal voltage of peak 283 V and frequency 50 Hz is applied to a series LCR circuit in when R = 3ohm, L is 25.48mH and C is 796 microF. Find the power dissipated in the circuit.

The book says,
Power = I^2 R
Where I = rms current.
Now, I = 283/5 $\sqrt{2}$ = 40 A
So, power = 40×40×3 = 4800 W.

Why didn't they include 'the power factor' while calculating the power?
Also, isn't what they have calculated is a special case? I mean they've taken Z = R, ( because power = I^2 Z cos θ) which is only possible at resonance. (Cosθ = power factor, which they have taken as 1)
This can't be a general solution, right?

 

[ehild]

Hello!


Why didn't they include 'the power factor' while calculating the power?
Also, isn't what they have calculated is a special case? I mean they've taken Z = R, ( because power = I^2 Z cos θ) which is only possible at resonance. (Cosθ = power factor, which they have taken as 1)
This can't be a general solution, right?

The reactive elements (capacitor, inductor) do not dissipate power, as the phase difference between the voltage and current is ±90 degrees on them, so cosθ=0. Only the resistances dissipate power, where a power factor is 1.

To get the dissipated power, calculate the rms current I_rms=U_rms/|Z| and P = I_rms²R.

ehild

 

[Prashasti]

That means cosθ for a series rlc circuit is always 1?

 

[Vibhor]

That means cosθ for a series rlc circuit is always 1?

No .

P = i_rms²Zcosθ , where Z is the impedance of the circuit and cosθ is the power factor . But Zcosθ = R . So we have P = i_rms²R .

This is essentially what ehild has explained that power is dissipated only within the resistor .