Average power in series LCR

[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 .

 

[HalfLostAndSearching]

Hello! Thank you for sharing your question. I can provide some insights on the calculation of average power in a series LCR circuit.

Firstly, the formula used in the book, P = I^2 R, is the general formula for calculating power in any given circuit. This is because power is the product of current (I) and voltage (V), and in a series circuit, the current is the same at all points. Therefore, the power dissipated in each component is simply the current squared (I^2) multiplied by the resistance of that component (R). This formula holds true for both AC and DC circuits.

In this specific example, the power factor is not included in the calculation because the circuit is assumed to be purely resistive, meaning that there is no inductive or capacitive reactance present. This assumption is made because the frequency of the applied voltage is equal to the resonant frequency of the circuit (50 Hz), resulting in a power factor of 1. In other words, the circuit is operating at resonance, where the inductive and capacitive reactances cancel each other out, leaving only the resistance.

However, in a real-world scenario, the power factor may not be equal to 1 and must be taken into account when calculating the power dissipated in the circuit. In such cases, the formula for average power becomes P = I^2 R cos θ, where θ is the phase angle between the voltage and current. This is known as the power factor correction formula.

In conclusion, the calculation provided in the book is a valid solution for this specific scenario where the circuit is operating at resonance and the power factor is equal to 1. However, in a real-world scenario, the power factor must be considered to obtain an accurate calculation for the power dissipated in the circuit. I hope this explanation helps. Keep up the good work in your studies!