# Homework Help: Average power in series LCR

1. Sep 17, 2014

### Prashasti

Hello!
1. The problem statement, all variables and given/known data
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?

2. Sep 17, 2014

### ehild

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 Irms=Urms/|Z| and P = Irms2R.

ehild

3. Sep 18, 2014

### Prashasti

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

4. Sep 18, 2014

### Vibhor

No .

P = irms2Zcosθ , where Z is the impedance of the circuit and cosθ is the power factor . But Zcosθ = R . So we have P = irms2R .

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

Last edited: Sep 18, 2014
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