# Integration for Power in an RL ac circuit

#### Tera-Scale

Joined Jan 1, 2011
164
Does anybody have some good links on where I can read about deriving Power in an ac cycle for an LR circuit by integration. I have equations for instantaneous current and instantaneous voltage and even the impedance.

v=315sin(100pi*t)Volts
i=3.665sin(100pi*t-0.374)Amps
z= (80 + j10*pi)Ω

I manage to obtain the perfect graps for current voltage and power but the thing is that when I am integrating (definite) I am getting zero all the time :s

#### edgetrigger

Joined Dec 19, 2010
133
integrate product of Vrms and Irms from 0 to 2∏, you'll get a finite non zero value.

#### Tera-Scale

Joined Jan 1, 2011
164
I got a strange number :s ... what does it represent exactly? It's nowhere near P=(Vrms squared)/Z or P=(Irms squared) * Z

Since I have to comment between the different results obtained. The difference is huge from the result you told me (3626.89) and the P's i mentioned of 577.25Watts.

thanks

Brandon

#### t_n_k

Joined Mar 6, 2009
5,448
As a check the average power would be Vrms*Irms*cos(θ)

Vrms=315/√2=222.74
Irms=3.665/√2=2.59

Hence

Pav=222.74*2.59*0.931=537W

#### Tera-Scale

Joined Jan 1, 2011
164
Yes, I considered that check but now I am stuck on what and how to integrate. I am asked to find the power in the first cycle and what I did was finding the area of voltage of the first half(0-0.01) and then the 2nd(0.01-0.02) And the same for the current but everything cancels out. :s Why is it that i have a negative area of power on the second half when i plot the graph it wasn't that way.

#### t_n_k

Joined Mar 6, 2009
5,448
Check this ...