Power Calculation of general electric circuit

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SUMMARY

The discussion focuses on calculating the average power of a general electric circuit where voltage leads current by a phase difference, denoted as ##\theta##. The instantaneous power is defined as ##p = V_m I_m \sin(\omega t + \theta) \sin(\omega t)##. The average power over one cycle is calculated using the formula ##P_{avg} = \frac{P_{1cycle}}{Duration of 1 Cycle}##, with the duration being ##2\pi##. The integration can be performed either with respect to time or angle, both yielding the same average power value.

PREREQUISITES
  • Understanding of sinusoidal waveforms and phase differences in AC circuits
  • Knowledge of instantaneous power calculations in electrical engineering
  • Familiarity with integration techniques in calculus
  • Basic concepts of average power in alternating current (AC) systems
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  • Study the derivation of average power in AC circuits using integration techniques
  • Learn about the power factor and its implications in electrical systems
  • Explore the differences in power calculations for resistors, inductors, and capacitors
  • Investigate the use of phasors in analyzing AC circuits
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Electrical engineers, physics students, and anyone involved in the analysis of alternating current circuits will benefit from this discussion.

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TL;DR
I want to calculate the power of general circuit with voltage leading the current by phase difference of Theta
I am trying to calculate the power calculation of a general circuit with voltage leading the current by a phase difference of ##\theta##.
The instantaneous voltage is given by ##v = V_m\sin(\omega t +\theta) ; i = I_m\sin(\omega t) ##. The instantaneous power is then
##p = V_m I_m \sin(\omega t + \theta) * \sin(\omega t) ## ---> 1. The first doubt is if i draw the waveform
1606059793294.png


The average power is the power calculated for one cycle ##P_{avg} = \frac {P_{1cycle}} {Duration of 1 Cycle}## where ##P_{1cycle}## is the power for one cycle of the signal. Is my understanding correct? The duration of 1Cycle is ##2\pi##?
2. The 1 cycle can be either from 0 to 2##\pi## represented by 2a to 2b and hence the limits of integration are 0 to 2##\pi## or it can be from 1a to 1b i.e the limits of integration from ##-\theta## to ##2\pi - \theta##?
3. The integral will be
##\int_0^{2\pi} p d\theta ## or ##\int_0^{2\pi} p dt## ? How do i know which one to use? (p is from eq1)
 
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It means equation 1.
 
The derivation is
##
\int_0^{2\pi} V_mI_m\sin(\omega t + \theta)*sin(\omega t) dt
##
##
\int_0^{2\pi} V_m I_m [\sin(\omega t) \cos(\theta) + \cos(\omega t) \sin(\theta)] \sin(\omega t) dt
##
##
\int_0^{2\pi} V_m I_m[\sin^2(\omega t) \cos(\theta) dt + \cos(\omega t) \sin(\omega t) \sin(\theta) dt]
##
##
P_{avg} = \frac{V_m I_m \pi \cos\theta} {2\pi} = \frac{V_mI_m\cos\theta} 2
##
It does seem to match for resistor R ##\theta = 0 ; P_{avg} = V_{rms}I_{rms} ##
for pure inductor or capacitor ##\theta = 90; P_{avg} = 0##
 
You actually have two options:

1) express ωt as an angle (let's say φ), and integrate wrt φ, with limits 0 to 2π or
2) express θ in seconds, and integrate wrt t, with limits 0 to 1/period (which would be (2π/ω))

both would give you the same average power value over 1 period, though you may want to stick with angles for the x-axis, assuming you want to arrive at the formula with power factor.
 

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