Power Calculation of general electric circuit

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Discussion Overview

The discussion revolves around the calculation of power in a general electric circuit where voltage leads current by a phase difference. Participants explore the instantaneous power, average power over a cycle, and the appropriate limits and variables for integration in the context of alternating current (AC) analysis.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the instantaneous voltage and current equations and questions the calculation of average power over one cycle.
  • Another participant suggests integrating with respect to time but seeks clarification on the notation used in the initial post.
  • A later reply clarifies that "----> 1" refers to equation 1 in the context of the discussion.
  • Further elaboration on the integration process is provided, including the breakdown of the integral for average power calculation.
  • One participant proposes two options for integration: using angle or time, noting that both methods yield the same average power value over one period.

Areas of Agreement / Disagreement

Participants generally agree on the approach to calculating average power but present differing views on the integration limits and whether to integrate with respect to time or angle. The discussion remains unresolved regarding the best method to use.

Contextual Notes

Some assumptions regarding the definitions of variables and the context of integration limits are not fully clarified, leading to potential ambiguity in the integration process.

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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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