Resistive Power Dissipation:Consider initially power dissipation in a resistor when supplied by a dc battery as shown in figure below. In this case, the voltage across resistor is constant and thus as well is the current flowing through it. As a result, the power dissipation, given as the product of voltage and current is constant and invariant with the time.
Figure: Power Dissipation in a DC Driven Resistive Circuit
The Power dissipated in the load is given as:P = VI = V2/R = I2RIn case of an ac voltage source as applied to a circuit as shown in figure below, the situation is somewhat different. Voltage across the resistor differs with time and as an outcome therefore does the current flowing via it. Though, in a resistor the current and voltage are in phase with one other and hence the waveforms of each are similar. That is, both contain the same function of time, as can be seen in case of sinusoidal source shown. This can as well be seen that the power, stated as the product of voltage and current, is as well a function of time, varying in a sinusoidal way. V(t) = Vm sin ωt i(t) = Im sin ωt Im = Vm/RPi(t) = V(t) i(t)
Figure: An AC Driven Resistive Circuit
Figure: Waveforms showing Power Dissipation in an AC driven Resistive Circuit
Instantaneous and Average Power:From the waveforms as shown in figure above for an ac voltage source driving a resistive load, it can be observe that as the current and voltage have exactly similar phase relationship and that the resultant power waveform is always positive. This signifies that power is continuously dissipated in the load, even although it differs as a function of time from zero to certain maximum value. This can as well be seen that the power waveform differs at twice the frequency of either voltage or current. The way of variation of power on a short-term cyclical basis is of seldom significant interest and it is the longer term power delivered to the load which is of interest, that is, the average power. Since the power variation is cyclical and thus repetitive, this is possible to compute the average value over one cycle of excitation and this hence symbolizes the long-term value with time. It as well corresponds to the equal amount of constant or dc-type power that would be delivered by a battery driving similar load.Instantaneous Power: Instantaneous power is the product of instantaneous voltage across and instantaneous current flowing via a load and is thus a function of time.Pi(t) = V(t) i(t) at the loadIf V(t) = Vm cos ωt and i(t) = Im cos ωtThen, Pi = Vm Im cos2 ωt And hence, Instantaneous Power Pi = (Vm Im/2) (1 + cos2 ωt)
Average Power:It is the long-term or average value of the instantaneous power. For a periodic source it is computed over one full cycle of the source delivering it to the load. This is an equivalent value of the constant power.
Remember the RMS value of a sinusoidal waveform, that is, the root of mean of the square.By definition:
This can be seen from the prior relationship for average powers that if:Im = Vm/R then PAVE = Vm Im/2 = (Vm/√2) (Im/√2) = VRMS IRMSBy definition this is the idea behind an rms value. rms value of an ac sinusoidal source voltage is that value of voltage that delivers similar average power to a load as a dc supply of similar value.PAVE = VRMS IRMS = V2RMS /R = I2RMS RAs well: Note that for sinusoidal source the instantaneous power differs between zero, whenever V(t) = 0, i(t) = 0 and a maximum that takes place when V(t) = Vm and i(t) = Im. This signifies that for a sinusoidal source, the average power is half of peak power. Power in a Purely Inductive Load:
In an inductor, the current lags the voltage by 90o as can be observe from the waveforms shown in figure below. When the source voltage is sinusoidal, then the current is as well sinusoidal however shifted in phase. The instantaneous power, stated as the product of instantaneous current and voltage, can as well be seen to be sinusoidal in time. Though, in contrast to resistive load, the instantaneous power in the inductor goes negative for the part of cycle of the source driving it. The average power can be determined in an identical way to that for the resistive load.
Figure: The Power related with a Purely Inductive Circuit
Note: The instantaneous power alternates negative and positive at twice the frequency of the source supplying it. Instantaneous Power:Pi = V(t) i(t) = - Vm sin ωt Im cos ωtPi = (-Vm Im /2) [sin 2ωt + sin 0]Pi = (-Vm Im /2) (sin 2ωt)Average Power:
PAVE = - 1/2T o∫T Vm Im Sin 2ωt
PAVE = - (Vm Im)/(2T) o∫T Sin 2ωt
PAVE = (Vm Im)/(2T2ω) |Cos 2ωt|oT
PAVE = (Vm Im)/(2T2ω)(Cos 4π - Cos 0)
PAVE = (Vm Im)/4ωT |1 - 1| = 0
We summed up that the average power dissipated in a pure inductance is zero. Though, it can be observe that the instantaneous power is not zero, apart from at zero crossings of the time axis. This signifies that power is drawn from source. What occurs is that whenever the instantaneous power is positive, energy is drawn from the source and stored in inductor for a quarter of cycle. If the instantaneous power is negative, then this stored energy is returned to the source that reabsorbs it. Though, the source should still have the capacity to give the power needed by the inductor, even although this is not consumed or dissipated.Power in a Purely Capacitive Load:In a capacitor, the current leads the voltage by 90o as can be observe from the waveforms as shown in figure below. When the source voltage is sinusoidal, then the current and instantaneous power is as well sinusoidal. Again, in contrary to the resistive load, the instantaneous power in the capacitor goes negative for the part of cycle of the source driving it, alternating between negative and positive phases twice per cycle. Average power dissipated by the capacitor can again be recognized by the integration of instantaneous power.
Figure: The Power related with a Purely Capacitive Circuit
Note: Since in the prior case, the instantaneous power alternates between negative and positive phases at twice the frequency of source supplying it. Instantaneous Power:Pi = V(t) i(t) = Vm sin ωt Im cos ωtPi = (Vm Im/2)[ sin 2ωt + sin 0]Pi = (Vm Im/2)[ sin 2ωt]
PAVE = - (VmIm/2T) 0∫T Sin 2ωt = 0
Thus, as was the case for inductor, the average power dissipated in a capacitor is zero. Since with the inductor, power is drawn from supply and stored as energy in the capacitor for the quarter of cycle and then returned to the source throughout the following quarter cycle.
Imaginary Power:The power transferred to the capacitor and inductor in reactive circuits is energy that is temporarily stored and then returned to the source. This power is not dissipated and can thus be considered as the imaginary power.
Consider the inductor as shown in figure below:
Figure: A Purely Inductive Circuit
But,i(t) = v(t)/ZL = v(t)/jωL = - j [v(t)/ωL] Then, Pi = (-1/2) Vm Im sin 2ωtAnd hence,Pi = j [(Vm)2/2 ωL] (sin 2 ωt) that is purely imaginary Consider the capacitor:
Figure: A Purely Capacitive Circuit
But, Then,Pi = (1/2) Vm Im sin 2 ωt And hence,Pi = j (1/2) V2m ωC sin 2 ωt that is purely imaginary This signifies that both the inductance and capacitor can be thought of as a consuming imaginary power. Though, they do not disperse the energy or power which they draw from the supply.
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