Potenza elettrica in corrente alternata sinusoidale
Electric power
in sinusoidal AC circuits
1.1 - General definition of power
We call power the rate at which any device can produce or use energy in the form of work, that is:
\[{P = \frac{\Delta L}{\Delta t}} \]
where ΔL, measured in joules, is the amount of work done or absorbed in the time interval Δt, in seconds. The unit of power is the "watt", which is dimensionally equivalent to [J]/[s].
1.2 - Instantaneous electric power
In electric systems, and in particular in direct current, where voltage and current are constant in time, power is given by:
\[P = \frac{ \Delta Q\cdot V }{\Delta t}= V\frac{ \Delta Q }{\Delta t} = V I\]
in this case P is also constant in time.
In alternating current we have to take into account that voltage and current depend on time according to suitable laws:
\[{v(t) = V_M \;sin(\omega t + \alpha )} \] \[{i(t) = I_M \;sin(\omega t + \beta)}\]
which are sinusoidal functions of time.
Activity 1
We then define the instantaneous power as the product:
\[p(t) = v(t) i(t)\]
which, as you can see, is still a sine wave, but with twice the frequency of voltage and current.
Depending on the kind of load (resistive, inductive, capacitive, etc.), voltage and current will have different phase relations, and this has important consequences for power.
Activity 2
For a purely resistive load (resistance only) the voltage and current curves are in phase: zeros, maxima and minima coincide. In this case the instantaneous power looks like a sine wave of twice the frequency, shifted so that it is entirely positive.
If the load is purely inductive (inductor only), the current lags the voltage by a quarter period (quadrature). The instantaneous power is still a sine wave of twice the frequency, but it is alternately positive and negative.
With a purely capacitive load (capacitor only), the current leads the voltage by a quarter period, and the instantaneous power behaves like in the previous case, but with the opposite sign.
Two sine waves are in quadrature when they are shifted by π/2 (90°) with respect to each other. As a result, the zeros of one coincide with the maxima (minima) of the other.
1.3 - Power and energy
Look again at the power curve. We said that \(P = \frac{\Delta L}{\Delta t}\), and therefore:
\[\Delta L = P \cdot \Delta t\]
A time interval Δt defines a rectangular area under the sine wave, with base Δt and height equal to the instantaneous power P. This area, P·Δt, is nothing but the energy, or work, exchanged in the time interval Δt.
Activity 3
Adding up the intervals until they cover the whole period (with Δt very small) we obtain the area of the region bounded by the horizontal axis t and the sine wave. It corresponds to the amount of energy exchanged (produced, used) in one period T.
1.4 - Active power
We have seen that the area under the curve p(t) of the instantaneous power corresponds to the energy that, for example, is transferred from the generator to the load. The average power is the value of P, constant in time, that transfers the same amount of energy (area of the rectangle, divided by ω if the plot is a function of the angle ωt) as the instantaneous power.
Notice that the average power decreases as the phase shift between voltage and current grows. In other words the average power, and therefore the energy transferred, goes from a maximum, when V and I are in phase, down to zero when they are in quadrature.
Activity 4
With current and voltage in phase it is easy to work out the average power Pm, because the rectangle with the same area bounded by the
sine wave (and therefore the same energy) has a height equal to half the peak value of the power \[ P_m = \frac {V_M I_M} {2} \]
Recalling that the RMS (Root Mean Square) values of voltage and current are defined as
\[V = \frac{V_M}{\sqrt{2}}\;\; \;\;\;\;\; I = \frac{I_M}{\sqrt{2}} \]
we can say that the active power, when (remember) voltage and current are in phase, is given by:
\[P = P_m = V I \]
However, the phase shift between voltage and current can take any value between -90° and +90°.
In general the active power is given by:
\[ P = V I cos \varphi \]
an expression for which, using well-known trigonometric identities, there is an
analytical proof
The \(cos \varphi \) that appears in the power formula is called the power factor (PF), or simply "cos phi".
The power factor takes values between 0 and 1 when the angle ranges from -90° to +90°, so it is always positive. Therefore the numerical value of the PF alone does not tell us the kind of load, except of course when \(cos \varphi = 1\) (current and voltage in phase). The kind of load can be specified by adding "lagging" for current lagging the voltage (resistive-inductive load), or "leading" for current leading the voltage (resistive-capacitive load).
Activity 5
Going back to the formula for active power:
\[ P = V I cos \varphi \]
we see that the power factor is the ratio between the actual active power P delivered by the system and the maximum power that could be obtained with V and I in phase, that is with \(cos \varphi = 1\), which would give \(P_{MAX} = V I \):
\[cos \varphi = \frac{P}{V I} \]
This product is usually written
\[S = V I \]
and is called the apparent power (see below).
For example, cos(φ) = 0.8 means that you can obtain an active power equal to 80% of the apparent power.
1.5 - Reactive power
Look at the power sine wave: in some stretches it is negative.
If we now also plot voltage and current on the same graph, we can see that:
It is easy to see that the area corresponding to the useful energy equals that of the rectangle whose height is the average power.
Looking carefully at the figure, we will probably notice that the average power has decreased by exactly the amount represented by the negative area.
This means that part of the energy, during some time intervals, flows from the generator (G) to the load (positive area), and during others from the load (U) back to the generator (negative area).
The share of power that allows this two-way exchange of energy is called reactive power.
Reactive power cannot be used to do work because, as you can see, the positive energy (G --> U) is exactly equal to the negative energy (U --> G), so its average over the period is zero.
This energy is needed by the systems themselves to build the magnetic or electric fields that the devices require in order to work.
Reactive power is measured in var, volt-amperes reactive.
The point becomes even clearer with purely inductive or purely capacitive loads.
a - purely inductive load
During the first quarter period the current is negative (U --> G) and goes from its maximum to zero: in this phase the magnetic field of the inductor is decreasing until it reaches zero. The energy stored in the field is returned to the generator (negative power).
In the next quarter period the current rises to its maximum, the power is positive, the energy flows toward the load (G --> U) and builds up the magnetic field of the inductor. In the following quarters the same events repeat with the roles reversed.
b - purely capacitive load
In the first quarter period the capacitor charges (the current goes from its maximum to zero), so the electric field between the plates must build up. In fact energy and power are positive (G --> U). In the next quarter, instead, the capacitor discharges, the electric field gradually vanishes and its energy must be returned to the generator. Power and energy are therefore negative (U --> G). Here too, in the following quarters the same events repeat with reversed polarity.
For reactive power we cannot use the average over the period, as we did for active power, because it is zero.
Reactive power is therefore defined as:
\[Q = V I sin \varphi \]
(see the proof for more details)
Reactive power has a sign: it is taken as positive for inductive or resistive-inductive loads, and negative for capacitive or resistive-capacitive loads.
1.6 - Apparent power and the power triangle
Apparent power has no real physical meaning other than the one given above: it represents the maximum active power that an alternating (sinusoidal) current electric system can deliver.
This happens when voltage and current are in phase.
Apparent power is defined by:
\[S = V I \]
it is measured in volt-amperes [VA] and, among other things, it is the power used to rate electric devices.
The relations between apparent, active and reactive power are:
\begin{align}S &= V I\\P & = V I cos \varphi \\Q & = V I sin \varphi \\ \end{align}
They can be seen as relations between the sides of a right triangle with hypotenuse S and legs P and Q, so that:
\begin{align}S &= \sqrt{P^2 + Q^2} \\P & = S cos \varphi \\Q & = S sin \varphi \\ \end{align}
From these relations we can build a triangle called the
power triangle.
1.7 - Complex power
To follow what comes next you need at least a basic knowledge of the theory of complex numbers (page in Italian).
If \(\alpha \) and \(\beta \) are the phases of voltage and current respectively, their phasors correspond to complex numbers, written here in rectangular form:
\begin{align} \mathbf V & = V cos \alpha + j V sin \alpha = V \angle \alpha \\ \mathbf I &= I cos \beta + j I sin \beta = I \angle \beta \\ \end{align}
and polar form:
\begin{align} \mathbf V & = V \angle \alpha \\ \mathbf I &= I \angle \beta \\ \end{align}
From the theory of complex numbers we know that multiplying two of them, in exponential or polar form, gives the product of the magnitudes and the algebraic sum of the angles.
If we took the product \(V \angle \alpha \cdot I \angle \beta\) we would get \( V I \angle (\alpha + \beta) \).
If instead we take the complex conjugate of the current:
\[\mathbf {I^{*}} = I cos \beta - j I sin \beta = I \angle-\beta\]
the product of the voltage and the complex conjugate of the current is:
\[\mathbf {V \cdot I^{*}} = V I \angle (\alpha-\beta)\]
but the angle \(\varphi\) between voltage and current is the algebraic difference of the two phases, that is:
\[\varphi = \alpha - \beta\]
and therefore:
\[\mathbf {V \cdot I^{*}} = V I \angle \varphi \]
For this reason the complex power is defined as precisely the product
\[\mathbf {S = V \cdot I^{*}} = V I \angle \varphi \]
which, in rectangular form, gives:
\[\mathbf {S = V \cdot I^{*}} = V I cos \varphi + j V I sin \varphi \]
that is:
\[\mathbf {S = V \cdot I^{*}} = P + j Q \]
If you look for a trigonometric justification of the above, notice that:
\[\mathbf {S = V \cdot I^{*}} = (V cos \alpha + j V sin \alpha) (I cos \beta - j I sin \beta) \]
Multiplying out, collecting common factors and recalling that \(j^2 = -1\):
\[\mathbf {S = V \cdot I^{*}} = V I (cos \alpha cos \beta + sin \alpha sin \beta) +j V I (sin \alpha cos \beta-cos \alpha sin \beta) \]
the first expression in parentheses is the cosine difference formula, while the second is the sine difference formula, so that:
\[\mathbf {S = V \cdot I^{*}} = V I cos (\alpha-\beta) + j V I sin (\alpha-\beta) \]
and once again we get:
\[\mathbf {S = V \cdot I^{*}} = V I cos \varphi + j V I sin \varphi \]
and
\[\mathbf {S = V \cdot I^{*}} = P + j Q \]
1.8 - Additivity of power: Boucherot's theorem
This important theorem, which follows from general principles such as the conservation of energy and of electric charge, holds for power.
The total complex power absorbed by the impedances of a network equals the sum of the complex powers absorbed by each impedance
\[\mathbf{S = S_1 + S_2 + S_3 + ... + S_n}\]
Written out:
\begin{split}\mathbf S &= \mathbf{S_1 + S_2 + S_3 + ... + S_n =}\\ &= ( P_1 \pm j Q_1) + (P_2 \pm j Q_2) + (P_3 \pm j Q_3) + ...+ (P_n \pm j Q_n) =\\&= (P_1 + P_2 + P_3 + ...+ P_n) + j (Q_1 \pm Q_2 \pm Q_3 \pm ...\pm Q_n) \end{split}
This translates into a few calculation rules that always apply and can be stated as follows:
a) In any electric network active powers add arithmetically:
\[{P = P_1 + P_2 + P_3 + ... + P_n}\]
b) In any electric network reactive powers add algebraically:
\[{Q = Q_1 \pm Q_2 \pm Q_3 \pm ... \pm Q_n}\]
the sign of each reactive power depends on the kind of load. It is positive for inductive (or resistive-inductive) loads and negative for capacitive (or resistive-capacitive) loads.
1.9 - Related topics, books and further reading
1.1 - General definition of power
We call power the rate at which any device can produce or use energy in the form of work, that is:
\[{P = \frac{\Delta L}{\Delta t}} \]
where ΔL, measured in joules, is the amount of work done or absorbed in the time interval Δt, in seconds. The unit of power is the "watt", which is dimensionally equivalent to [J]/[s].
1.2 - Instantaneous electric power
In electric systems, and in particular in direct current, where voltage and current are constant in time, power is given by:
\[P = \frac{ \Delta Q\cdot V }{\Delta t}= V\frac{ \Delta Q }{\Delta t} = V I\]
in this case P is also constant in time.
In alternating current we have to take into account that voltage and current depend on time according to suitable laws:
\[{v(t) = V_M \;sin(\omega t + \alpha )} \] \[{i(t) = I_M \;sin(\omega t + \beta)}\]
which are sinusoidal functions of time.
Activity 1
We then define the instantaneous power as the product:
\[p(t) = v(t) i(t)\]
which, as you can see, is still a sine wave, but with twice the frequency of voltage and current.
Depending on the kind of load (resistive, inductive, capacitive, etc.), voltage and current will have different phase relations, and this has important consequences for power.
Activity 2
For a purely resistive load (resistance only) the voltage and current curves are in phase: zeros, maxima and minima coincide. In this case the instantaneous power looks like a sine wave of twice the frequency, shifted so that it is entirely positive.
If the load is purely inductive (inductor only), the current lags the voltage by a quarter period (quadrature). The instantaneous power is still a sine wave of twice the frequency, but it is alternately positive and negative.
With a purely capacitive load (capacitor only), the current leads the voltage by a quarter period, and the instantaneous power behaves like in the previous case, but with the opposite sign.
Two sine waves are in quadrature when they are shifted by π/2 (90°) with respect to each other. As a result, the zeros of one coincide with the maxima (minima) of the other.
1.3 - Power and energy
Look again at the power curve. We said that \(P = \frac{\Delta L}{\Delta t}\), and therefore:
\[\Delta L = P \cdot \Delta t\]
A time interval Δt defines a rectangular area under the sine wave, with base Δt and height equal to the instantaneous power P. This area, P·Δt, is nothing but the energy, or work, exchanged in the time interval Δt.
Activity 3
Adding up the intervals until they cover the whole period (with Δt very small) we obtain the area of the region bounded by the horizontal axis t and the sine wave. It corresponds to the amount of energy exchanged (produced, used) in one period T.
1.4 - Active power
We have seen that the area under the curve p(t) of the instantaneous power corresponds to the energy that, for example, is transferred from the generator to the load. The average power is the value of P, constant in time, that transfers the same amount of energy (area of the rectangle, divided by ω if the plot is a function of the angle ωt) as the instantaneous power.
Notice that the average power decreases as the phase shift between voltage and current grows. In other words the average power, and therefore the energy transferred, goes from a maximum, when V and I are in phase, down to zero when they are in quadrature.
Activity 4
With current and voltage in phase it is easy to work out the average power Pm, because the rectangle with the same area bounded by the
sine wave (and therefore the same energy) has a height equal to half the peak value of the power \[ P_m = \frac {V_M I_M} {2} \]
Recalling that the RMS (Root Mean Square) values of voltage and current are defined as
\[V = \frac{V_M}{\sqrt{2}}\;\; \;\;\;\;\; I = \frac{I_M}{\sqrt{2}} \]
we can say that the active power, when (remember) voltage and current are in phase, is given by:
\[P = P_m = V I \]
However, the phase shift between voltage and current can take any value between -90° and +90°.
In general the active power is given by:
\[ P = V I cos \varphi \]
an expression for which, using well-known trigonometric identities, there is an
analytical proof
The \(cos \varphi \) that appears in the power formula is called the power factor (PF), or simply "cos phi".
The power factor takes values between 0 and 1 when the angle ranges from -90° to +90°, so it is always positive. Therefore the numerical value of the PF alone does not tell us the kind of load, except of course when \(cos \varphi = 1\) (current and voltage in phase). The kind of load can be specified by adding "lagging" for current lagging the voltage (resistive-inductive load), or "leading" for current leading the voltage (resistive-capacitive load).
Activity 5 Going back to the formula for active power:
\[ P = V I cos \varphi \]
we see that the power factor is the ratio between the actual active power P delivered by the system and the maximum power that could be obtained with V and I in phase, that is with \(cos \varphi = 1\), which would give \(P_{MAX} = V I \):
\[cos \varphi = \frac{P}{V I} \]
This product is usually written
\[S = V I \]
and is called the apparent power (see below).
For example, cos(φ) = 0.8 means that you can obtain an active power equal to 80% of the apparent power.
1.5 - Reactive power
Look at the power sine wave: in some stretches it is negative.
If we now also plot voltage and current on the same graph, we can see that:
- power is negative when voltage and current have opposite signs (directions)
- the average power is lower than when V and I are in phase
- the value of the average power is equal to \(V I cos(\varphi)\)
It is easy to see that the area corresponding to the useful energy equals that of the rectangle whose height is the average power.
Looking carefully at the figure, we will probably notice that the average power has decreased by exactly the amount represented by the negative area.
This means that part of the energy, during some time intervals, flows from the generator (G) to the load (positive area), and during others from the load (U) back to the generator (negative area).
The share of power that allows this two-way exchange of energy is called reactive power.
Reactive power cannot be used to do work because, as you can see, the positive energy (G --> U) is exactly equal to the negative energy (U --> G), so its average over the period is zero.
This energy is needed by the systems themselves to build the magnetic or electric fields that the devices require in order to work.
Reactive power is measured in var, volt-amperes reactive.
The point becomes even clearer with purely inductive or purely capacitive loads.
a - purely inductive load
During the first quarter period the current is negative (U --> G) and goes from its maximum to zero: in this phase the magnetic field of the inductor is decreasing until it reaches zero. The energy stored in the field is returned to the generator (negative power).
In the next quarter period the current rises to its maximum, the power is positive, the energy flows toward the load (G --> U) and builds up the magnetic field of the inductor. In the following quarters the same events repeat with the roles reversed.
b - purely capacitive load
In the first quarter period the capacitor charges (the current goes from its maximum to zero), so the electric field between the plates must build up. In fact energy and power are positive (G --> U). In the next quarter, instead, the capacitor discharges, the electric field gradually vanishes and its energy must be returned to the generator. Power and energy are therefore negative (U --> G). Here too, in the following quarters the same events repeat with reversed polarity.
For reactive power we cannot use the average over the period, as we did for active power, because it is zero.
Reactive power is therefore defined as:
\[Q = V I sin \varphi \]
(see the proof for more details)
Reactive power has a sign: it is taken as positive for inductive or resistive-inductive loads, and negative for capacitive or resistive-capacitive loads.
1.6 - Apparent power and the power triangle
Apparent power has no real physical meaning other than the one given above: it represents the maximum active power that an alternating (sinusoidal) current electric system can deliver.
This happens when voltage and current are in phase.
Apparent power is defined by:
\[S = V I \]
it is measured in volt-amperes [VA] and, among other things, it is the power used to rate electric devices.
The relations between apparent, active and reactive power are:
\begin{align}S &= V I\\P & = V I cos \varphi \\Q & = V I sin \varphi \\ \end{align}
They can be seen as relations between the sides of a right triangle with hypotenuse S and legs P and Q, so that:
\begin{align}S &= \sqrt{P^2 + Q^2} \\P & = S cos \varphi \\Q & = S sin \varphi \\ \end{align}
From these relations we can build a triangle called the
power triangle.
1.7 - Complex power
To follow what comes next you need at least a basic knowledge of the theory of complex numbers (page in Italian).
If \(\alpha \) and \(\beta \) are the phases of voltage and current respectively, their phasors correspond to complex numbers, written here in rectangular form:
\begin{align} \mathbf V & = V cos \alpha + j V sin \alpha = V \angle \alpha \\ \mathbf I &= I cos \beta + j I sin \beta = I \angle \beta \\ \end{align}
and polar form:
\begin{align} \mathbf V & = V \angle \alpha \\ \mathbf I &= I \angle \beta \\ \end{align}
From the theory of complex numbers we know that multiplying two of them, in exponential or polar form, gives the product of the magnitudes and the algebraic sum of the angles.
If we took the product \(V \angle \alpha \cdot I \angle \beta\) we would get \( V I \angle (\alpha + \beta) \).
If instead we take the complex conjugate of the current:
\[\mathbf {I^{*}} = I cos \beta - j I sin \beta = I \angle-\beta\]
the product of the voltage and the complex conjugate of the current is:
\[\mathbf {V \cdot I^{*}} = V I \angle (\alpha-\beta)\]
but the angle \(\varphi\) between voltage and current is the algebraic difference of the two phases, that is:
\[\varphi = \alpha - \beta\]
and therefore:
\[\mathbf {V \cdot I^{*}} = V I \angle \varphi \]
For this reason the complex power is defined as precisely the product
\[\mathbf {S = V \cdot I^{*}} = V I \angle \varphi \]
which, in rectangular form, gives:
\[\mathbf {S = V \cdot I^{*}} = V I cos \varphi + j V I sin \varphi \]
that is:
\[\mathbf {S = V \cdot I^{*}} = P + j Q \]
If you look for a trigonometric justification of the above, notice that:
\[\mathbf {S = V \cdot I^{*}} = (V cos \alpha + j V sin \alpha) (I cos \beta - j I sin \beta) \]
Multiplying out, collecting common factors and recalling that \(j^2 = -1\):
\[\mathbf {S = V \cdot I^{*}} = V I (cos \alpha cos \beta + sin \alpha sin \beta) +j V I (sin \alpha cos \beta-cos \alpha sin \beta) \]
the first expression in parentheses is the cosine difference formula, while the second is the sine difference formula, so that:
\[\mathbf {S = V \cdot I^{*}} = V I cos (\alpha-\beta) + j V I sin (\alpha-\beta) \]
and once again we get:
\[\mathbf {S = V \cdot I^{*}} = V I cos \varphi + j V I sin \varphi \]
and
\[\mathbf {S = V \cdot I^{*}} = P + j Q \]
1.8 - Additivity of power: Boucherot's theorem
This important theorem, which follows from general principles such as the conservation of energy and of electric charge, holds for power.
The total complex power absorbed by the impedances of a network equals the sum of the complex powers absorbed by each impedance
\[\mathbf{S = S_1 + S_2 + S_3 + ... + S_n}\]
Written out:
\begin{split}\mathbf S &= \mathbf{S_1 + S_2 + S_3 + ... + S_n =}\\ &= ( P_1 \pm j Q_1) + (P_2 \pm j Q_2) + (P_3 \pm j Q_3) + ...+ (P_n \pm j Q_n) =\\&= (P_1 + P_2 + P_3 + ...+ P_n) + j (Q_1 \pm Q_2 \pm Q_3 \pm ...\pm Q_n) \end{split}
This translates into a few calculation rules that always apply and can be stated as follows:
a) In any electric network active powers add arithmetically:
\[{P = P_1 + P_2 + P_3 + ... + P_n}\]
b) In any electric network reactive powers add algebraically:
\[{Q = Q_1 \pm Q_2 \pm Q_3 \pm ... \pm Q_n}\]
the sign of each reactive power depends on the kind of load. It is positive for inductive (or resistive-inductive) loads and negative for capacitive (or resistive-capacitive) loads.
1.9 - Related topics, books and further reading