The electrical power consumed by a resistance in an AC circuit differs from the power consumed by a reassurance, since the reactenance does not dissipating the energy.
In a DC circuit, the power consumed is the multiplication of the DC current of the DC voltage given in watts.But for AC circuits with reactive components, we need to calculate the consumed power differently.
Electric power is the "ratio" of energy consumed in a circuit, and therefore all electrical and electronic components and devices have a limit on the amount of electrical power they can safely handle.For example, a resistance of 1/4 watt or an amplifier of 20 watts.
Electrical power may vary by time in dc quantity or AC quantity.At any given time, the amount of power in a circuit is called instantaneous power, and the well-known power equals the voltage multiplied amperage (P = V*I) relationship.That is, one watt (the rate of energy expenditure at one joule per second), a volt times will equal the volt-amperage product of an amperage.
Next, the power absorbed or supplied by a circuit element is the I product of the V voltage that passes through the element and the current passing through it.Therefore, if we had an "R" ohm resistant DC circuit, the power expended in watts by resistance is given with any of the following general formulas:
Where: V is dc voltage, I is dc current, and R is the value of resistance.
Therefore, the power in an electrical circuit is only available when both voltage and current are present, this is not open circuit or closed circuit conditions.Consider a simple example of a standard resistant DC circuit below:
DC Resistant Circuit
Electrical Power in AC Circuit
In a DC circuit, voltages and currents are usually constant; this does not change over time, since there is no sinusoidal waveform associated with feeding.However, in an AC circuit, voltage, current and therefore instantaneous values of power are constantly changed by being affected by feeding.So we cannot calculate the power in AC circuits in the same way as we do in DC circuits, but we can still say that the power (p) is equal to the voltage (v) times amperage (i).
Another important point is that AC circuits contain reacques, so there is a power component as a result of the magnetic and/or electrical fields created by the components.As a result, unlike a fully resistant component, this power is stored, and then returned to the source as the sinusoidal waveform passes through a full periodic cycle.
Thus, the average power absorbed by a circuit is the sum of the stored power and the power returned over a full cycle.Therefore, the average power consumption of a circuit will be the average of instantaneous power over a full cycle with instantaneous power, p , defined as the multiplication of instantaneous voltage, v by instantaneous current, i .Note that since the sinus function is periodic and continuous, the average power given at all times will be exactly the same as the average power given in a single cycle.
Let's say that the waveforms of tension and current are both sinusoidal, so let's remember this:
Sinusoidal Voltage WaveForm
Because instantaneous power is power at any time in time:
If we apply the trigonometric product-total identification of:
and ε = ε v – ε (phase difference between voltage and current waveforms) in the above equation gives:
Where the V and I sinusoidal waveforms are the phase difference between the two waves, the values are v and I, respectively, in root-mean-square (rms).Therefore, we can express instant power as follows:
Instant AC Power Equation
This equation shows us that instant AC power has two different parts and therefore the sum of these two terms.The second term is a sinusoid that changes over time, the frequency of which is equal to twice the angular frequency of the feed due to the 2ω part of the term.However, the first term is a constant whose value depends only on the phase difference between voltage (V) and current (I).
This makes measurement difficult, as instantaneous power constantly changes with the profile of the sinusoid over time.Therefore, it is more convenient and easier in mathematics to use the average or average value of power.Thus, over a fixed number of cycles, the average value of the instantaneous power of sinusoid is simply given as follows:
Where V and I are rms values of sinusoids and ε (Theta) is the phase angle between voltage and current.Power units are in watts (W).
AC Power spent in a circuit can also be found from the impedance (Z) of the circuit using the voltage, V rms or current, I rms, which are transferred through the circuit as shown.
AC Power Question Sample 1
The voltage and current values of a 50Hz sinusoidal feed are given as v t = 240 sin(ωt +60 o )Volt and i t = 5 sin(ωt -10 o )Amps, respectively.Find the instantaneous power and average power values taken by the circuit.
From above, the instantaneous power absorbed by the circuit is given as follows:
Applying the trigonometric id rule from above gives:
The average power is then calculated as follows:
You may have noticed that the average power value of 205.2 watts is also the first term value of instantaneous power p (t), since this first term is the constant value, the average or average rate of energy change between source and load.
AC Power in a Fully Resistant Circuit
We have seen so far that in a DC circuit, power equals the product of voltage and current, and this relationship also applies to a fully resistant AC circuit.Resistors are energy-consuming electrical devices, and the power in a resistance is supplied with p = VI = I 2 R = V 2 /R.This power is always positive.
Consider the following fully resistant circuit (i.e. infinite capacitance, C = ∞ and zero inductancy, L = 0) circuit with resistance connected to an AC source, as shown.
Fully Resistant Circuit
When a pure resistance is connected to a sinusoidal voltage source, the current passing through the resistance will change in proportion to the supply voltage, that is, the voltage and current waveforms are in the "phase" of each other.Since the phase difference between the voltage waveform and the current waveform is 0 o, the phase angle resulting in cos 0 o will be equal to 1.
Then the electrical power consumed by the resistance is given as follows:
Electric Power in Pure Resistance
Since voltage and current waveforms are in the same phase, that is, both waveforms reach peak values at the same time and pass through zero at the same time, the above power equation is reduced only to: V*I.Therefore, power at any time can be found by multiplying two waveforms to give the volt-amperage product.This is called "Real Power", measured in ( P ) watts, (W), Kilowatts (kW), Megawatts (MW), etc.
AC Power WaveForms for Pure Resistance
The diagram shows the voltage, current, and corresponding power waveforms.Since both voltage and current waveforms are in the same phase, during a positive half-cycle, the current is also positive when the voltage is positive, so the power is positive, the positive times equals one positive positive.During a negative half-cycle, the voltage is negative, that is, the current that causes the power to be positive, the negative multiplication is negative equal to the positive.
Then, in a fully resistant circuit, ALL electrical power is consumed during the time the current passes through the resistance and is given as follows: P = V*I = I 2 R watts.Note that both V and I can have rms values: V = I*R and I = V/R
AC Power in Fully Inductive Circuit
In the completely inductive (i.e. infinite capacitance, C = ∞ and zero resistance, R = 0) circuit of L Henries, voltage and current waveforms are not in the same phase.When a voltage is applied to a completely inductive coil, a "back" imp is produced by the coil due to its self-inductive induction.This self-induced inducta resistsand limits any change in the current flowing in the coil.
These back impediment effects are that the current cannot increase immediately along the coil in the same phase as the applied voltage, causing the current waveform to reach its peak or maximum value some time after the voltage.The result is that in a completely inductive circuit, the voltage is always 90 o (π/2) behind, as shown by the current (ELI).
Fully Inductive Circuit
The above waveforms show us instantaneous voltage and instantaneous current along a fully inductive coil as a function of time.The maximum current, I max, consists of the full quarter (90 o) of a cycle after the maximum (peak) value of the voltage.Here, the current is indicated by its negative maximum value at the beginning of the voltage cycle, and when the voltage waveform reaches its maximum value at 90o, itpasses through zero, rising to the positive maximum value.
Thus, voltage and current waveforms are no longer rising together, but instead a phase shift of 90 o (π/2) occurs in the coil, so the voltage and current waveforms are "out of phase" with each. the other directsthe voltage current to 90.Since the phase difference between the voltage waveform and the current waveform is 90o, the phase angle results in cos 90o = 0.
Therefore, the electric power stored by a pure inductor, Q L is given as follows:
Real Power in Pure Inductor
Clearly then, a pure inductor does not consume or distribute any real power, but since we have both voltage and current, the use of cos(ε) in the expression: P = V*I*cos(ε) for a pure inductor is no longer valid.In this case, the product of current and voltage is the imaginary power, which is usually called "Reactive Power", (Q) volt-amp reagent, (VAr), Kilo-voltamper reagent (KVAr), etc.
Voltamper should not be confused with reagent, VAr, wattage used for real power, (W).VAr represents the multiplication of volts and amps that are 90 o-phases with each other.The sine function is used to mathematically determine the reactive average strength.Then the average reactive power equation in an inductor is as follows:
Reactive Power in Pure Inductor
Real power (P), like reactive power (Q), also depends on voltage and current, but also depends on the phase angle between them.Therefore, it is the product of the applied voltage and the component part of the current that is 90 o-phase with the voltage shown.
AC Power WaveForms for Pure Inductor
In the positive half of the voltage waveform between 0 o and 90 o, the inductor current is negative while the supply voltage is positive.Therefore, the volt and amperage product gives a negative power as negative times equals negative times positive.Between 90 o and 180 o, both current and voltage waveforms are positive in the value resulting in positive force.This positive power indicates that the coil consumes electrical energy without supply.
In the negative half of the voltage waveform between 180 o and 270 o, there is a negative voltage and positive current indicating a negative power.This negative power indicates that the coil returns the stored electrical energy to the source.Between 270 o and 360 o, both the inductor current and the supply voltage are negative, which leads to a positive power period.
Then, during a full cycle of the voltage waveform, we have two identical positive and negative power pulses with an average value of zero, so real power is not used, as the power alternately flows to the source.This means that the total power received by a pure inductor in a full cycle is zero, so the reactive power of an inductor does not do any real work.
AC Power in Fully Capacitive Circuit
The fully capacitive (i.e. zero inducing, L = 0 and infinite resistance, R = ∞) C Farads circuit has the ability to delay changes in voltage on it.Capacitors store electrical energy in the form of an electric field in dielectric, so that a pure capacitor does not dispense any energy, but instead stores it.
In a fully capacitive circuit, the voltage cannot increase in the same phase as the current, since it must first "load" the capacitor plates.This causes the voltage waveform to peak or reach its maximum value some time after the current.The result is that in a fully capacitive circuit, the voltage is always 90 o (ω/2) "direction" (ICE), as shown by the current.
Fully Capacitive Circuit
The waveform shows us the voltage and current on a pure capacitor as a function of time.The maximum current, Im, occurs up to a full quarter of a cycle (90 o) before the maximum (peak) value of the voltage.Here, the current is indicated by its positive maximum value at the beginning of the voltage cycle and passes through zero, the voltage waveform decreases to a negative maximum value while at 90 o.
Thus, for a fully capacitive circuit, the phase angle is ε = -90 o and the average reactive power equation in a capacitor is as follows:
Reactive Power in Pure Capacitor
Here – V*I*sin(ε) is a negative sine wave.In addition, the symbol of capacitive reactive power is QCwith volt-amperage reagent (VAR), which is the same unit of measure as that of the inductor.Then, like the completely inductive circuit above, we can see that a pure capacitor does not consume or dispense any real power, P.
AC Power WaveForms for Pure Capacitor
In the positive half of the voltage waveform between 0 o and 90 o, the value of both current and voltage waveforms is positive, which leads to the consumption of positive force.Between 90 o and 180 o, the capacitor current is negative and the supply voltage is still positive.Therefore, the volt-amperage product gives a negative power as negative times equals positive.This negative power indicates that the coil returns the stored electrical energy to the source.
In the negative half of the voltage waveform between 180 o and 270 o, both the capacitor current and the supply voltage are negative, which leads to a positive power period.This positive power time indicates that the coil consumes electrical energy without supply.Between 270 o and 360 o , once again there is a negative voltage and positive current indicating a negative power.
Then, during a full cycle of the voltage waveform, the same situation exists with the fully inductive circuit, since we have two identical positive and negative power pulses with an average value of zero.Thus, the power transmitted from the source to the capacitor is exactly equal to the power returned to the source by the capacitor, so no real power is used as the power alternately flows to the source.This means that the total power received by a pure capacitor in a full cycle is zero, so the reactive power of capacitors does not do any real work.
AC Power Question Sample 2
A solenoid coil with 30 ohm resistant and 200 mH induced is connected to the 230VAC, 50Hz feed.(a) calculate the solenoid impedance, (b) the current consumed by the solenoid, (c) the phase angle between the current and the voltage applied, and (d) the average power consumed by the solenoid.
The data given is: R = 30Ω, L = 200mH, V = 230V and ε = 50Hz.
(a) Impedance of the solenoid coil (Z):
(b) Current consumed by the solenoid coil (I):
(c) Phase angle, ε:
(d) The average AC power consumed by the solenoid coil:
AC Electric Power Summary
Here we found that the voltage and current flowing in a completely passive circuit in AC circuits are normally out of phase and as a result cannot be used to perform any real work.We also found that electrical power in a direct current (DC) circuit is equal to the multiplication of voltage and current, or P = V*I, but we cannot do this as we should calculate it as in AC circuits. Be sure to consider any phase difference.
In a fully resistant circuit, the current and voltage are in the same phase, and all electrical power is consumed by resistance, usually as heat.As a result, none of the electrical power is returned to its source or circuit.
However, in a completely inductive or fully capacitive circuit containing (X), the reactatement will take the voltage forward or back exactly 90 o (phase angle), so that the power is both stored and returned to the source.Thus, the average power calculated over a full periodic cycle will be equal to zero.
The electrical power (R) consumed by a resistance is called real power and is simply achieved by multiplying the rms voltage by the rms current.The power stored by a reactax (X) is called reactive power and is obtained by multiplying the sine of the voltage, current and phase angle between them.
The symbol of the phase angle is ε (Theta) and represents the inefficiency of the AC circuit in terms of total reactive impedance (Z) that corresponds to the current flow in the circuit.