Resistors in AC circuits are the only components that truly consume power (as heat), unlike reactances (capacitors and inductors) which just store and release it.
Resistors convert electrical energy into heat (they dissipate energy), but inductors and capacitors store and release energy throughout the AC cycle, meaning they don’t actually consume power overall.
In contrast to DC circuits, where power consumption may be calculated simply by multiplying voltage by current, reactive components in AC circuits need an alternative methodology.
This is because, in contrast to resistors, which release energy as heat, reactive components like inductors and capacitors store and release energy throughout the cycle.
Electronic devices with power ratings, such as a 20 watt amplifier or a 1/4 watt resistor, indicate how much energy they can manage or handle.
Electrical power can be either continuous (DC) or continually changing (AC).
Instantaneous power refers to the power consumed at any particular time.
Power is determined using the same formula as in DC circuits: P = V x I. This indicates that one watt (the rate of energy consumption at one joule per second) is equivalent to one volt multiplied by one amp.
Electrical Power
Consider electrical power as the flow of energy in a circuit.
The quantity of energy flowing at any one place is determined by two factors: the voltage that drives the current (such as water pressure) and the current itself.
DC Circuit with a Resistor
In a basic direct current circuit with a resistor (such as a light bulb), this energy is converted into heat. Any of the following formulae may be used to calculate the rate at which this occurs and the power wasted by the resistor as heat.
P = V * I = V2 / R = I2 * R (watts)
- These formulas have three crucial values:
V (dc voltage): This is the electrical pressure that forces the current through the circuit. - I (direct current): This is the amount of current flowing (picture water running through a conduit).
- R (resistance): This symbolizes the opposition to the current flow, just how a narrow conduit slows water flow.
There is one significant caveat: power occurs only in circuits where both voltage and current are present.
This indicates that no power flows in an open circuit (broken connection), where current cannot flow, or in a short circuit (direct connection), where voltage is zero.
To demonstrate this in action, consider a basic DC circuit with a resistor.
- P = V * I = 9 * 1 = 9 watts
- P = V2 / R = 92 / 9 = 9 watts
- P = I2 * R = 12 * 9 = 9 watts
Calculating Power in AC Circuits
In DC circuits, things are simple: voltage and current are constant, and power consumption is just a function of voltage times current.
However, alternating current circuits are another issue. Because voltage and current vary direction and magnitude in a wave-like manner (sine wave), so does instantaneous power.
This means that we can’t apply the same basic formula from DC circuits (P = VI) to calculate total power in an AC circuit.
But, the idea of power (rate of energy consumption) is still applicable. We’ll need other tools to examine AC power, which we’ll go over in future conversations.
Reactive components (inductors and capacitors) make alternating current circuits even more intriguing.
These components, unlike resistors, do not simply drain power. Instead, they store energy in magnetic fields (inductors) or electric fields (capacitors) for a portion of the AC cycle before returning it to the circuit later.
This back-and-forth exchange occurs throughout each cycle of the sine wave.
Unlike resistors, which continually dissipate power as heat, reactive components borrow and return power, resulting in a different total power usage than a basic DC circuit.
To understand the overall power utilized, we must examine the average of the continually changing “instantaneous power” (power at any given time).
Instantaneous power
Instantaneous power is still determined by multiplying the voltage (v) by the current (i) at any given time, same as in DC circuits. However, with AC, both voltage and current are always changing, therefore instantaneous power fluctuates.
The good news is that, while instantaneous power varies, the average power throughout a whole cycle of the sine wave (imagine one full wave of the voltage/current graph) is typical of total power use. This is because the sine function is cyclical, therefore it repeats.
We’ll assume that both the voltage and current waveforms are sinusoidal, as is normal in alternating current circuits. In following conversations, we will look at how to determine average power.
Voltage Waveform of a Sinusoidal Supply
- e1 = Vmsin(0°) = Vm * 0 = 0V or e1 = Vmsin(180°) = Vm * 0 = 0V or e1 = Vmsin(360°) = Vm * 0 = 0V
- e2 = Vmsin(90°) = Vm * (+1) = +Vm
- e3 = Vmsin(270°) = Vm * (-1) = -Vm
In AC circuits, instantaneous power is the power at any point of time, so:
- p = v * i
- where: v = Vm sin(ωt + θv), and
- i = Im sin(ωt + θi)
- p = [Vm sin(ωt + θv)] * [Im sin(ωt + θi)]
- ∴ p = VmIm[sin(ωt + θv)sin(ωt + θi)]
We may apply a mathematical method to examine AC power. This method, known as the “trigonometric product-to-sum identity,” converts the product of two sines (voltage and current in our example) into a function that is easier to calculate averages.
sinA sinB = 1/2[cos(A – B) – cos(A + B)]
Substituting this identity, into the above equation (where θ represents the phase difference between voltage and current waveforms, θ = θv – θi), we get a new equation:
- p = VmIm / 2 [cosθ – cos(2ωt + θ)]
- Since, we have VmIm / 2 = (Vm / √2) * (Im / √2)
- = Vrms * Irms (Watts)
Here, V and I represent the typical values (root-mean-square) of the voltage (v) and current (i) waveforms, and θ is the angle difference between their peaks.
Thus, the instantaneous power in AC circuits can be expressed as given below:
p = V * I * cos(θ) – V * I * cos(2ωt + θ)
This equation breaks-down instantaneous power in AC circuits into two parts.
The first term a constant, depends only on the phase difference (θ) between voltage (V) and current (I).
The second term is a fluctuating component that changes with time like a sine wave. Interestingly its frequency is double that of the original AC supply because of the factor 2ω.
Since instantaneous power constantly varies, directly measuring it can be tricky.
Thankfully.. for calculations, we can use the average power instead. This is the mean value of the instantaneous power over a specific number of cycles. The formula to find the average power of the sinusoid is much simpler, as shown below:
p = V * I * cos(θ)
where V and I are the root-mean-square (rms) values of the voltage and current waveforms, and θ is the phase difference between them. This formula, gives us power in watts (W).
There is an another way to find power in AC circuits. If you know the circuits impedance (Z) you can use either the rms voltage (Vrms) or the rms current (Irms) along with the following equations:
- Z = √[R2 + (XL – XC)2]
- θ = cos-1 = R / Z, or sin-1 = XL / Z, or tan-1 = XL / R
- ∴ P = (V2 / Z) cos(θ) or P = I2Zcos(θ)
Solving an Instantaneous Power Problem
You are given the voltage (vt) and current (it) of a 50Hz AC circuit represented by sine waves. The voltage equation is vt = 220 sin(ωt + 60°) volts, and the current equation is it = 3 sin(ωt – 10°) amps. Your task is to find the instantaneous power and the average power absorbed by the circuit
- p = v * i = 220(sinωt + 60°) * 3(sinωt – 10°)
- p = 220 * 3[sin(314.2t + 60°)sin(314.2t – 10°)]
- ∴p = 660[sin(314.2t + 60°)sin(314.2t – 10°)]
If we apply the trigonometric identity rule from the above equation we get:
- sinA sinB = 1/2 [cos(A – B) – cos(A + B)]
- Subsequently, we get:
- p = VmIm/2[cosθ – cos(2ωt + θ)]
- = 220 * 3 / 2[cos(60 – (-10)) – cos(2 * 314.2t + 60 + (-10))]
- ∴p = 330[cos(70°) – cos(628.4t + 50°)]
- = 330[0.34202014 – cos(628.4t + 50°)]
- = 112.86 – 330[cos(628.4t + 50°)] Watts
We can calculate the average power as:
- P(avg) = VmIm / 2 [cos(θv – θi)]
- P(avg) = 220 * 3 / 2 [cos(60° – (-10°)]
- ∴P(avg) = 330 cos(70°) = 112.86 Watts
The constant term, you are seeing in the expression for instantaneous power (P(t)) actualy represents the average power of 112.86 watts.
This term is constant, because it reflects the average rate at which energy flows (changes), between the power source and the device, using the power (the load).
Circuits with Only Resistors
We learned that power in a DC circuit equals voltage times current. This relationship holds true for circuits with just resistors even when using alternating current (AC). Resistors are components, that use up electrical energy. The power dissipated in a resistor can be calculated in three ways:
- P = VI (voltage multiplied by current
- P = I²R (current squared times resistance)
- P = V²/R (voltage squared divided by resistance)
No matter how you calculate it… the power will always be a positive value.
The Example Circuit
Let’s look at this simple circuit, having only a resistor connected to an AC power source. In this ideal scenario, the capacitance (C) is infinitely large, and the inductance (L) is zero. This means that, the circuit behaves purely based on resistance.
In-Phase Current and Power in a Resistive Circuit
As shown above, imagine connecting a resistor directly to an AC power source that follows a sine wave pattern (sinusoidal).
Because the resistor only opposes current flow, the current through it will also exactly follow the ups and downs of the voltage.
In other words the voltage and current waveforms are perfectly aligned, with no lag or lead between them. We call this being “in-phase.”
Since there’s no delay between the voltage and current peaks, the phase difference between their waveforms is zero degrees (0°). This translates to a cosine of zero degrees (cos 0°) which is equal to 1.
Since voltage and current are perfectly in sync in a resistive circuit, the way they interact to determine power consumption becomes quite straightforward. Now, we can discuss the formula for the electrical power, as below:
Calculating Power Consumption in a Pure Resistor Circuit
- P = V * I * cosθ
- cos(0o) = 1
- ∴ P = V * I * 1 = V * I (Watts)
Real Power in a Resistive Circuit
Because voltage and current in a resistive circuit are perfectly aligned… they reach their highs and lows at exactly the same time. This means that the power equation simplifies to just P = V x I.
Imagine multiplying the voltage and current values at any given instant. This product which is called the volt-ampere product, represents the real power (P) consumed by the resistor, measured in watts (W). We can also use larger units like kilowatts (kW), or megawatts (MW), depending on the power level.
Visualizing AC Power in a Resistor
The diagram illustrates the voltage, current, and resulting power waveforms. Since these waveforms are in phase they rise and fall together. Let’s break it down, as below:
- Positive Half-Cycle: During this time both voltage and current are positive. When you multiply two positive values, the resultant (power) is also positive.
- Negative Half-Cycle: Here the voltage and current are both negative. However when multiplying two negatives results in a positive value for power.
Continuous Power Consumption:
In a purely resistive circuit, this in-phase behavior means the resistor consumes electrical power throughout the entire cycle, whenever current flows. This power consumption can be calculated using the formula:
P = V * I (watts)
Using RMS Values:
We can also use the root-mean-square (rms) values of voltage and current in the equation. Please recall the relationship between voltage, current, and resistance in a resistive circuit: V = I * R and I = V / R.
AC Circuits with Only Inductors (Coils)
Imagine a circuit containing only an inductor (coil) having an infinite capacitance (C = ∞), and zero resistance (R = 0). In this ideal scenario, the circuit behaves purely based on inductance (L, measured in Henries). Unlike resistors, inductors dont like changes in current flow.
The Opposing Force of Back EMF
When you apply a changing AC voltage to this purely inductive circuit, the coil fights back! The coil generates its own voltage called a “back EMF,” which now opposes the original voltage. This resistance to variations keeps the current from rising immediately and aligning with the voltage.
The Lagging Current
As a result of the back EMF the current in the circuit can’t reach its peak at the same time as the voltage. Instead, the current “lags behind” the voltage by a full 90 degrees (π/2 radians), as shown in the diagram.
Thats why we use the term “ELI” (Electromotive force Lags Inductance) as a mnemonic to remember that electromotive force lags behind inductance.
Understanding Voltage and Current in an Purely Inductive Circuit
The waveforms show voltage and current over time in a purely inductive circuit. We can see key differences compared to resistive circuits:
- Lagging Current: The current peak (Imax) arrives much later than the voltage peak. It reaches its maximum value at a quarter cycle (90°) after the voltage peak.
- Opposite Extremes: At the beginning of the voltage cycle (when voltage is most positive), the current is at its most negative value. As the voltage reaches its peak, the current is zero and then increases to its positive peak.
- Out-of-Phase Waveforms: Because of this lag, the voltage and current waveforms are no longer synchronized. Theres a phase shift of 90 degrees (π/2) between them. We say, they are “out-of-phase” with the voltage leading the current by 90°.
- Phase Angle and Cosine: This 90° phase difference translates to a cosine of 90° (cos 90°) which equals zero.
Hence the electrical power stored in a pure inductor, denoted as QL, can be expressed by:
Real Power in a Pure Inductor
- P = V * I * cosθ
- cos(90°) = 0
- ∴ P = V * I * 0 = 0 (Watts)
Even though a perfect inductor doesnt use up real power (watts), it does have voltage and current. But in an inductor, these waveforms are out of phase by 90 degrees. This means the simple power formula (P = V x I) with cosine (cos(θ)) no longer applies directly.
Imaginary Power vs. Real Power
In this situation, the product of voltage and current becomes something called reactive power (Q), measured in volt-amperes reactive (VAr) or kilo-VAr (kVAr).
It’s important to distinguish VAr from watts (W) used for real power. VAr represents the product of voltage and current that are 90° out of phase.
Calculating Reactive Power in Inductors
To calculate the average reactive power in an inductor.. we use the sine function (sin) instead of cosine. The formula for average reactive power in an inductor becomes:
- QL = V * I * sinθ
- sin(+90o) = +1
- QL = V * I + 1 = V * I (VAr)
Reactive Power and the Phase Shift
Just like real power, reactive power (Q) also depends on both voltage, and current. However unlike real power, which considers only the in-phase components, reactive power factors in the phase angle between them.
In an inductor, the current lags the voltage by 90 degrees. This means that, the reactive power is calculated using the product of the applied voltage and the specific portion of the current that’s out-of-phase with the voltage, as shown in the diagram.
Out-of-Phase Waveforms in a Pure Inductor
Energy Exchange in a Pure Inductor
The voltage and current waveforms in an inductor depicts an interesting exchange of energy, rather than true power consumption. Let’s analyze each half cycle:
- Positive Half-Cycle (0° to 90°): The voltage is positive but the current is negative. Their product is negative but this doesn’t represent power consumption. The negative current signifies the inductor resisting the change in voltage and storing energy in its magnetic field.
- Negative Half-Cycle (90° to 180°): Both voltage and current are positive, resulting in a positive product. However this doesn’t indicate real power use either. Here, the inductor is returning the stored energy back to the source, not consuming it.
Similar Pattern in the Second Half:
The cycle repeats in the second half:
- Negative Half-Cycle (180° to 270°): Negative voltage with positive current leads to a negative product. This again reflects the inductor opposing a change and building its magnetic field, without using any power.
- Positive Half-Cycle (270° to 360°): Positive voltage and current create a positive product. The inductor is releasing the stored energy back to the source, not consuming power.
Reactive Power and No Real Power Consumption:
Over a full cycle (0° to 360°), we experience two positive and two negative power pulses which cancel each other out. The average value is zero, signifying no real power (watts) is used. This is because, an ideal inductor doesn’t convert electrical energy to heat (unlike a resistor).
The Role of Reactive Power:
Instead of real power consumption, an inductor exhibits reactive power exchange (VAR). This exchange represents the transfer of energy between the inductors magnetic field and the source, without any net power being consumed.
Circuits with Only Capacitors
Imagine a circuit containing only a capacitor (with zero inductance, L = 0, and infinite resistance, R = ∞). With this ideal situation, the circuit behaves purely based on capacitance (C, measured in Farads).
Unlike resistors, capacitors love to store electrical energy in an electric field between their plates. This simply means that a perfect capacitor doesnt use up any power, but rather acts like a temporary storage unit for electrical energy.
The Lagging Voltage
In a purely capacitive circuit there’s a twist compared to resistive circuits. The voltage across the capacitor can’t rise immediately in sync with the current.
It takes some time to “charge up” the plates first. This delay to charge up, causes the voltage waveform to reach its peak after the current peak.
Leading Current (Remember ICE!)
As a result of this lag, the current in a purely capacitive circuit reaches its peak, before the voltage does. We say the current “leads” the voltage by 90 degrees (π/2 radians), as shown in the diagram. This is a helpful memory aid: ICE (Inductance is Compensated for by Electromotive force). In capacitors, the opposite is true – the current leads the voltage.
The graph above shows, how voltage and current change over time in a circuit with a pure capacitor. The current reaches its maximum (Im) first, a quarter cycle (90 degrees) ahead of the voltage peak.
This means that the current waveform starts at its highest point when the voltage is just beginning to rise.
The current then goes to zero and reaches its most negative value when the voltage reaches its peak (at the 90-degree mark).
This is the opposite of what happens in inductive circuits, where the voltage leads the current. Because of this phase shift, the phase angle (θ) in a purely capacitive circuit is -90 degrees.
This difference in phase angle affects the way the circuit consumes reactive power, as shown by the equation below:
Reactive Power Consumption in a Purely Capacitive Circuit
- QC = V * I * sinθ
- sin(-90°) = -1
- ∴ QC = V * I * -1
- = – V * I (VAr)
The term “-V*I*sin(θ)” represents a negative sine wave which signifies the reactive power (denoted by QC) consumed by a capacitor. Just like inductors, this reactive power is measured in Volt-Ampere Reactive (VAR). However unlike inductors, a pure capacitor doesn’t consume any real power (P).
Visualizing AC Power in a Capacitor
During the positive half cycle (0° to 90°) of the voltage waveform, both the current and voltage are positive. This results in a positive product (V x I) and hence positive power consumption.
However between 90° and 180°, the capacitor current becomes negative while the voltage remains positive. The negative current, when multiplied by the positive voltage, produces a negative power value (V x I < 0). This signifies that the capacitor is returning the stored energy back to the source.
In the same way, in the negative half cycle (180° to 270°) with both voltage and current negative, the power term (V x I) becomes positive again, indicating the capacitor absorbing energy from the source.
The final half cycle (270° to 360°) repeats the scenario from 90° to 180°, with a negative voltage and positive current resulting in negative power (energy returning to the source).
Over a complete cycle (360°), we can see identical positive and negative power pulses, canceling each other out. The average power delivered by the source to the capacitor (positive power) is exactly equal to the power returned by the capacitor (negative power). As a result, the net real power consumption (P) is zero.
Therefore the capacitor’s reactive power (QC), despite its cyclic exchange with the source, does not contribute to any actual work performed in the circuit.
Solving a Series RL Circuit with an AC Voltage
We have a solenoid coil connected to a power source. Here’s what we know about the circuit:
- Resistance (R): 32 ohms (Ω)
- Inductance (L): 210 millihenrys (mH) (converted to Henrys [H] later for calculations)
- Voltage (V): 220 volts (V) (AC – Alternating Current)
- Frequency (f): 50 Hertz (Hz)
We need to find:
- (a) Impedance (Z) of the solenoid: This value combines the effects of resistance and inductance on current flow.
- (b) Current (I) consumed by the solenoid: This will depend on the impedance and applied voltage.
- (c) Phase angle (θ) between the current and voltage: This angle indicates the time difference between their peaks.
- (d) Average power (P) consumed by the solenoid: This tells us how much real power the solenoid uses on average.
Note: We’ll need to convert the inductance from millihenrys to Henrys for calculations since most impedance formulas use Henrys.
Impedance Z of solenoid can be found as given below:
- XL = 2πfL = 2π * 50 * 210 * 10-3 = 65.94 Ω
- Z = √(R2 + X2L)
- = √(322 + 65.942)
- = 73.29 Ω
The solenoids current consumption (I) can be calculated as below:
- V = I * Z
- ∴ I = V / Z = 220 / 73.29
- = 3.00 Amps RMS.
We can calculate the phase angle θ, as given below:
- cosθ = R / Z, or sinθ = XL / Z, or tanθ = XL / R
- ∴ cosθ = R / Z = 32 / 73.29 = 0.43
- cos-1 (0.43) = 64.53∘ (lagging)
Now, to find how much Average AC power the solenoid would consume, we use the following calculations:
- P = V * I * cosθ
- = 220 * 3 * cos(64.53∘)
- = 220 * 3 * 0.430
- = 283.80 watts
Conclusion
Understanding Power in AC Circuits
Unlike DC circuits, where power is simply voltage times current (P = VI), things get a bit more complex in AC circuits. Here’s why:
- Phase Shift: In AC circuits, voltage and current waveforms can be out of sync (not in phase) due to components like capacitor and inductors. This phase difference affects how power is used.
Types of Power in AC Circuits:
- Real Power (P): This is the actual power used to perform work in a circuit, typically dissipated as heat in resistors. It’s calculated using the root mean square (rms) voltage and current (Prms = VrmsIrms).
- Reactive Power (Q): This power is not used for work, but gets stored and returned between the circuit and reactive components (capacitors and inductors) during each cycle. It’s calculated using voltage, current, and the sine of the phase angle between them (Q = VrmsIrms sin(θ)).
Impact of Circuit Components:
- Resistor: These consume real power, converting it to heat (θ = 0°, current and voltage are in phase).
- Inductor and Capacitor: They store and return reactive power causing a phase shift (θ = 90°, current leads/lags voltage by 90°). The average real power used over a full cycle is zero.
Phase Angle (θ):
- This angle represents the difference in timing between voltage and current peaks. It indicates, how efficiently the circuit uses real power, with a higher phase angle due to reactive components signifying lower efficiency.
- A higher total impedance (Z) in the circuit which includes resistance and reactance, can also lead to inefficiencies.
In short, AC circuits deal with both real power that does work and reactive power that gets exchanged, but doesn’t contribute to actual work. Understanding these concepts helps us analyze and optimize AC power systems.
References: Power in AC Circuits
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