this is a light bulb which we can assume to be a purely resistive load when we apply direct current to this light bulb it turns on we can depict the voltage and current across the light bulb over time in a graph like this since the voltage and current are constant we can calculate the power used by the light bulb by simply multiplying the voltage and current it's a straightforward calculation if we neglect the wire resistance we can confidently state that all the power supplied by the power source is consumed by the light bulb now let's
consider another circuit containing the same light bulb but powered using an alternating current power source as we know the current and voltage both change sinusoidally with time let's focus on one cycle if both voltage and current change constantly how can we rate this power source in alternating current systems we calculate the root mean Square voltage and current values the RMS values can be calculated by dividing the peak value by the square root of two we won't delve deeply into RMS values in this video for a better understanding of RMS voltage and current you can watch
our previous video I'll provide the link in the description we typically calculate the power of an AC circuit by multiplying the RMS values of voltage and current alternatively we can multiply each instantaneous voltage and current to graph the instantaneous power of the circuit the power Curve will look like this it's important to note that negative voltage and negative current will result in a positive power therefore the power is always positive the area under the power curve represents the transferred energy if we calculate the average power of this graph it will be equal to the power
calculated using RMS values still we can say that all the power supplied by the power source is consumed by the light bulb right now we have basic understanding about how we calculate the power of AC and DC circuits things start to behave differently when a reactive component is connected to the Circuit it's important to note that reactants can only be observed in AC powered circuits for a more in-depth understanding of reactance and resistance check out our dedicated video I'll provide the link in the description let's introduce an inductive reactant to this Circuit by connecting an
inductor in series with the light bulb now what will happen we know that the inductor does not like to change the current across it so instead of changing with the voltage the current will change after the voltage in this scenario we can say the voltage leads the current or the current lags the voltage if you are not familiar with inductance I highly recommend watching our inductance video due to this lag there are instances where voltage is positive but current is negative and vice versa let's focus on one cycle what has happened to the power now
we can calculate the instantaneous power over one cycle as before we can observe negative power regions where the current is negative while the voltage is positive and vice versa this negative power doesn't imply that the load is generating energy instead it simply means that during those brief moments the power flow reverses Direction with the stored energy in the magnetic field of the inductor temporarily returning to the source let's attempt to explain this more clearly here is the voltage and current graph of our circuit now let's draw the power graph again as mentioned earlier there are
negative power regions and positive power regions at the beginning power is positive indicating that the source delivers energy to the Circuit part of this energy is used to create the magnetic field around the inductor and the the remaining energy is used by the light bulb the negative part represents the energy stored in the magnetic field of the inductor now the magnetic field is discharging and the energy is flowing back to the source however this energy does not go in a useful way it only exchanges back and forth between the source and the inductor during this
period the light bulb does not receive energy this process repeats now let's compare the two circuits the resistance of the the light bulb is the same for both circuits the RMS voltage of the power source is equal in both let's assume the RMS current for both circuits is equal voltage and current for both circuits can be plotted like this now let's draw the power graph in circuit a the power is always positive indicating that all the energy supplied by the power source is consumed by the light bulb however in circuit B there are negative power
regions only the positive regions represent the energy transferred from the power source to the light bulb let's focus solely on the energy transferred from the source to the load in circuit B it's evident that due to the effect of the inductor the amount of power received by the light bulb has decreased however the power source still supplies the same RMS voltage and current let's calculate the average power consumed by the load in these circuits as we discussed earlier for circuit a the average power is equal to its RMS voltage multiplied by RMS current however it's
clear that the average power of circuit B is less than its RMS volt voltage multiplied by RMS current now let's focus on circuit B the load requires this much power to operate properly and the source is supposed to supply that amount however due to the inductive reactant of the inductor the load only receives a reduced amount of power in order to receive the required power at the light bulb as before the load has to supply more power than earlier therefore the load has to increase the current let's compare this with circuit a even though the
load receives the same power in both circuits in circuit B the load has to supply more so it seems some amount of energy goes to waste in circuit B let's summarize in alternating current circuits if there is a net reactance the circuit's current will lead or lag the voltage when we draw the power graph based on this VI graph we observe both negative and positive power flows this occurs because the power is divided part of it energizes the load while some power is stored by the reactive components of the circuit this stored power in the
reactive components flows back to the power source with each cycle therefore it's evident that reactive components do not consume power instead they store and release power back we will only focus on the power coming from the power source to the Circuit disregarding the negative power flow now it appears that the energy from the power source is consumed by both the load and the inductor this can be Illustrated in the power graph like this by averaging these values we can illustrate the energy distribution the energy transferred to the reactive components doesn't perform useful work so we
refer to this power as reactive power we can calculate the reactive power of a circuit if we know the reactant voltage or current only the energy transferred to the resistive load performs useful work known as real power or active power we can calculate the active power of a circuit if we know the resistance voltage or current the amount of power expected based solely on the RMS voltage and current of the power source is called apparent power it can be calculated by knowing two of voltage current and impedance however apparent power might be higher than the
real power dissipation because of the lagging or leading between voltage and current which creates reactive power contributing to apparent power this is why it's called apparent normally we measure power using Watts however here we have mentioned three types of power if we measure all three using Watts it will be confusing therefore we only measure the active or real power using Watts because it is the power actually doing the work as discussed before reactive power does not actually do any work it oscillates back and forth between the power source and reactive components however it causes an
increase in current in the circuit so it is important to measure this power reactive power is measured using volt ampers reactive or V apparent power represents the total power seemingly available in the circuit since apparent power includes both real and reactive power measuring it in Watts wouldn't provide a complete picture it would only represent the real power component omitting the important information about the reactive power present therefore it is measured using volt ampers VA these three types of power true reactive and apparent are interrelated and can be represented graphically by what we call the power
triangle the power triangle is a right angled triangle triangle the horizontal or adjacent side represents the circuit's real power P the vertical or opposite side represents the circuit's reactive power q and the hypotenuse represents the resulting apparent power s the angle between the real power and apparent power signifies the impedance phase angle by applying trigonometric principles we can calculate the length of any side given the lengths of the other two sides or the length of one side and an angle when powering up any type of load it's crucial to assess how much of the supplied
power is effectively utilized to perform work in a circuit Engineers utilize the ratio of actual real power to the apparent power termed as the power factor of a circuit power factor serves as an expression of Energy Efficiency ranging from 0 to 1 typically represented as a percentage the lower the power factor the less efficient the power usage if the power factor is equal to one that means active power and apparent power is equal hence there is no reactive power all the power supplied by the power source is going to the load to do work if
the power factor is less than one it indicates that the apparent power is larger than the active power this implies that some of the power is wasted as reactive power a lower power factor indicates that more power is wasted as reactive power rendering such circuits highly ineffective they draw an unnecessary amount of current this reactive power necessitates generators transmission lines and Transformers to be oversized to handle the additional current despite not performing any productive work this oversized infrastructure increases capital and maintenance costs for the power company additionally rea active power causes voltage fluctuations and energy
losses in the transmission lines further reducing efficiency by charging a penalty for low power factor power companies encourage customers to improve their power factor and reduce their Reliance on reactive power what causes the low power factor most of the time a low power factor is caused by inductive loads connected to the power lines inductive loads like Motors Transformers solenoid valves and relays draw reactive power which creates a magnetic field but does not contribute to actual work how to improve the power factor as mentioned earlier inductive loads such as Motors cause current lag however capacitors behave
oppositely to inductors they lead the current to counteract the effect of inductors we can install capacitors that lead the reactive power thus offsetting the lagging reactive power from inductive loads this effectively cancels out the lagging power and improves the overall power factor capacitor banks are relatively inexpensive and require minimal maintenance I believe you now have a clearer understanding of what occurs when there is reactants in a circuit let's proceed with some power factor calculations to reinforce these Concepts imagine we have connected an induction motor to our 120 volt 60 HZ main power supply if we
only consider the resistance we can draw the circuit like this the resistance of the motor is 60 ohms from there we can simply calculate the RMS current through the circuit which will be 2 Amp now let's calculate the power supplied by the power source this represents the active power of the circuit since we're neglecting the reactants the apparent power is equal to the active or real power therefore the power factor is equal to one indicating that all the power supplied by the power source is effectively utilized to do the work things may seem straightforward but
unfortunately this method is incorrect indeed neglecting inductance for inductive loads like Motors is not accurate let's consider both resistance and inductance now the inductance of this motor is 160 m Henry now the calculations are more complex than before we need to calculate the power factor which requires knowing both the apparent power and active power the apparent power can be calculated using the impedence of the circuit and current while the active power can be calculated using resistance and current resistance is given however we don't know the current and impedance of the circuit to calculate the current
we also need the impedance and voltage voltage is given to calculate the impedance we need to know both the resistance and reactance of this circuit we're given the resistance but to calculate the reactance we require both the inductive and capacitive reactants since there is no capacitance in this circuit the capacitive reactance is zero therefore we only need to calculate the inductive reactants to do this we require the inductance and frequency both of which are known this is where we should start our calculations as you can see it's a more complex process process right let's do
the math first we need to calculate the inductive reactants given that the resistance is provided and the capacitive reactance is zero with the inductive reactance determined we can then calculate the impedance of this circuit with the voltage provided we can calculate the current after obtaining the current we can calculate the apparent power using the impedance and current then the active or real power can be determined using the current and resistance Now we move to the final step calculating the power factor it seems the power factor of this system is 0.705 Which is less than one
that means only 70.5% of the power is utilized for work while the rest is lost within the circuit this indicates inefficiency to improve this we can connect a capacitor to increase the power factor let's see how it's done the poor power factor is primarily due to the inductive reactants to counteract this we need to add a capacitor in parallel to our example circuit let's determine the value of the capacitor we need to connect the capacitor is meant to counteract the inductive reactive power of the circuit therefore our first step is to determine the inductive reactive
power of this circuit which we aim to minimize now we have to connect a parallel capacitor that generates an equal and opposite capacitive reactive power to cancel out the total reactive power so the capacitive reactive power should also be 11 19.9 2 V now we need to calculate the correct capacitive reactants to produce an equivalent amount of capacitive reactive power in this circuit it's important to note that adding a capacitor will alter the current therefore we can't employ the same current value to calculate the capacitive reactive power as this capacitor will be directly connected in
parallel with the source we'll utilize the power formula which begins with voltage and reactants we have determined the reactance of the capacitor now let's find the capacitance required to achieve this reactance in the circuit Eureka we have calculated the capacitance to correct the power factor we need to install a 22.09 microfarads capacitor Bank in parallel with this circuit that's all for today if you think my contents are valuable to the world you are welcome to join my patreon Community like And subscribe to Professor mad for more interesting videos
