Understanding AC Circuit Components and Analysis

Good day everyone and welcome back. Last class we learned how to represent a complex phasor quantity from a trigonometrical time varying function. Now we are adequately prepared to start the topic AC circuit analysis. Firstly we are going to look at the components we would typically find in AC circuits. We'll take a look at how they work or behave in AC circuits. how they are quantified and ultimately how they are used in phasor calculations for AC circuit analysis. We have resistance which we covered before in a DC context, then I'll introduce inductance and capacitor. Then overall we'll look at impedance and how they behave in different circuit connections. And finally we'll take a look at the method we can employ for AC circuit analysis. In Lecture 2, Basic Concepts and Laws, you were introduced to Ohm's Law, B is equal to IR. This form of Ohm's Law is specific to DC circuits where resistance R was the only passive component in the DC circuit. For AC circuits, however, we have different passive components, which are collectively called impedance and denoted by Z at an angle. Also remember for AC circuits, which are now alternating or changing with time, we will be using phasors instead of constants, which have both a magnitude and an angle. So each of these quantities have angles and Ohm's law for AC circuits is now a complex calculation. AC loads are passive components quantified by the impedance. Impedance Z is the ratio of the phasor voltage to the phasor current measured in ohms, which is ohms log V on I is equal to Z. The load type influences the phase angle of the current flowing through it. There are different types of loads, resistors, inductors, and capacitors, and they may cause a phase shift in the current, meaning they may affect the current angle theta. So these types of loads in a circuit are quantified collectively as a complex number z phi in polar form. In rectangular form it is r plus j x. The real component quantifies the resistance and the imaginary or reactive component quantifies inductance and capacitance. Revisiting resistors, we know its resistance is dependent on the material's resistivity, length, and cross-sectional area. And we know it obeys Ohm's law V is equal to IR. Let's say we have an AC voltage source supplying a resistance in a closed circuit. Applying Ohm's law specific to AC circuits where the current is at time varying trigonometrical function and the resistance is a constant. The voltage across the resistor, Vr according to Ohm's law, is Vm sin omega t. From this analysis, we can see that the angles of the voltage across the resistor and the current through it is the same. Therefore, the voltage is in phase with the current. This means the waveform coincides with each other. as shown in the time domain graph. From the calculation, the AC resistance is given by the peak voltage divided by the peak current. The time domain calculation is the function Vr over the function Ir as shown. Let's do the calculation in the phasor domain. Here we have a complex calculation V phi. on I phi. And since the angles are the same, remember the angle of the phasor is the phase shift. That is where the waveform starts. Both are zero degrees. So a V rms on I rms will give us the same resistance value, which is a real number. So for resistors in an AC circuit, we have the formula in the time domain. and the waveforms that are in phase. And in the frequency or phasor domain, we have the formula using RMS values that are complex numbers but the angles are zero, and the phasor quantities on the phasor diagram which are also in phase. A capacitor is a passive element that has the ability to store energy or charge in its electric field. It consists of two conductive parallel plates with a dielectric between them. We have the symbols here for capacitors, unpolarized, polarized, and variable. Some properties of capacitors are the voltage across it cannot change instantaneously. The current through it only exists while the capacitor voltage is changing. It ideally does not dissipate energy and it acts like an open circuit for DC sources. This last one we will explore later after fully understanding how the capacitor works. So we have two conductive plates and a dielectric between them. and the capacitance is the ability of the capacitor to store a charge between its plates. We may have remembered when a voltage is applied to the terminals or plates, electrolysis occurs and the dielectric breakdown into positive and negative charges on the plates occur. This is how the capacitor charges and stores charge. The greater the capacitance, the greater the charge stored for an applied voltage. This capacitance value depends on the permittivity, the distance between the plates, and cross-sectional area of the plates. In case you all don't recall, permittivity is the ability of a material, the dielectric, to store electrical potential energy under the influence of an electric field. It's also called the dielectric constant. When the voltage applied across the plate changes, the current flowing through the capacitor deposits or removes charge from its plates, with the amount of charge being proportional to the applied voltage. And we have the formula Q is equal to CV. So this charging or discharging current is given by the formula I is equal to dq dt or C dv dt. And we also have formulae for the voltage across the capacitor and the work done to charge the capacitor. To explore charging and discharging of a capacitor further, let's look at a basic DC circuit. with a DC voltage source supplying a resistor and a capacitor in series. The source is turned on and supplies e volts for some time and then is turned off. From zero to T2 when the supply is on e is applied to the circuit components. Initially the voltage across the capacitor was zero. But now he is applying a voltage across its plates. Remember the capacitor does not instantaneously change its voltage and must charge to match this voltage. Charges must accumulate on the plates to create this electric potential. Initially the current flow is high and then it decreases to zero as the voltage The current magnitude is dictated by the formula I is equal to C dV dt, meaning the greater the change in voltage across the capacitor from its initial state, the more current will flow to charge the capacitor to achieve this applied voltage. From T on 2 to T, When the supply is off, the voltage applied across the plates is zero. At Tion 2, the voltage across the capacitor was not zero. It was fully charged. Again, the capacitor does not instantaneously change its voltage and must discharge to match the zero voltage. Charges must be removed on the plates to lose electrical potential. Initially, The current flow is high but in the opposite direction since it is discharging and then it decreases as the voltage reaches zero. Again the current magnitude is dictated by the formula I is equal to C dV dt. This time the change is decreasing voltage and so the current will flow in the opposite direction. discharge the capacitor to achieve this zero voltage. Like before, more current will flow initially since the change in voltage is largest at the start. So let's say we have an AC voltage source supplying a capacitor in a closed circuit. Applying the capacitor formula where the voltage is a time varying trigonometrical function and capacitance is a constant. When we substitute and differentiate the current through the capacitor IC is IM plus omega T or IM sine omega T plus 90. From this analysis we can see that the angles of the voltage across the capacitor and the current through it is different. The current is 90 degrees ahead of the voltage. Therefore, the current leads the voltage. This means the waveforms do not coincide with each other as shown in the time domain graph. There is a 90 degree shift in current ahead of the voltage. Take note here that ahead is earlier in time in the graph. So the current would start first and then the voltage. From the calculation, the peak voltage divided by the peak current gives the expression 1 on omega c. This is called the capacitive reactance. The time domain calculation is the function vc on ic. We can represent the current as sine omega t plus 90 since this is the same as cos omega t. This will be rigorous to solve in the time domain. So let's do the calculation in the phasor domain. We can convert the phasors to get rms v rms at 0 degrees over i rms at 90 degrees. We know 90 degrees is the j operator. So we now have V on J I. From before, we know V on I is 1 on omega C. This derivation shows us that the capacitive reactance is a negative imaginary value when being used in phasor calculations. Because 1 on omega C is X C. And when we bring J. from the denominator to the numerator it will become negative j. So for capacitors in an AC circuit we have the formula in the time domain and the waveforms that are out of phase by 90 degrees where the current occurs before the voltage it leads. And in the frequency of phasor domain, we have the formula using the RMS values and the phasor quantities on the phasor diagram showing a 90 degree leading current since anti-clockwise is the positive direction. I have only explained how the capacitor manages to shift the current through it ahead by 90 degrees mathematically. What physically happens to our capacitor throughout the whole cycle of the sinusoidal voltage will take a while to explain, more time than we have in this class. Therefore, I will create a class forum post explaining this. So if you all are interested in understanding exactly how a capacitor achieves this phase shift, you can look out for that explanation on my learning. Some practical examples of capacitors are variable or fixed capacitor banks used for power factor correction, a topic we'll soon cover, motor starting capacitors used to start single phase AC motors, and synchronous motors which have a capacitive movement which I will cover with the mechanical students in my year three course. We can also use capacitors to model any conductive surfaces with a dielectric in the middle. Examples of this are transmission lines, cables, and windings. For example, transmission lines and one transmission line and ground or another transmission line would be the plates and the dielectric would be the air between them. So this can be modeled as a capacitor. If we have capacitors in series, how can we quantify the equivalent capacitors? We can derive this using the formula we derived in the phasor domain. In a series circuit applying Kirchhoff's voltage law, the supply voltage V is equal to the sum of the voltages across the series capacitors in the loop. So substituting for the capacitor voltages in the phasor using the phasor formula I on j omega c and since I is the same through all the capacitors in a series circuit and j omega are constants we can factor out I on j omega After making it the subject of the equation, J omega V on I is equal to the sum of the reciprocals of all the series capacitors. Rearranging the formula, we can see that J omega V on I is also equal to the reciprocal of the equivalent capacitance 1 on CEQ. Therefore, 1 on CEQ is also equal to the sum of the reciprocals of all the series capacitance. If we have capacitors in parallel however, how can we quantify the equivalent capacitance? We can also derive this using our phasor domain formula. In a parallel circuit applying Kirchhoff's current law, the supply current I is equal to the sum of the currents through the parallel capacitors in the circuit. Substituting for the capacitor currents j omega vc and since v is the same across all capacitors in a parallel circuit and j omega are constants we can factor out these After making it the subject of the equation, I over j omega v is also equal to the equivalent capacitance, CEQ. Therefore, CEQ is also equal to the sum of the parallel capacitance. An inductor is a passive element that has the ability to store energy in its magnetic field. It consists of a coil of conducting wire, also known as a coil or choke. We have some symbols for air core, iron core and variable inductors. Some properties of inductors are the current through it cannot change instantaneously. The voltage across it only exists while the current is changing, it ideally does not dissipate energy. A practical non-ideal inductor has some resistance, Rw, and a capacitance, Cw. It acts like a short circuit for DC sources and we will explore that later after fully understanding how the inductor works. We have the model for the practical non-ideal inductor. So we have a wire wound in a coil around a piece of material, a solenoid. And the inductance is the ability of the coil or solenoid to produce magnetic field lines when current flows through it. So when a current flows through the coil, a magnetic field is developed. This is how the inductor stores and releases energy. The greater the inductance, the greater the field developed for a given current. This inductance value depends on the permeability, the number of turns of the coil, the length of the coil, and cross-sectional area of the coil. In case you all don't recall permeability, It is the measure of the ability of a material to support the formation of a magnetic field within itself. It is the degree of magnetization that a material obtains in response to an applied magnetic field. So this inductor obeys Faraday's law. When the current through the coil changes, an electromotive force is induced in the coil. proportional to the rate of change of current. This developed EMF of voltage is given by the formula VL is equal to minus L GIDT. We also have formulae for the current through the inductor and the work done to develop the field in the inductor. To explore storage and dissipation of magnetic energy and inductor further, let's look at the basic DC circuit with a DC voltage source applied supplying a resistor and an inductor in series. Like before, the source is turned on and supplies eVolts for some time and then is turned off. From zero to T on 2, when the supply is on, E is applied to the circuit. Initially, the current through the inductor was zero, but now E is applying a voltage causing a current to flow, but not instantaneously change its current. and the current slowly increases. Initially the voltage across the coil is high to prevent an increase in current and then it decreases to zero as the current If voltage magnitude is dictated by the formula V is equal to L di dt, meaning the greater the change in current through the inductor from its initial state, the more voltage will be induced across the coil. From T on 2 to T, when the supply voltage is off, the voltage applied to the coil is at zero. At T on 2, the voltage across the inductor was zero since the current was constant. Again, the inductor does not instantaneously change its current and slowly decreases the voltage spikes but in opposite direction and then it decreases as the current reaches. Again, the voltage magnitude is dictated by the formula V is equal to LgId. This time, the change is decreasing currently. so the voltage will develop to prevent the decrease in current. Like before, more voltage will be induced initially since the change in current is largest at the beginning. So let's say we have an AC voltage source supplying an inductor in a closed circuit. Applying the inductor formula Faraday's law, where the voltage is a time varying trigonometrical function, the inductance is also a constant, when we substitute and differentiate, the voltage across the inductor VL is Vm cos omega t or Vm sine omega t plus 90 degrees. From this analysis, we can see that the angles of the voltage across the inductor and the current through it is different. The voltage is 90 degrees ahead of the current. Therefore, we can say that the current lags the voltage. This means, like the case for the capacitor, the waveforms do not coincide with each other and as shown in the time domain graphs. However, this time There is a 90 degree shift in the current behind of the voltage. Take note that behind is later in time on this graph. So the current starts after the voltage. It lags the voltage. From the calculation, the peak voltage divided by the peak current gives the expression omega L. This is called the inductive reactor. The time domain calculation is the function VL on IL. We can represent the voltage as a sine omega t plus 90 since this is the same as cos omega t. This again will be rigorous to solve in the time domain. So let's do it in the phasor domain. We can convert to phasors. to get V rms at positive 90 degrees on I rms at zero degree. We know 90 degrees is the J operator so we now have J V on the I. From before we know V on I is omega L so substituting gives J omega L. This derivation shows us that the inductive reactance is a positive imaginary value when being used in phasor calculations. So for inductors in an AC circuit, we have the formula in the time domain and the waveforms that are out of phase by 90 degrees, where the current occurs after the voltage it lags, and in the frequency of phasor domain, we have the formula using RMS values and the phasor quantities on the phasor diagram show a 90 degree lag in current since clockwise is the negative direction. I have only explained how the inductor manages to shift the current through it behind by 90 degrees mathematically. What physically happens to an inductor throughout the whole cycle of sinusoid current will take a while to explain. Therefore, I will create a class forum post explaining this as well. So if you all are interested in understanding exactly how an inductor achieves this phase shift, you can look out for that explanation on my website. Some practical examples of inductive loads in appliances are motos and these motos are typically found in air conditioners, washers and dryers, refrigerators and water pumps. All these appliances uses motors for compressors, pumps, agitators. All motors contains multiple coils of we also have inductors and transformers are found in microwaves electric eaters and welding plants These appliances uses transformers which are sets of coils wound on ferromagnetic materials. We will do this topic in detail after AC circuit analysis. So if we have inductors in series, how can we quantify the equivalent inductance? We can also derive this using our phasor domain formula. In a series circuit applying Kirchhoff's voltage law, The supply voltage V is equal to the sum of the voltages across the series inductor. By substituting for the inductor voltages using the phasor formula, G omega I L, and since I is the same throughout all the inductors in a series circuit, and G omega are constants, we could factor out G omega I. After making it the subject of the equation, V on j omega i is equal to the sum of all the series inductance. Rearranging the formula, we could see V on j omega i is also equal to the equivalent inductance L eq. Therefore, L eq is also equal to the sum of the series inductance. If we have inductors in parallel however, how can we quantify the equivalent inductor? We can also derive this using our phasor domain formula. In a parallel circuit applying Kirchhoff's current law, the supply current is equal to the sum of the currents to the parallel inductors in the circuit. Substituting for the inductor currents using the phasor formula V on j omega L, And since v is the same across all inductors in a parallel circuit and j omega are constants, we can factor out v on j omega. After making it the subject of the equation, j omega L on v, j omega i on v is equal to the sum of the parallel, the reciprocal, sorry, the sum of the reciprocal of the parallel inductors. Rearranging the formula we can see that j omega l v is also equal to the reciprocal of the equivalent inductance 1 on L eq. Therefore 1 on L eq is also equal to the sum of the reciprocals of all the parallel inductors. So we have derived how inductors are represented by a positive imaginary number and capacitors are represented by a negative imaginary number for impedance in the phasor domain. This x value is reactance and can be different polarities and values based on the reactive component of the circuit. If the reactive components are in series, they can be added to find the resultant reactants as shown in the diagram. Again, this can be expressed in polar form as the magnitude z at an angle phi. This calculation simply shows how to convert from polar to rectangular form. So again, the benefit of representing time varying values as phases is to use the same simple method of analysis. Although we would be working with complex numbers, we can still use the voltage and current divider theorem as we saw in the last couple slides. Just to be clear because it isn't shown here, all these variables are complex, meaning they have both magnitudes and angles. So for a series circuit, we can find the current by applying Ohm's law and dividing the source voltage by the sum of the equivalent of the parallel impedances. The voltage divider theorem can also be used to find the voltages across each of the individual impedances z1 and z2. And for a parallel connection, we can also apply Ohm's law to find the currents and voltages. The current divider theorem can be used to find the currents through each of the individual impedances z1 and z2. This slide shows a similar derivation like we did for resistance, inductance, and capacitance, but this time for impedance, which is the complex representation of an AC load which may consist of real and or reactive components. Again, Ohm's law applied to AC circuits gives us impedance, Z, if we divide the phase, the phasor voltage by the phasor current. In a series circuit applying Kirchhoff's voltage law, the supply voltage V is equal to the sum of voltages across the impedances in the loop, substituting for the individual voltages across each impedance using the phasor formula IZ. And since I is the same throughout the series circuit, we can factor these variables. After making it the subject of the equation, V on i is equal to the sum of all the series impedances. As we know, V on i is also equal to the equivalent impedance, Z e cube. Therefore, Z e cube is equal to the sum of all the series impedances in the circuit. In a parallel circuit applying Kirchhoff's current law, the supply current I is equal to the sum of currents through the parallel impedances. Substituting for the individual currents in each branch using the phasor formula V on Z, and since V is the same across all impedances in a parallel circuit, we can factor this variable out. After making it the subject of the equation I on V, we can see this is equal to the sum of the reciprocals of the parallel impedances. As we know, I on V is also equal to the reciprocal of the equivalent impedance. Therefore, 1 on ZQ is equal to the sum of the reciprocal of all the parallel impedances. impedance in the circuit. If we are considering only two parallel impedances at a time, we can simplify this formula. Factorizing, we get the sum of the impedances over their product. This gives us 1 on Z eq, so Z eq is the reciprocal. In conclusion, the equivalent impedance of two impedances in parallel. given by their product over their sum. If the impedances are more than two but are equal, the equivalent is given by the value of the impedance divided by the number of them in parallel. This can be proven of course by using the general formula but I'll let you do that on your own. So let's do a simple example to help your understanding of the topic so far. We have a simple series circuit with an 8 ohm resistor, J18 ohm inductor, and minus J12 ohm capacitor. The voltage source supplies 100 volts and its reference is 0 degrees. Let's find the source current, I. Using Ohm's law, V on Z, we have 100 degrees. 100 volts at zero degrees divided by the resistance eight plus the inductive reactance j18 plus the inductive the capacitive reactance minus j12 the equivalent impedance here is 8 plus j6 Expressed in polar form it is 10 at 36.87 degrees and the current was calculated to be 10 at minus 36.87 degrees. Since the imaginary value of the equivalent impedance is positive, the resultant reactance of the circuit is inductive and this brings about a negative lagging current. which is drawn in the phasor diagram as shown as a lagging a current behind the voltage below the voltage which is the reference line i have it here drawn on a proper phasor diagram and also as waveforms in the time domain. Notice the green waveform which is the current starts after or lags the voltage in red. A lagging waveform occurs later in time. Remember that. So you've had a lot to take in in this lesson so I have a summary sheet prepared. In conclusion if an AC circuit has a strictly resistive load i.e resistances the current will be in phase with the voltage and if the ac circuit has an inductive load meaning the resistances and inductor and inductances in the system the current will lag the voltage if an ac circuit has a capacitive load meaning there are resistances and capacitors in the system the current will lead the voltage. Although I want to encourage all students to remember the physical operation of the components and the mathematical derivation of their effects, this may not be practical. So a lot of you all often use the anagram civil. If the circuit has a capacitive load, I comes before V. that is it would be leading and if the circuit has an inductive load v comes before i meaning the current will lag the voltage. On to some more complex examples we have a not too complex ac circuit and we need to determine the input impedance of the circuit in 50 radians per second. Now 50 radians per second is simply omega, the angular frequency of the phasors. Let's call the reactants of the two microfarad capacitor Z1. We utilize the equation for the capacitive reactants one on j omega c. to get minus 10 J ohms. Z2 is the 3 ohm for the resistance plus the capacitive reactance of the 10 microfarad minus J 2 ohms. And finally Z3 is the 8 ohm for the resistance Plus using the formula for inductive reactants j omega L, the reactants of the 0.2 Henry inductor is j 10 ohm. Now figuring out the configuration of the components is the most difficult part. The input impedance of the circuit means we want to find the equivalent impedance from these two input terminals. So what you do is start from the furthest end of from the terminals you are looking through and work towards them. So from the far right we can see Z3 is in parallel with Z2 and the resultant of that will be in series with Z1. Using product over sum the two impedances in parallel Z2 two and three can be resolved and we obtain 3.22 minus j 1.07 so i advise you do this calculation in a calculator using complex mode we then add that to the series reactants minus j10 which is z1 and the final answer is 3. 22 minus J11.070. So you all are free to find the equivalent capacitance of that capacitive reactance. You all know the formula is 1 on J omega C is equal to XC. Using the same process and equations we covered, try to find the input impedance of this circuit at home. Note that omega this time is 10 radians per second. I have provided the final answer for comparison with yours. Feel free to message on the class forum if you have any questions on this example. Example 2. asks for voltage across the 5 Henry inductor in the given circuit. The voltage source is given as a trig function, so we must first obtain its phasor representation. The magnitude of the voltage phasor would be the RMS of the sinusoid, given by the peak voltage, 20, divided by root 2, to give 14.14. at an angle of minus 15 degrees given by the phase shift of the waveform. The function as is gives us the omega, which in this case is four radians per second. The reactances of the inductor and the capacitor must then be formed. using the relevant formulae as shown. Now to find the voltage across the 5 Henry inductor, we must first find the parallel combination of the inductor and the capacitor. They are in parallel with each other and only this resultant impedance will give us the voltage across them. Let us call this impedance Z2 and Z1. So the equivalent impedance is Z2 and Z1 is the 60 ohm resistor. So Z2 was calculated using the product of the two reactances over the sum. So the answer was. obtained to be J100. Now we have a series circuit. We can use the voltage divider theorem to find the voltage across Z2. This voltage is given by the supply voltage Z2 sorry the supply voltage multiplied by z2 over the sum of z1 and z2. One thing we must remember is to include all j operators in these equations since we are dealing with phasors. The phasor voltage across the inductor was calculated to be 12.12 at 15.96. degrees. Finally, if we are asked to revert to the time domain representation, the peak voltage would be 12.12 multiplied by root 2 to get 17.15 and the phase shift is positive 15.96 degrees. And we express the function in the typical form. So we've gone through a lot today with some examples. I hope you all understood most of it. If you all have any issues, feel free to post your problems on the class forum or bring them up in the feedback session.

We'll take a look at how they work or behave in AC circuits. how they are quantified and ultimately how they are used in phasor calculations for AC circuit analysis. We have resistance which we covered before in a DC context, then I'll introduce inductance and capacitor. Then overall we'll look at impedance and how they behave in different circuit connections.

And finally we'll take a look at the method we can employ for AC circuit analysis. In Lecture 2, Basic Concepts and Laws, you were introduced to Ohm's Law, B is equal to IR. This form of Ohm's Law is specific to DC circuits where resistance R was the only passive component in the DC circuit. For AC circuits, however, we have different passive components, which are collectively called impedance and denoted by Z at an angle.

Also remember for AC circuits, which are now alternating or changing with time, we will be using phasors instead of constants, which have both a magnitude and an angle. So each of these quantities have angles and Ohm's law for AC circuits is now a complex calculation. AC loads are passive components quantified by the impedance. Impedance Z is the ratio of the phasor voltage to the phasor current measured in ohms, which is ohms log V on I is equal to Z. The load type influences the phase angle of the current flowing through it.

There are different types of loads, resistors, inductors, and capacitors, and they may cause a phase shift in the current, meaning they may affect the current angle theta. So these types of loads in a circuit are quantified collectively as a complex number z phi in polar form. In rectangular form it is r plus j x.

The real component quantifies the resistance and the imaginary or reactive component quantifies inductance and capacitance. Revisiting resistors, we know its resistance is dependent on the material's resistivity, length, and cross-sectional area. And we know it obeys Ohm's law V is equal to IR.

Let's say we have an AC voltage source supplying a resistance in a closed circuit. Applying Ohm's law specific to AC circuits where the current is at time varying trigonometrical function and the resistance is a constant. The voltage across the resistor, Vr according to Ohm's law, is Vm sin omega t.

From this analysis, we can see that the angles of the voltage across the resistor and the current through it is the same. Therefore, the voltage is in phase with the current. This means the waveform coincides with each other.

as shown in the time domain graph. From the calculation, the AC resistance is given by the peak voltage divided by the peak current. The time domain calculation is the function Vr over the function Ir as shown.

Let's do the calculation in the phasor domain. Here we have a complex calculation V phi. on I phi.

And since the angles are the same, remember the angle of the phasor is the phase shift. That is where the waveform starts. Both are zero degrees.

So a V rms on I rms will give us the same resistance value, which is a real number. So for resistors in an AC circuit, we have the formula in the time domain. and the waveforms that are in phase. And in the frequency or phasor domain, we have the formula using RMS values that are complex numbers but the angles are zero, and the phasor quantities on the phasor diagram which are also in phase. A capacitor is a passive element that has the ability to store energy or charge in its electric field.

It consists of two conductive parallel plates with a dielectric between them. We have the symbols here for capacitors, unpolarized, polarized, and variable. Some properties of capacitors are the voltage across it cannot change instantaneously. The current through it only exists while the capacitor voltage is changing.

It ideally does not dissipate energy and it acts like an open circuit for DC sources. This last one we will explore later after fully understanding how the capacitor works. So we have two conductive plates and a dielectric between them. and the capacitance is the ability of the capacitor to store a charge between its plates.

We may have remembered when a voltage is applied to the terminals or plates, electrolysis occurs and the dielectric breakdown into positive and negative charges on the plates occur. This is how the capacitor charges and stores charge. The greater the capacitance, the greater the charge stored for an applied voltage.

This capacitance value depends on the permittivity, the distance between the plates, and cross-sectional area of the plates. In case you all don't recall, permittivity is the ability of a material, the dielectric, to store electrical potential energy under the influence of an electric field. It's also called the dielectric constant.

When the voltage applied across the plate changes, the current flowing through the capacitor deposits or removes charge from its plates, with the amount of charge being proportional to the applied voltage. And we have the formula Q is equal to CV. So this charging or discharging current is given by the formula I is equal to dq dt or C dv dt. And we also have formulae for the voltage across the capacitor and the work done to charge the capacitor.

To explore charging and discharging of a capacitor further, let's look at a basic DC circuit. with a DC voltage source supplying a resistor and a capacitor in series. The source is turned on and supplies e volts for some time and then is turned off.

From zero to T2 when the supply is on e is applied to the circuit components. Initially the voltage across the capacitor was zero. But now he is applying a voltage across its plates. Remember the capacitor does not instantaneously change its voltage and must charge to match this voltage. Charges must accumulate on the plates to create this electric potential.

Initially the current flow is high and then it decreases to zero as the voltage The current magnitude is dictated by the formula I is equal to C dV dt, meaning the greater the change in voltage across the capacitor from its initial state, the more current will flow to charge the capacitor to achieve this applied voltage. From T on 2 to T, When the supply is off, the voltage applied across the plates is zero. At Tion 2, the voltage across the capacitor was not zero.

It was fully charged. Again, the capacitor does not instantaneously change its voltage and must discharge to match the zero voltage. Charges must be removed on the plates to lose electrical potential. Initially, The current flow is high but in the opposite direction since it is discharging and then it decreases as the voltage reaches zero.

Again the current magnitude is dictated by the formula I is equal to C dV dt. This time the change is decreasing voltage and so the current will flow in the opposite direction. discharge the capacitor to achieve this zero voltage. Like before, more current will flow initially since the change in voltage is largest at the start.

So let's say we have an AC voltage source supplying a capacitor in a closed circuit. Applying the capacitor formula where the voltage is a time varying trigonometrical function and capacitance is a constant. When we substitute and differentiate the current through the capacitor IC is IM plus omega T or IM sine omega T plus 90. From this analysis we can see that the angles of the voltage across the capacitor and the current through it is different. The current is 90 degrees ahead of the voltage. Therefore, the current leads the voltage.

This means the waveforms do not coincide with each other as shown in the time domain graph. There is a 90 degree shift in current ahead of the voltage. Take note here that ahead is earlier in time in the graph. So the current would start first and then the voltage. From the calculation, the peak voltage divided by the peak current gives the expression 1 on omega c.

This is called the capacitive reactance. The time domain calculation is the function vc on ic. We can represent the current as sine omega t plus 90 since this is the same as cos omega t. This will be rigorous to solve in the time domain.

So let's do the calculation in the phasor domain. We can convert the phasors to get rms v rms at 0 degrees over i rms at 90 degrees. We know 90 degrees is the j operator. So we now have V on J I. From before, we know V on I is 1 on omega C. This derivation shows us that the capacitive reactance is a negative imaginary value when being used in phasor calculations.

Because 1 on omega C is X C. And when we bring J. from the denominator to the numerator it will become negative j. So for capacitors in an AC circuit we have the formula in the time domain and the waveforms that are out of phase by 90 degrees where the current occurs before the voltage it leads. And in the frequency of phasor domain, we have the formula using the RMS values and the phasor quantities on the phasor diagram showing a 90 degree leading current since anti-clockwise is the positive direction.

I have only explained how the capacitor manages to shift the current through it ahead by 90 degrees mathematically. What physically happens to our capacitor throughout the whole cycle of the sinusoidal voltage will take a while to explain, more time than we have in this class. Therefore, I will create a class forum post explaining this.

So if you all are interested in understanding exactly how a capacitor achieves this phase shift, you can look out for that explanation on my learning. Some practical examples of capacitors are variable or fixed capacitor banks used for power factor correction, a topic we'll soon cover, motor starting capacitors used to start single phase AC motors, and synchronous motors which have a capacitive movement which I will cover with the mechanical students in my year three course. We can also use capacitors to model any conductive surfaces with a dielectric in the middle.

Examples of this are transmission lines, cables, and windings. For example, transmission lines and one transmission line and ground or another transmission line would be the plates and the dielectric would be the air between them. So this can be modeled as a capacitor. If we have capacitors in series, how can we quantify the equivalent capacitors? We can derive this using the formula we derived in the phasor domain.

In a series circuit applying Kirchhoff's voltage law, the supply voltage V is equal to the sum of the voltages across the series capacitors in the loop. So substituting for the capacitor voltages in the phasor using the phasor formula I on j omega c and since I is the same through all the capacitors in a series circuit and j omega are constants we can factor out I on j omega After making it the subject of the equation, J omega V on I is equal to the sum of the reciprocals of all the series capacitors. Rearranging the formula, we can see that J omega V on I is also equal to the reciprocal of the equivalent capacitance 1 on CEQ.

Therefore, 1 on CEQ is also equal to the sum of the reciprocals of all the series capacitance. If we have capacitors in parallel however, how can we quantify the equivalent capacitance? We can also derive this using our phasor domain formula.

In a parallel circuit applying Kirchhoff's current law, the supply current I is equal to the sum of the currents through the parallel capacitors in the circuit. Substituting for the capacitor currents j omega vc and since v is the same across all capacitors in a parallel circuit and j omega are constants we can factor out these After making it the subject of the equation, I over j omega v is also equal to the equivalent capacitance, CEQ. Therefore, CEQ is also equal to the sum of the parallel capacitance. An inductor is a passive element that has the ability to store energy in its magnetic field.

It consists of a coil of conducting wire, also known as a coil or choke. We have some symbols for air core, iron core and variable inductors. Some properties of inductors are the current through it cannot change instantaneously.

The voltage across it only exists while the current is changing, it ideally does not dissipate energy. A practical non-ideal inductor has some resistance, Rw, and a capacitance, Cw. It acts like a short circuit for DC sources and we will explore that later after fully understanding how the inductor works. We have the model for the practical non-ideal inductor. So we have a wire wound in a coil around a piece of material, a solenoid.

And the inductance is the ability of the coil or solenoid to produce magnetic field lines when current flows through it. So when a current flows through the coil, a magnetic field is developed. This is how the inductor stores and releases energy. The greater the inductance, the greater the field developed for a given current. This inductance value depends on the permeability, the number of turns of the coil, the length of the coil, and cross-sectional area of the coil.

In case you all don't recall permeability, It is the measure of the ability of a material to support the formation of a magnetic field within itself. It is the degree of magnetization that a material obtains in response to an applied magnetic field. So this inductor obeys Faraday's law. When the current through the coil changes, an electromotive force is induced in the coil. proportional to the rate of change of current.

This developed EMF of voltage is given by the formula VL is equal to minus L GIDT. We also have formulae for the current through the inductor and the work done to develop the field in the inductor. To explore storage and dissipation of magnetic energy and inductor further, let's look at the basic DC circuit with a DC voltage source applied supplying a resistor and an inductor in series.

Like before, the source is turned on and supplies eVolts for some time and then is turned off. From zero to T on 2, when the supply is on, E is applied to the circuit. Initially, the current through the inductor was zero, but now E is applying a voltage causing a current to flow, but not instantaneously change its current. and the current slowly increases. Initially the voltage across the coil is high to prevent an increase in current and then it decreases to zero as the current If voltage magnitude is dictated by the formula V is equal to L di dt, meaning the greater the change in current through the inductor from its initial state, the more voltage will be induced across the coil.

From T on 2 to T, when the supply voltage is off, the voltage applied to the coil is at zero. At T on 2, the voltage across the inductor was zero since the current was constant. Again, the inductor does not instantaneously change its current and slowly decreases the voltage spikes but in opposite direction and then it decreases as the current reaches.

Again, the voltage magnitude is dictated by the formula V is equal to LgId. This time, the change is decreasing currently. so the voltage will develop to prevent the decrease in current.

Like before, more voltage will be induced initially since the change in current is largest at the beginning. So let's say we have an AC voltage source supplying an inductor in a closed circuit. Applying the inductor formula Faraday's law, where the voltage is a time varying trigonometrical function, the inductance is also a constant, when we substitute and differentiate, the voltage across the inductor VL is Vm cos omega t or Vm sine omega t plus 90 degrees.

From this analysis, we can see that the angles of the voltage across the inductor and the current through it is different. The voltage is 90 degrees ahead of the current. Therefore, we can say that the current lags the voltage. This means, like the case for the capacitor, the waveforms do not coincide with each other and as shown in the time domain graphs. However, this time There is a 90 degree shift in the current behind of the voltage.

Take note that behind is later in time on this graph. So the current starts after the voltage. It lags the voltage. From the calculation, the peak voltage divided by the peak current gives the expression omega L. This is called the inductive reactor.

The time domain calculation is the function VL on IL. We can represent the voltage as a sine omega t plus 90 since this is the same as cos omega t. This again will be rigorous to solve in the time domain. So let's do it in the phasor domain.

We can convert to phasors. to get V rms at positive 90 degrees on I rms at zero degree. We know 90 degrees is the J operator so we now have J V on the I. From before we know V on I is omega L so substituting gives J omega L.

This derivation shows us that the inductive reactance is a positive imaginary value when being used in phasor calculations. So for inductors in an AC circuit, we have the formula in the time domain and the waveforms that are out of phase by 90 degrees, where the current occurs after the voltage it lags, and in the frequency of phasor domain, we have the formula using RMS values and the phasor quantities on the phasor diagram show a 90 degree lag in current since clockwise is the negative direction. I have only explained how the inductor manages to shift the current through it behind by 90 degrees mathematically.

What physically happens to an inductor throughout the whole cycle of sinusoid current will take a while to explain. Therefore, I will create a class forum post explaining this as well. So if you all are interested in understanding exactly how an inductor achieves this phase shift, you can look out for that explanation on my website.

Some practical examples of inductive loads in appliances are motos and these motos are typically found in air conditioners, washers and dryers, refrigerators and water pumps. All these appliances uses motors for compressors, pumps, agitators. All motors contains multiple coils of we also have inductors and transformers are found in microwaves electric eaters and welding plants These appliances uses transformers which are sets of coils wound on ferromagnetic materials. We will do this topic in detail after AC circuit analysis. So if we have inductors in series, how can we quantify the equivalent inductance?

We can also derive this using our phasor domain formula. In a series circuit applying Kirchhoff's voltage law, The supply voltage V is equal to the sum of the voltages across the series inductor. By substituting for the inductor voltages using the phasor formula, G omega I L, and since I is the same throughout all the inductors in a series circuit, and G omega are constants, we could factor out G omega I.

After making it the subject of the equation, V on j omega i is equal to the sum of all the series inductance. Rearranging the formula, we could see V on j omega i is also equal to the equivalent inductance L eq. Therefore, L eq is also equal to the sum of the series inductance.

If we have inductors in parallel however, how can we quantify the equivalent inductor? We can also derive this using our phasor domain formula. In a parallel circuit applying Kirchhoff's current law, the supply current is equal to the sum of the currents to the parallel inductors in the circuit.

Substituting for the inductor currents using the phasor formula V on j omega L, And since v is the same across all inductors in a parallel circuit and j omega are constants, we can factor out v on j omega. After making it the subject of the equation, j omega L on v, j omega i on v is equal to the sum of the parallel, the reciprocal, sorry, the sum of the reciprocal of the parallel inductors. Rearranging the formula we can see that j omega l v is also equal to the reciprocal of the equivalent inductance 1 on L eq.

Therefore 1 on L eq is also equal to the sum of the reciprocals of all the parallel inductors. So we have derived how inductors are represented by a positive imaginary number and capacitors are represented by a negative imaginary number for impedance in the phasor domain. This x value is reactance and can be different polarities and values based on the reactive component of the circuit.

If the reactive components are in series, they can be added to find the resultant reactants as shown in the diagram. Again, this can be expressed in polar form as the magnitude z at an angle phi. This calculation simply shows how to convert from polar to rectangular form.

So again, the benefit of representing time varying values as phases is to use the same simple method of analysis. Although we would be working with complex numbers, we can still use the voltage and current divider theorem as we saw in the last couple slides. Just to be clear because it isn't shown here, all these variables are complex, meaning they have both magnitudes and angles.

So for a series circuit, we can find the current by applying Ohm's law and dividing the source voltage by the sum of the equivalent of the parallel impedances. The voltage divider theorem can also be used to find the voltages across each of the individual impedances z1 and z2. And for a parallel connection, we can also apply Ohm's law to find the currents and voltages.

The current divider theorem can be used to find the currents through each of the individual impedances z1 and z2. This slide shows a similar derivation like we did for resistance, inductance, and capacitance, but this time for impedance, which is the complex representation of an AC load which may consist of real and or reactive components. Again, Ohm's law applied to AC circuits gives us impedance, Z, if we divide the phase, the phasor voltage by the phasor current. In a series circuit applying Kirchhoff's voltage law, the supply voltage V is equal to the sum of voltages across the impedances in the loop, substituting for the individual voltages across each impedance using the phasor formula IZ.

And since I is the same throughout the series circuit, we can factor these variables. After making it the subject of the equation, V on i is equal to the sum of all the series impedances. As we know, V on i is also equal to the equivalent impedance, Z e cube. Therefore, Z e cube is equal to the sum of all the series impedances in the circuit. In a parallel circuit applying Kirchhoff's current law, the supply current I is equal to the sum of currents through the parallel impedances.

Substituting for the individual currents in each branch using the phasor formula V on Z, and since V is the same across all impedances in a parallel circuit, we can factor this variable out. After making it the subject of the equation I on V, we can see this is equal to the sum of the reciprocals of the parallel impedances. As we know, I on V is also equal to the reciprocal of the equivalent impedance.

Therefore, 1 on ZQ is equal to the sum of the reciprocal of all the parallel impedances. impedance in the circuit. If we are considering only two parallel impedances at a time, we can simplify this formula. Factorizing, we get the sum of the impedances over their product. This gives us 1 on Z eq, so Z eq is the reciprocal.

In conclusion, the equivalent impedance of two impedances in parallel. given by their product over their sum. If the impedances are more than two but are equal, the equivalent is given by the value of the impedance divided by the number of them in parallel.

This can be proven of course by using the general formula but I'll let you do that on your own. So let's do a simple example to help your understanding of the topic so far. We have a simple series circuit with an 8 ohm resistor, J18 ohm inductor, and minus J12 ohm capacitor.

The voltage source supplies 100 volts and its reference is 0 degrees. Let's find the source current, I. Using Ohm's law, V on Z, we have 100 degrees. 100 volts at zero degrees divided by the resistance eight plus the inductive reactance j18 plus the inductive the capacitive reactance minus j12 the equivalent impedance here is 8 plus j6 Expressed in polar form it is 10 at 36.87 degrees and the current was calculated to be 10 at minus 36.87 degrees.

Since the imaginary value of the equivalent impedance is positive, the resultant reactance of the circuit is inductive and this brings about a negative lagging current. which is drawn in the phasor diagram as shown as a lagging a current behind the voltage below the voltage which is the reference line i have it here drawn on a proper phasor diagram and also as waveforms in the time domain. Notice the green waveform which is the current starts after or lags the voltage in red. A lagging waveform occurs later in time. Remember that.

So you've had a lot to take in in this lesson so I have a summary sheet prepared. In conclusion if an AC circuit has a strictly resistive load i.e resistances the current will be in phase with the voltage and if the ac circuit has an inductive load meaning the resistances and inductor and inductances in the system the current will lag the voltage if an ac circuit has a capacitive load meaning there are resistances and capacitors in the system the current will lead the voltage. Although I want to encourage all students to remember the physical operation of the components and the mathematical derivation of their effects, this may not be practical. So a lot of you all often use the anagram civil. If the circuit has a capacitive load, I comes before V. that is it would be leading and if the circuit has an inductive load v comes before i meaning the current will lag the voltage.

On to some more complex examples we have a not too complex ac circuit and we need to determine the input impedance of the circuit in 50 radians per second. Now 50 radians per second is simply omega, the angular frequency of the phasors. Let's call the reactants of the two microfarad capacitor Z1.

We utilize the equation for the capacitive reactants one on j omega c. to get minus 10 J ohms. Z2 is the 3 ohm for the resistance plus the capacitive reactance of the 10 microfarad minus J 2 ohms. And finally Z3 is the 8 ohm for the resistance Plus using the formula for inductive reactants j omega L, the reactants of the 0.2 Henry inductor is j 10 ohm. Now figuring out the configuration of the components is the most difficult part.

The input impedance of the circuit means we want to find the equivalent impedance from these two input terminals. So what you do is start from the furthest end of from the terminals you are looking through and work towards them. So from the far right we can see Z3 is in parallel with Z2 and the resultant of that will be in series with Z1.

Using product over sum the two impedances in parallel Z2 two and three can be resolved and we obtain 3.22 minus j 1.07 so i advise you do this calculation in a calculator using complex mode we then add that to the series reactants minus j10 which is z1 and the final answer is 3. 22 minus J11.070. So you all are free to find the equivalent capacitance of that capacitive reactance. You all know the formula is 1 on J omega C is equal to XC.

Using the same process and equations we covered, try to find the input impedance of this circuit at home. Note that omega this time is 10 radians per second. I have provided the final answer for comparison with yours.

Feel free to message on the class forum if you have any questions on this example. Example 2. asks for voltage across the 5 Henry inductor in the given circuit. The voltage source is given as a trig function, so we must first obtain its phasor representation.

The magnitude of the voltage phasor would be the RMS of the sinusoid, given by the peak voltage, 20, divided by root 2, to give 14.14. at an angle of minus 15 degrees given by the phase shift of the waveform. The function as is gives us the omega, which in this case is four radians per second. The reactances of the inductor and the capacitor must then be formed. using the relevant formulae as shown.

Now to find the voltage across the 5 Henry inductor, we must first find the parallel combination of the inductor and the capacitor. They are in parallel with each other and only this resultant impedance will give us the voltage across them. Let us call this impedance Z2 and Z1.

So the equivalent impedance is Z2 and Z1 is the 60 ohm resistor. So Z2 was calculated using the product of the two reactances over the sum. So the answer was. obtained to be J100. Now we have a series circuit.

We can use the voltage divider theorem to find the voltage across Z2. This voltage is given by the supply voltage Z2 sorry the supply voltage multiplied by z2 over the sum of z1 and z2. One thing we must remember is to include all j operators in these equations since we are dealing with phasors.

The phasor voltage across the inductor was calculated to be 12.12 at 15.96. degrees. Finally, if we are asked to revert to the time domain representation, the peak voltage would be 12.12 multiplied by root 2 to get 17.15 and the phase shift is positive 15.96 degrees.

And we express the function in the typical form. So we've gone through a lot today with some examples. I hope you all understood most of it. If you all have any issues, feel free to post your problems on the class forum or bring them up in the feedback session.

Transcript for:Understanding AC Circuit Components and Analysis

Transcript for:
Understanding AC Circuit Components and Analysis