Understanding AC Circuit Analysis Techniques

Good day everyone. Today we are going to look at some introductory material for AC circuit analysis. For AC circuits we no longer do simple algebraic calculation, but we need to employ complex numbers in the phasor domain. So we are going to look at average and RMS values, then we will specifically look at average and RMS of sinusoid waveforms. And finally, I will introduce the phasor though. From the last lecture, we saw that DC values are constants. They do not vary with time and are unidirectional. And DC voltages are produced by a battery or a DC generator. And there we have it expressed graphically. AC values do vary with time. and are represented by periodic waveforms. There are different patterns of alternating waveforms as shown. We have sinusoids which you should be familiar with, square waves, triangular waves, and complex waves which are non-uniform and chaotic every half cycle but are still periodic. Voltages of a sinusoidal profile are produced by an AC generator or an alternator. AC values have the same peak and average value. AC values on the other hand, the average can be found simply by calculating the average value of one complete cycle of the waveform. We have the simple graphical method where we can take samples or mid-ordinates, sum them, and then divide by the number of samples taken. Obviously here The more samples we take, the more accurate the calculated average would be. And then we have the analytical method where we can calculate the area under the curve by integrating its function and dividing by its period. This method obtains a more accurate average with less calculation. To understand average parameters in an electric circuit and their usefulness, Let's consider an AC circuit. Here we are going to create manually the alternating square wave using a DC source and a switch. The switch is put on for five seconds and taken off for another five seconds. The voltage applied to the circuit therefore will look like this. and the power dissipated by the resistor will look like this. 10 watts was calculated by v squared on R. Since this example is simple we can use the graphical method to easily find the average of the voltage waveform. The only two samples we can take is 10 volts when on and zero volts when off. 10 plus 0 divided by the number of samples 2 gives us an average of 5 volts. Similarly we can use Ohm's law to find the current flowing in the circuit. It would be 10 volts on 10 ohms when on and 0 volts on 10 ohms when off. This gives an average current of 0.5 amps. Again using the graphical method, the average power would be 10 on 2, 5 watts. Therefore, the total energy delivered over one period of 10 seconds is 50 joules. If we had applied 5 watts continuously for the whole 10 seconds of the period, we'd have delivered the same 50 joules. Applying 10 watts for half the time. 5 seconds dissipated 50 joules. Remember energy is power over a period of time. There are some salient points we need to highlight from this experiment. The effective or heating power of our circuit is 5 watts. which is the average power. However, the average voltage times the average current does not give us this average power. Multiplying those two we get 2.5 watts which is not the 5 watt. Hence, we need some other measure of voltage and current to calculate average power. without first finding the instantaneous power. So we need a voltage and a current we can use to calculate the average power because in much more complex waveforms and circuits we cannot find the instantaneous power as easily. So we would like to use a voltage and a current to calculate it. So let's first find the RMS. of the voltage. RMS is a mathematical function like average that reduces a complex function to a single value. It is the root of the mean of the square of the function. So finding the RMS of the voltage, first we find the square of the voltage which is shown in this plot on the right. It would be 100 volts. squared, 10 squared, when it's on, and 0 volts squared when it's off. Now we find the mean or average using the same graphical method, which would just be half, 50 volts squared. Finally, we find the square root, which gives us 7.071. Similarly, we can find the rms of the current. The average of 1 amp squared is 0.5 amps squared and the square root gives us 0.7071. Now if we multiply the RMS values we find that we will get the average power 5 watts. Therefore, we can deduce that the importance of RMS voltage and current are that they can be directly used to calculate the average power or equivalent power. Similarly, we can derive other formulae to calculate the average power. Average power is the average squared voltage over resistance. Since resistance is a constant value, its average is that same value. This gives us the average of V squared over R. Now, VRMS is the root mean square of V. Squaring both sides, we have VRMS squared is equal to the average of V squared. Substituting, we have VRMS squared on R. Similarly, we can derive... the formula I rms squared by r. These two formulae can be used to find the useful average power. Recall that the power was 10 watts peak. Squaring it we would get 100 watts squared. the average of the square of the pole would then be 50 watts squared and then the square root of that average is 7.07 we found out earlier that the equivalent heating power of our circuit the average power was 5 watt not 7.07 so therefore The RMS value of power is not the equivalent heating power. In fact, it doesn't represent anything useful. In conclusion, V average by I average is not the average power. We have to find and multiply the RMS voltage and current to find the average power. And this power, as well as the RMS power, doesn't represent anything useful. the rms and average values of nearly all ac waveforms are completely Commercial power generation is a sinusoidal wave. So since commercial power on an electricity grid is sinusoidal, we will consider the sinusoidal wave in detail. Using the analytical method, we can find the average of a sinusoidal wave. We integrate the function of the voltage waveform, Vp sine omega t, and divide by the period. Here I divided by pi, which is half the period, because I wanted to obtain a value which is 0.637 vp or 63.7% of the peak volt. The average of a full sinusoidal cycle, however, 2 pi would be 0 volts. Now how do we calculate the RMS of this waveform using the analytical method? The average is the integral of the function divided by the period. So the RMS is the root of the integral of the square of the function divided by that period. From this we obtain 0.707 VP. or 70.7% of the peak voltage or simply V peak on route 2. In summary, the equivalent heating power of our waveform is the average power. The RMS value of the power doesn't represent anything. The RMS values of voltage and current are useful because they can be used to calculate the average power. A few extra points to note for your information are 120 volts AC refers to rms voltage of a sinusoidal waveform delivered to your home and the AC meters volt meters and ammeters measure rms values of time varying quantities. Phasor analysis is a mathematical tool. A phasor is a rotating vector with a scaled line whose length represents an AC quantity that has both magnitude, RMS amplitude, and direction, also called phase, which is frozen at some point in time. This helps characterize the two parameters of the sinusoidal wave. the magnitude and the phasor. Phasor domain is a mathematical transformation from the time domain to the frequency domain. It drops the time component. It is used for steady state analysis of circuits and electrical systems. So using We have the equations here in the time domain of voltage and current. They are sinusoidal functions. And using Euler's formula, we can express the trigonometric time-dependent functions of voltage and current as exponential terms. Vm cos theta is just the real component of this exponential form, since the entire thing is cos omega t. plus sine omega t. When going to the phasor domain, we drop the time dependent component. We can then express in polar or rectangular form. So all we have there now is the magnitude and the phase angle. Here I have a diagram and animation of a function e of t expressed in the time and phasor domain. This better shows that a phasor is a rotating vector that represents a sinusoid oscillating in time. We can see how the angles coincide. As the time varying value oscillates forward, the phase rotates anticlockwise. We can also note that the phasor does not vary in magnitude. It has a fixed magnitude, which is the RMS of the sinusoidal function. v peak on root 2. This slide serves to remind you of the complex number plots both in rectangular form and polar form. If z theta is equal to x plus j y, this means that the real magnitude is x and imaginary is y. Recall that the imaginary axis is 90 degrees from the real, so the j operator is used to signify a rotation of 90 degrees anticlockwise. If we plot these coordinates x, y on the graph, we will obtain this point. And in polar form, we can represent this complex number as the magnitude of the line z. at an angle theta from the reference line. Please be advised that I have posted an online resource on the course page to refresh and familiarize you with complex numbers and other topics in mathematics relevant to the course. So you may ask at this point why? Why am I teaching you to and hence convert our trig function to our phasor. Well this is all in preparation for AC circuit analysis. If we are to do calculations on a circuit with time varying parameters they will be mathematically intensive as shown. You'd have to differentiate them. This AC circuit has an AC voltage source in series with a resistor inductor and capacitor which are some AC passive components. If we have to implement Kirchhoff's voltage law in this circuit, we will have to multiply trig functions as well as differentiate and integrate. So to simplify this calculation, we can move to the phasor domain, calculate it there in complex numbers, and then convert back to the time. So the question is now, how do we convert voltages and currents from the... time domain to the phasor domain. The magnitude of a phasor is the rms value of the sinusoidal function. Therefore, its peak is divided by root 2. The phase displacement of the sinusoid gives the position on the phasor diagram. It is the same angle from the reference line on the phasor diagram. For example here, phi is where the waveform starts and so it is the angle of the phasor. Vm is the peak. Vm on root 2 which we derive to be the rms is the magnitude of the phasor. So we have the voltage waveform which starts at zero is on the zero reference line in the phasor domain. And the current waveform which starts five degrees later is minus five on the phasor deck. And the magnitudes of these phasors would be the peaks Vm and Im on root two. vr i rms usually v is at reference and unless i specify otherwise always assume it to be at reference this diagram once again shows our voltage and current value in the time and phaser domain if you have any questions on the phaser domain feel free to post them in the class forum on my learning or ask me in the feedback session

So we are going to look at average and RMS values, then we will specifically look at average and RMS of sinusoid waveforms. And finally, I will introduce the phasor though. From the last lecture, we saw that DC values are constants.

They do not vary with time and are unidirectional. And DC voltages are produced by a battery or a DC generator. And there we have it expressed graphically. AC values do vary with time. and are represented by periodic waveforms.

There are different patterns of alternating waveforms as shown. We have sinusoids which you should be familiar with, square waves, triangular waves, and complex waves which are non-uniform and chaotic every half cycle but are still periodic. Voltages of a sinusoidal profile are produced by an AC generator or an alternator. AC values have the same peak and average value.

AC values on the other hand, the average can be found simply by calculating the average value of one complete cycle of the waveform. We have the simple graphical method where we can take samples or mid-ordinates, sum them, and then divide by the number of samples taken. Obviously here The more samples we take, the more accurate the calculated average would be.

And then we have the analytical method where we can calculate the area under the curve by integrating its function and dividing by its period. This method obtains a more accurate average with less calculation. To understand average parameters in an electric circuit and their usefulness, Let's consider an AC circuit. Here we are going to create manually the alternating square wave using a DC source and a switch. The switch is put on for five seconds and taken off for another five seconds.

The voltage applied to the circuit therefore will look like this. and the power dissipated by the resistor will look like this. 10 watts was calculated by v squared on R. Since this example is simple we can use the graphical method to easily find the average of the voltage waveform. The only two samples we can take is 10 volts when on and zero volts when off.

10 plus 0 divided by the number of samples 2 gives us an average of 5 volts. Similarly we can use Ohm's law to find the current flowing in the circuit. It would be 10 volts on 10 ohms when on and 0 volts on 10 ohms when off. This gives an average current of 0.5 amps. Again using the graphical method, the average power would be 10 on 2, 5 watts.

Therefore, the total energy delivered over one period of 10 seconds is 50 joules. If we had applied 5 watts continuously for the whole 10 seconds of the period, we'd have delivered the same 50 joules. Applying 10 watts for half the time. 5 seconds dissipated 50 joules.

Remember energy is power over a period of time. There are some salient points we need to highlight from this experiment. The effective or heating power of our circuit is 5 watts. which is the average power.

However, the average voltage times the average current does not give us this average power. Multiplying those two we get 2.5 watts which is not the 5 watt. Hence, we need some other measure of voltage and current to calculate average power. without first finding the instantaneous power. So we need a voltage and a current we can use to calculate the average power because in much more complex waveforms and circuits we cannot find the instantaneous power as easily.

So we would like to use a voltage and a current to calculate it. So let's first find the RMS. of the voltage. RMS is a mathematical function like average that reduces a complex function to a single value.

It is the root of the mean of the square of the function. So finding the RMS of the voltage, first we find the square of the voltage which is shown in this plot on the right. It would be 100 volts. squared, 10 squared, when it's on, and 0 volts squared when it's off.

Now we find the mean or average using the same graphical method, which would just be half, 50 volts squared. Finally, we find the square root, which gives us 7.071. Similarly, we can find the rms of the current.

The average of 1 amp squared is 0.5 amps squared and the square root gives us 0.7071. Now if we multiply the RMS values we find that we will get the average power 5 watts. Therefore, we can deduce that the importance of RMS voltage and current are that they can be directly used to calculate the average power or equivalent power.

Similarly, we can derive other formulae to calculate the average power. Average power is the average squared voltage over resistance. Since resistance is a constant value, its average is that same value. This gives us the average of V squared over R.

Now, VRMS is the root mean square of V. Squaring both sides, we have VRMS squared is equal to the average of V squared. Substituting, we have VRMS squared on R. Similarly, we can derive... the formula I rms squared by r.

These two formulae can be used to find the useful average power. Recall that the power was 10 watts peak. Squaring it we would get 100 watts squared.

the average of the square of the pole would then be 50 watts squared and then the square root of that average is 7.07 we found out earlier that the equivalent heating power of our circuit the average power was 5 watt not 7.07 so therefore The RMS value of power is not the equivalent heating power. In fact, it doesn't represent anything useful. In conclusion, V average by I average is not the average power.

We have to find and multiply the RMS voltage and current to find the average power. And this power, as well as the RMS power, doesn't represent anything useful. the rms and average values of nearly all ac waveforms are completely Commercial power generation is a sinusoidal wave.

So since commercial power on an electricity grid is sinusoidal, we will consider the sinusoidal wave in detail. Using the analytical method, we can find the average of a sinusoidal wave. We integrate the function of the voltage waveform, Vp sine omega t, and divide by the period. Here I divided by pi, which is half the period, because I wanted to obtain a value which is 0.637 vp or 63.7% of the peak volt.

The average of a full sinusoidal cycle, however, 2 pi would be 0 volts. Now how do we calculate the RMS of this waveform using the analytical method? The average is the integral of the function divided by the period.

So the RMS is the root of the integral of the square of the function divided by that period. From this we obtain 0.707 VP. or 70.7% of the peak voltage or simply V peak on route 2. In summary, the equivalent heating power of our waveform is the average power.

The RMS value of the power doesn't represent anything. The RMS values of voltage and current are useful because they can be used to calculate the average power. A few extra points to note for your information are 120 volts AC refers to rms voltage of a sinusoidal waveform delivered to your home and the AC meters volt meters and ammeters measure rms values of time varying quantities. Phasor analysis is a mathematical tool.

A phasor is a rotating vector with a scaled line whose length represents an AC quantity that has both magnitude, RMS amplitude, and direction, also called phase, which is frozen at some point in time. This helps characterize the two parameters of the sinusoidal wave. the magnitude and the phasor. Phasor domain is a mathematical transformation from the time domain to the frequency domain. It drops the time component.

It is used for steady state analysis of circuits and electrical systems. So using We have the equations here in the time domain of voltage and current. They are sinusoidal functions.

And using Euler's formula, we can express the trigonometric time-dependent functions of voltage and current as exponential terms. Vm cos theta is just the real component of this exponential form, since the entire thing is cos omega t. plus sine omega t.

When going to the phasor domain, we drop the time dependent component. We can then express in polar or rectangular form. So all we have there now is the magnitude and the phase angle.

Here I have a diagram and animation of a function e of t expressed in the time and phasor domain. This better shows that a phasor is a rotating vector that represents a sinusoid oscillating in time. We can see how the angles coincide.

As the time varying value oscillates forward, the phase rotates anticlockwise. We can also note that the phasor does not vary in magnitude. It has a fixed magnitude, which is the RMS of the sinusoidal function.

v peak on root 2. This slide serves to remind you of the complex number plots both in rectangular form and polar form. If z theta is equal to x plus j y, this means that the real magnitude is x and imaginary is y. Recall that the imaginary axis is 90 degrees from the real, so the j operator is used to signify a rotation of 90 degrees anticlockwise. If we plot these coordinates x, y on the graph, we will obtain this point. And in polar form, we can represent this complex number as the magnitude of the line z.

at an angle theta from the reference line. Please be advised that I have posted an online resource on the course page to refresh and familiarize you with complex numbers and other topics in mathematics relevant to the course. So you may ask at this point why?

Why am I teaching you to and hence convert our trig function to our phasor. Well this is all in preparation for AC circuit analysis. If we are to do calculations on a circuit with time varying parameters they will be mathematically intensive as shown. You'd have to differentiate them. This AC circuit has an AC voltage source in series with a resistor inductor and capacitor which are some AC passive components.

If we have to implement Kirchhoff's voltage law in this circuit, we will have to multiply trig functions as well as differentiate and integrate. So to simplify this calculation, we can move to the phasor domain, calculate it there in complex numbers, and then convert back to the time. So the question is now, how do we convert voltages and currents from the...

time domain to the phasor domain. The magnitude of a phasor is the rms value of the sinusoidal function. Therefore, its peak is divided by root 2. The phase displacement of the sinusoid gives the position on the phasor diagram.

It is the same angle from the reference line on the phasor diagram. For example here, phi is where the waveform starts and so it is the angle of the phasor. Vm is the peak.

Vm on root 2 which we derive to be the rms is the magnitude of the phasor. So we have the voltage waveform which starts at zero is on the zero reference line in the phasor domain. And the current waveform which starts five degrees later is minus five on the phasor deck.

And the magnitudes of these phasors would be the peaks Vm and Im on root two. vr i rms usually v is at reference and unless i specify otherwise always assume it to be at reference this diagram once again shows our voltage and current value in the time and phaser domain if you have any questions on the phaser domain feel free to post them in the class forum on my learning or ask me in the feedback session

Transcript for:Understanding AC Circuit Analysis Techniques

Transcript for:
Understanding AC Circuit Analysis Techniques