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Plot the wave this one time showing its amplitude vs time. What is its shape? What is its max value. Let's see what you come up with.
Ratch
Some very interesting (but somewhat extraneous to the topic) discussion about ohms law has been moved to here.
The first part of the problem is to get V stated in terms of t.
V is in volts, t is in seconds.
The general form of the equation is:
V = a cos(b.t + c)
a, b, and c may be shown as different letters (Greek or otherwise), and either sin or cos may be used as the function.
You should know that both sin and cos are periodic functions, with a period of 360 degrees, or 2*pi radians. Either degrees or radians can be used (for simple problems), but I was taught to use radians.
Interestingly enough, AC (when not otherwise described) is generally assumed to be a sine wave. And this is what you get when you plot y = cos(x).
The trick is to make your function match the description of the AC signal you have been given.
Let's start with c. If you plot y = cos(x + c) for various values of c, you will notice that it seems to move the graph left and right. This is one interpretation (and it may be useful in some circumstances). Another interpretation (which could be argued as more correct) is that it changes the phase of the signal. You should know that the difference between sin and cos is a phase relationship. This, c is important where the phase of the signal is important. It also explains why either sin or cos can be used (and where the exact phase is not an issue, interchangeably)
Ok, so the original question doesn't mention anything like "the voltage at time 0 is x", so the phase of unimportant. So c can be anything (zero is best) and either sin or cos can be used.
So let's use cos, and if c=0, we can just eliminate it from the equation.
The next parameter is b. If you graph y = cos(b.x) for various values of b (b not equal to zero) you will notice that the period is shortened for b > 1 and lengthened for b < 1.
Now let's move on to a. If you graph y = a cos(x) for various values of a, you will immediately notice that a has a relationship to amplitude. Specifically, it determines the peak absolute deviation from 0. More simply, the graph varies between y = -a and +a.
Now we know that a affects amplitude, b affects frequency, and c affects phase.
I've already told you c can be ignored in your problem (and in many simple problems).
So, if our frequency is f (in hz), it is f cycles per second. Cos(x) has a period of 2.pi radians. If we want f hz, then we need f.2.pi radians per second, so, in terms of t, x = f.2.pi.t (or more commonly 2.pi.f.t). In our equation (which is in terms of t) b.t must be 2.pi.f.t.
We measure AC voltage in several ways, peak, peak to peak, and RMS (root mean square).
Peak is simply the furthest the voltage deviates from zero. If you look at the definition for a, this is it. In many texts, and in many simulation packages, this is how AC is defined unless otherwise specified.
Peak to peak is the sum of the maximum positive and negative peaks. For a waveform which is symmetric about zero, this is twice the peak voltage. In this case a would be HALF the peak to peak voltage.
The most common AC measurements (in terms of those you come across in real life) are RMS values. A simple description of RMS is that it expresses an arbitrary waveform (not necessarily even AC) as the equivalent DC value which would cause the same heating effect in a resistor.
You have been given an RMS value for the voltage. Because we have assumed the waveform is a sine wave, we can also assume that the peak voltage is the square root of 2 (1.414) times the RMS value.
Armed with all of this, you should be able to come up with the correct equation.
This is assumed knowledge in the question you have been asked. If you don't recall it, you need to go back over your notes and study it until you have internalized it.
The next part of the question (which yields a large part of the answer) involves calculating the first derivative of this equation. In general, this can require intermediate skills in calculus, but this specific case is a simple identity that may be referenced in your text without mention of calculus.
The first part of the problem is to get V stated in terms of t.
V is in volts, t is in seconds.
The general form of the equation is:
V = a cos(b.t + c)
a, b, and c may be shown as different letters (Greek or otherwise), and either sin or cos may be used as the function.
You should know that both sin and cos are periodic functions, with a period of 360 degrees, or 2*pi radians. Either degrees or radians can be used (for simple problems), but I was taught to use radians.
So what is a radian? It is simply the ratio of the arc distance to and the radius for the angle. More simply, if one assumes the radius is 1, it is the distance you would have to walk around the circumference of the circle to sweep out the angle. Because the circumference is 2.pi.r, and because a full revolution is 360 degrees, 360° = 2.pi radians.
Interestingly enough, AC (when not otherwise described) is generally assumed to be a sine wave. And this is what you get when you plot y = cos(x).
The trick is to make your function match the description of the AC signal you have been given.
Let's start with c. If you plot y = cos(x + c) for various values of c, you will notice that it seems to move the graph left and right. This is one interpretation (and it may be useful in some circumstances). Another interpretation (which could be argued as more correct) is that it changes the phase of the signal. You should know that the difference between sin and cos is a phase relationship. This, c is important where the phase of the signal is important. It also explains why either sin or cos can be used (and where the exact phase is not an issue, interchangeably)
Ok, so the original question doesn't mention anything like "the voltage at time 0 is x", so the phase of unimportant. So c can be anything (zero is best) and either sin or cos can be used.
So let's use cos, and if c=0, we can just eliminate it from the equation.
The next parameter is b. If you graph y = cos(b.x) for various values of b (b not equal to zero) you will notice that the period is shortened for b > 1 and lengthened for b < 1.
Ok remember I said sin and cos were interchangeable, and b should not be zero? Plotting y=sin(b.x) and y=cos(b.x) for b=0 will show a very important difference. Indeed, it is the reason that cos is normally used, but in this question it is not relevant.
So, in the graph of y=cos(b.x) for various values of b, you might see a clear indication that b has some relation to frequency in the equation V = a cos(b.x + c).
Now let's move on to a. If you graph y = a cos(x) for various values of a, you will immediately notice that a has a relationship to amplitude. Specifically, it determines the peak absolute deviation from 0. More simply, the graph varies between y = -a and +a.
Now we know that a affects amplitude, b affects frequency, and c affects phase.
I've already told you c can be ignored in your problem (and in many simple problems).
So, if our frequency is f (in hz), it is f cycles per second. Cos(x) has a period of 2.pi radians. If we want f hz, then we need f.2.pi radians per second, so, in terms of t, x = f.2.pi.t (or more commonly 2.pi.f.t). In our equation (which is in terms of t) b.t must be 2.pi.f.t.
You may be wondering why I decided to use radians rather than degrees. Ok, I said I had been taught that way, but really, I was first introduced to degrees. However, in more advanced math radians are used because there are relationships between transindental functions (sin, cos, tan,...) and exponential functions (ln and e^x) which become apparent when complex numbers are used. (e.g. -e^(-i.pi)). These relationships make the use of radians and natural logarithms far more simple and direct than the (perhaps) more intuitive degrees and base 10 logs/exp functions. The fact that you see 2.pi.f in many electronic equations that don't even use sin or cos is another indication.
We are now left with a. This seems so trivial that maybe you think I don't need to explain it at all. However, here be dragons!
We measure AC voltage in several ways, peak, peak to peak, and RMS (root mean square).
Peak is simply the furthest the voltage deviates from zero. If you look at the definition for a, this is it. In many texts, and in many simulation packages, this is how AC is defined unless otherwise specified.
Peak to peak is the sum of the maximum positive and negative peaks. For a waveform which is symmetric about zero, this is twice the peak voltage. In this case a would be HALF the peak to peak voltage.
The most common AC measurements (in terms of those you come across in real life) are RMS values. A simple description of RMS is that it expresses an arbitrary waveform (not necessarily even AC) as the equivalent DC value which would cause the same heating effect in a resistor.
You have been given an RMS value for the voltage. Because we have assumed the waveform is a sine wave, we can also assume that the peak voltage is the square root of 2 (1.414) times the RMS value.
Armed with all of this, you should be able to come up with the correct equation.
This is assumed knowledge in the question you have been asked. If you don't recall it, you need to go back over your notes and study it until you have internalized it.
The next part of the question (which yields a large part of the answer) involves calculating the first derivative of this equation. In general, this can require intermediate skills in calculus, but this specific case is a simple identity that may be referenced in your text without mention of calculus.
You guys confused me and I already understood it. He said he was a student at the stage of learning this. Recognize your audience. No disrespect intended.
Simply stated the name of the game is to find "things" which have interesting relationships between voltage and current and exploit them.
Resistance by definition is the ratio of voltage to current (period!). It's units are therefore Volts/Amp. Resistors are devices whose voltage is proportional to the current over a wide range of currents.
Capacitors are only a tiny bit more complicated. Capacitance is the ratio of current to the "change in voltage over time". A simpler definition of "change in voltage over time" is the SLOPE OF THE VOLTAGE. Devices with a constant ratio between the current through them to the SLOPE of the voltage across them are called capacitors. Slope has units of Volts/second. For instance if the voltage across a capacitor changes linearly from 0 to 5 volts in 2 seconds (i.e. a 5 Volt ramp with 2 second period) then the slope of the voltage across the capacitor is 5 Volts / 2 seconds = 2.5 V/s.
Now examine the equation they are teaching you. It will help to rearrange it so the capacitance is on the left.
C = Ic / dV/dt
dV/dt is nothing more than a complicated way of saying the slope of the voltage. It basically says the current through a capacitor at any instant in time is "proportional" to the SLOPE of the voltage across it. That's all there is to capacitors. If you place this 5 Volt ramp with 2 second period across a capacitor and you get 2.5 Amps of current flowing into the capacitor you know the capacitance is 1 Farad (the units given to capacitance). If you get 1.175 uAmps flowing through the capacitor you know it is a 0.47 uF capacitor (because 1.175 uA / 2.5 V/s) = 0.47 uF
C = 2.5 Amps / 2.5 V/s = 1 Farad
or
C = 1.175 uA / 2.5 V/s = 0.47 uFarad
Now of course in your case they gave you the capacitance and asked you to figure out the current but that is just basic algebra. Another example may help.
I have a ramp voltage which rises from 0 Volts to 5 Volts in 100 microseconds. Therefore the slope of this ramp is 5 Volts / 100 u-seconds = 50,000 V/s
I place this voltage across a 0.47 uF capacitor. What is the current through the capacitor while the voltage is ramping up?
Answer: Ic = 0.47 uF X 50/000 V/s = 0.0235 Amps
During the entire 100 microseconds in which the voltage is ramping up, how much charge is forced onto the capacitor plates?
0.0235 Amps X 100 useconds = 0.0235 Coulombs/second X 0.0001 seconds = 2.35 microCoulombs
of course you should have known this intuitively since the final voltage on the capacitor is 5 Volts. If this wasn't intuitive then understand the following.
What are the base units of a Farad? Well current is really Coulombs/second and we already know dV/dt is the slope which has units of Volt/second. So Farads must have units of
Coulombs/second
______________ = Coulombs/Volt (because the seconds cancel out).
Volts/second
To understand capacitors it may help to stop and really think about its units. What it tells you is the amount of charge on the plates of the capacitor determines how much voltage will be measured across its terminal. If I have 1 micro-Coulomb of charge (a reasonable amount) on a 1 uF capacitor I will measure 1 volt on its terminals. If I double the amount of charge to 2uC then I will measure 2 Volts on the 1uF capacitor's terminals. This makes sense since the more electrons you crowd onto the plates, the more forcefully the electrons are wanting to get out. The ratio of Charge on the capacitor to the voltage across its terminals is CAPACITANCE. So with a BFC (big @#$%^ capacitor) you can pump alot of charge onto its plates and the voltage doesn't rise much. A small capacitor it only takes a small amount of charge to increase the terminal voltage.
If you attach a 1 Amp current source to a 1 Farad capacitor the voltage across it's terminals will increase by a constant 1 Volt every second.
If you understand the above you will have little problem with capacitors.
Now to your problem. The problem they gave you seems more complicated for 2 reasons. First, they gave you an rms voltage. Second, they gave you a sine wave instead of a voltage with a constant slope. In reality there is a trick which actually makes the problem much simpler.
I won't confuse you with a long explanation of rms except to say it stands for root mean square and we engineers use it because two signals with the same rms voltage will supply the same amount of average power to a resistive load. So 5 VDC across a 1 ohm load will on average deliver 25 Watts of power. Likewise a 5 Vrms sine wave applied across a 1 ohm load will on average deliver 25 Watts of power. Since at times the sine wave is near zero you should expect the top of the 5 Vrms sine wave must be above 5 Volts. You would be correct. In order for a sine wave to deliver this same power its peak amplitude will be 5 X the square root of 2 Volts. The square root of 2 is 1.414 so a 5 Vrms sine wave will have a peak amplitude of 1.414 X 5 = 7.07 Volts. To know this it would help to learn calculus.
Sine waves have interesting mathematical properties. One of the interesting properties of sine waves is that the slope of a sine wave at every point in time is equal to the cosine. To visualize this examine the following display of a sine wave and a cosine wave.
Check a few points. The sine wave starts out with a slope of 1 V/s. Sure enough the cosine has a value of 1. At the peak the sine wave obviously has a slope of 0. Sure enough the cosine has a value of 0 at this same point in time. When the sine wave returns to the x-axis it has a slope of -1. Sure enough the cosine has a value of -1 here too. At the bottom of the sine wave again it has a slope of 0. Sure enough the cosine once again has a value of 0. Amazing huh!!!
It is this property of sine and cosine waves which gives rise to another important phenomenon which you are about to learn. Namely that all linear devices (and even combinations of linear devices) which are driven by sine wave sources will have voltages and currents across and through those components which are also sine waves at the same frequency as the driving voltage. They will have differing sine wave amplitudes and differing sine wave phases (meaning the sine wave across a particular component may have its peaks at different points in time) but all the currents and voltages will be SINE waves and those sine waves will be at the same frequency (only the amplitudes and the time at which the signals peak will differ). This is in fact why engineers use sine waves to study and completely define the behavior of AC circuits.
So to answer your problem, the amplitude of your 5 Vrms sine wave will be 7.07 Volts. The frequency of your sine wave means it will complete 2 pi radians in 10 ms = 628 rad/s = w. An equation for your sine wave is therefore 7.07 sin (628 t). Which has a slope of 7.07 X 628 cosine (628 t) = 4442 cos(628 t). Again study calculus to know this. Note this is in radians so don't forget to switch your calculator to radians. Engineers use radians.
Ic = C X dV/dt = 0.47 uF X 4442 cos (628 t)
Ic = 2.088 cos (628 t) mAmps
See LTCAD below:
I would like to add 1 last thing I think you should absorb if you really want to understand capacitors fully. Since the slope of the sine wave is at its greatest when the value of the sine wave is at 0 this means the peak current through the capacitor will occur when the voltage source is at zero. When the sine-wave voltage source is at it's peak it has a 0 slope. This means when the sine wave of the voltage source peaks the sine wave current through the capacitor is 0. Likewise when the voltage source falls through 0 Volts the capacitor is discharging at its maximum rate.
This is not AT ALL the way resistor behave!!!! With resistors the voltage and current peaks always occur at the same point in time. This is precisely why capacitors (and as you'll later learn inductors) have what is called phase shift. The current through a resistor is always "in-phase" with the voltage across it. In contrast, the current through a capacitor is always 90 degrees ahead of the voltage. For an inductor you will find the current is always 90 degrees behind the voltage. An easy way engineers remember this relationship is to use the phrase ELI the ICE man. ELI meaning voltage (E) leads current (I) in an inductor (L) and similarly current (I) leads voltage (E) in a capacitor (C).
What is truly amazing about sine waves is you can have sine waves at differing amplitudes and differing phases across each and every component and yet they can still always equal the value of the sine wave of the voltage source at every point in time. Another neat feature of sine waves is that any other periodic signal can be constructed from sine waves. For instance using a square wave is really just a bunch of sine waves at various amplitudes and frequencies added together. So understanding how a circuit responds to any and all sine waves completely defines how it will react to any signal.
Last point I would like to emphasize is that if you think about it, the higher the frequency of the sine wave, the more current flows into and out of a capacitor. This can be seen by noting that if you increase the frequency of the sine wave, it goes up and down (the slope) much quicker. Since higher slope means higher Ic this means more current flows as the frequency of the voltage source increases. This is again not at all like resistors. Resistors resist current the same no matter what the frequency. The resistance (impedance) of capacitors decreases as the frequency increases. Essentially, capacitors can be thought of as frequency dependent resistors. In fact it is this useful relationship between voltage and current in capacitors which we exploit when we use them.
It makes perfect sense to say the impedance of a resistor is say 100 Ohms. This is because the impedance doesn't ever change (or at least only be "tolerable" amounts). In contrast, it makes no sense at all to say the impedance of a capacitor is 100 Ohms. You have to specify what frequency you are referring when you claim a capacitor has a certain impedance. Something like the impedance of this capacitor is 100 Ohms at 100 KHz is perhaps more acceptable. Rather than do that, we engineers instead just state the capacitance of the capacitor which doesn't change and leave it up to the consumer of this information to determine what the impedance is at whatever frequency they are interested.
Such consumers are assumed to also understand the phase relationship of capacitors (and later inductors). If I have a 300 Ohm resistor and I have a capacitor I have determined to have 400 Ohms of impedance at my frequency of interest, the two impedances DO NOT add up to 700 Ohms. This is because the impedance of the capacitor is 90 degrees out-of-phase. Instead I should draw a right triangle in my head with one leg of the triangle 300 Ohms and the other 400 Ohms. The impedance of the resistor and capacitor combined will be equal to the length of the hypotenuse which in this case is 500 Ohm (not 700). Once I know this I can easily find the magnitude by dividing the amplitude of the sine wave voltage source by 500.
Simply stated the name of the game is to find "things" which have interesting relationships between voltage and current and exploit them.
Resistance by definition is the ratio of voltage to current (period!). It's units are therefore Volts/Amp. Resistors are devices whose voltage is proportional to the current over a wide range of currents.
Capacitors are only a tiny bit more complicated. Capacitance is the ratio of current to the "change in voltage over time". A simpler definition of "change in voltage over time" is the SLOPE OF THE VOLTAGE. Devices with a constant ratio between the current through them to the SLOPE of the voltage across them are called capacitors. Slope has units of Volts/second. For instance if the voltage across a capacitor changes linearly from 0 to 5 volts in 2 seconds (i.e. a 5 Volt ramp with 2 second period) then the slope of the voltage across the capacitor is 5 Volts / 2 seconds = 2.5 V/s.
Now examine the equation they are teaching you. It will help to rearrange it so the capacitance is on the left.
C = Ic / dV/dt
dV/dt is nothing more than a complicated way of saying the slope of the voltage. It basically says the current through a capacitor at any instant in time is "proportional" to the SLOPE of the voltage across it. That's all there is to capacitors. If you place this 5 Volt ramp with 2 second period across a capacitor and you get 2.5 Amps of current flowing into the capacitor you know the capacitance is 1 Farad (the units given to capacitance). If you get 1.175 uAmps flowing through the capacitor you know it is a 0.47 uF capacitor (because 1.175 uA / 2.5 V/s) = 0.47 uF
C = 2.5 Amps / 2.5 V/s = 1 Farad
or
C = 1.175 uA / 2.5 V/s = 0.47 uFarad
Now of course in your case they gave you the capacitance and asked you to figure out the current but that is just basic algebra. Another example may help.
I have a ramp voltage which rises from 0 Volts to 5 Volts in 100 microseconds. Therefore the slope of this ramp is 5 Volts / 100 u-seconds = 50,000 V/s
I place this voltage across a 0.47 uF capacitor. What is the current through the capacitor while the voltage is ramping up?
Answer: Ic = 0.47 uF X 50/000 V/s = 0.0235 Amps
During the entire 100 microseconds in which the voltage is ramping up, how much charge is forced onto the capacitor plates?
0.0235 Amps X 100 useconds = 0.0235 Coulombs/second X 0.0001 seconds = 2.35 microCoulombs
of course you should have known this intuitively since the final voltage on the capacitor is 5 Volts. If this wasn't intuitive then understand the following.
What are the base units of a Farad? Well current is really Coulombs/second and we already know dV/dt is the slope which has units of Volt/second. So Farads must have units of
Coulombs/second
______________ = Coulombs/Volt (because the seconds cancel out).
Volts/second
To understand capacitors it may help to stop and really think about its units. What it tells you is the amount of charge on the plates of the capacitor determines how much voltage will be measured across its terminal. If I have 1 micro-Coulomb of charge (a reasonable amount) on a 1 uF capacitor I will measure 1 volt on its terminals. If I double the amount of charge to 2uC then I will measure 2 Volts on the 1uF capacitor's terminals. This makes sense since the more electrons you crowd onto the plates, the more forcefully the electrons are wanting to get out. The ratio of Charge on the capacitor to the voltage across its terminals is CAPACITANCE. So with a BFC (big @#$%^ capacitor) you can pump alot of charge onto its plates and the voltage doesn't rise much. A small capacitor it only takes a small amount of charge to increase the terminal voltage.
If you attach a 1 Amp current source to a 1 Farad capacitor the voltage across it's terminals will increase by a constant 1 Volt every second.
If you understand the above you will have little problem with capacitors.
Now to your problem. The problem they gave you seems more complicated for 2 reasons. First, they gave you an rms voltage. Second, they gave you a sine wave instead of a voltage with a constant slope. In reality there is a trick which actually makes the problem much simpler.
I won't confuse you with a long explanation of rms except to say it stands for root mean square and we engineers use it because two signals with the same rms voltage will supply the same amount of average power to a resistive load. So 5 VDC across a 1 ohm load will on average deliver 25 Watts of power. Likewise a 5 Vrms sine wave applied across a 1 ohm load will on average deliver 25 Watts of power. Since at times the sine wave is near zero you should expect the top of the 5 Vrms sine wave must be above 5 Volts. You would be correct. In order for a sine wave to deliver this same power its peak amplitude will be 5 X the square root of 2 Volts. The square root of 2 is 1.414 so a 5 Vrms sine wave will have a peak amplitude of 1.414 X 5 = 7.07 Volts. To know this it would help to learn calculus.
Sine waves have interesting mathematical properties. One of the interesting properties of sine waves is that the slope of a sine wave at every point in time is equal to the cosine. To visualize this examine the following display of a sine wave and a cosine wave.
Check a few points. The sine wave starts out with a slope of 1 V/s. Sure enough the cosine has a value of 1. At the peak the sine wave obviously has a slope of 0. Sure enough the cosine has a value of 0 at this same point in time. When the sine wave returns to the x-axis it has a slope of -1. Sure enough the cosine has a value of -1 here too. At the bottom of the sine wave again it has a slope of 0. Sure enough the cosine once again has a value of 0. Amazing huh!!!
It is this property of sine and cosine waves which gives rise to another important phenomenon which you are about to learn. Namely that all linear devices (and even combinations of linear devices) which are driven by sine wave sources will have voltages and currents across and through those components which are also sine waves at the same frequency as the driving voltage. They will have differing sine wave amplitudes and differing sine wave phases (meaning the sine wave across a particular component may have its peaks at different points in time) but all the currents and voltages will be SINE waves and those sine waves will be at the same frequency (only the amplitudes and the time at which the signals peak will differ). This is in fact why engineers use sine waves to study and completely define the behavior of AC circuits.
So to answer your problem, the amplitude of your 5 Vrms sine wave will be 7.07 Volts. The frequency of your sine wave means it will complete 2 pi radians in 10 ms = 628 rad/s = w. An equation for your sine wave is therefore 7.07 sin (628 t). Which has a slope of 7.07 X 628 cosine (628 t) = 4442 cos(628 t). Again study calculus to know this. Note this is in radians so don't forget to switch your calculator to radians. Engineers use radians.
Ic = C X dV/dt = 0.47 uF X 4442 cos (628 t)
Ic = 2.088 cos (628 t) mAmps
See LTCAD below:
I would like to add 1 last thing I think you should absorb if you really want to understand capacitors fully. Since the slope of the sine wave is at its greatest when the value of the sine wave is at 0 this means the peak current through the capacitor will occur when the voltage source is at zero. When the sine-wave voltage source is at it's peak it has a 0 slope. This means when the sine wave of the voltage source peaks the sine wave current through the capacitor is 0. Likewise when the voltage source falls through 0 Volts the capacitor is discharging at its maximum rate.
This is not AT ALL the way resistor behave!!!! With resistors the voltage and current peaks always occur at the same point in time. This is precisely why capacitors (and as you'll later learn inductors) have what is called phase shift. The current through a resistor is always "in-phase" with the voltage across it. In contrast, the current through a capacitor is always 90 degrees ahead of the voltage. For an inductor you will find the current is always 90 degrees behind the voltage. An easy way engineers remember this relationship is to use the phrase ELI the ICE man. ELI meaning voltage (E) leads current (I) in an inductor (L) and similarly current (I) leads voltage (E) in a capacitor (C).
What is truly amazing about sine waves is you can have sine waves at differing amplitudes and differing phases across each and every component and yet they can still always equal the value of the sine wave of the voltage source at every point in time. Another neat feature of sine waves is that any other periodic signal can be constructed from sine waves. For instance using a square wave is really just a bunch of sine waves at various amplitudes and frequencies added together. So understanding how a circuit responds to any and all sine waves completely defines how it will react to any signal.
Last point I would like to emphasize is that if you think about it, the higher the frequency of the sine wave, the more current flows into and out of a capacitor. This can be seen by noting that if you increase the frequency of the sine wave, it goes up and down (the slope) much quicker. Since higher slope means higher Ic this means more current flows as the frequency of the voltage source increases. This is again not at all like resistors. Resistors resist current the same no matter what the frequency. The resistance (impedance) of capacitors decreases as the frequency increases. Essentially, capacitors can be thought of as frequency dependent resistors. In fact it is this useful relationship between voltage and current in capacitors which we exploit when we use them.
It makes perfect sense to say the impedance of a resistor is say 100 Ohms. This is because the impedance doesn't ever change (or at least only be "tolerable" amounts). In contrast, it makes no sense at all to say the impedance of a capacitor is 100 Ohms. You have to specify what frequency you are referring when you claim a capacitor has a certain impedance. Something like the impedance of this capacitor is 100 Ohms at 100 KHz is perhaps more acceptable. Rather than do that, we engineers instead just state the capacitance of the capacitor which doesn't change and leave it up to the consumer of this information to determine what the impedance is at whatever frequency they are interested.
Such consumers are assumed to also understand the phase relationship of capacitors (and later inductors). If I have a 300 Ohm resistor and I have a capacitor I have determined to have 400 Ohms of impedance at my frequency of interest, the two impedances DO NOT add up to 700 Ohms. This is because the impedance of the capacitor is 90 degrees out-of-phase. Instead I should draw a right triangle in my head with one leg of the triangle 300 Ohms and the other 400 Ohms. The impedance of the resistor and capacitor combined will be equal to the length of the hypotenuse which in this case is 500 Ohm (not 700). Once I know this I can easily find the magnitude by dividing the amplitude of the sine wave voltage source by 500.
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Incidentally, to determine the impedance of a capacitor at some frequency of interest the equation is
Zc = 1/wC
where w is the frequency of interest in radians/second.
You might ask, why not just take the voltage across the capacitor and divide it by the current determined using Ic = C dV/dt like we did earlier. Well if you wanted instantaneous resistance you could do exactly that and you would be correct (though I'm not sure it would mean much depending on your goals). This is because it isn't the voltage which is determining the amount of current so why would you care (actually there are very rare cases where it is of use (see finite element analysis)). Rather it is the change in voltage. The voltage source can be at a value of 1, 10, 50 or even 100 KV and still have identical capacitor currents as long as the slope of the voltages at those points are equal. And what of the problem of having the sine wave voltage at it's peak extremes (where the slope and current are zero). Since real resistance is Voltage/Amps this would give you an infinite result. Instead we normally care about the reactance of a capacitor which is similar to resistance but 90 degrees out-of-phase. Impedance is the combined effect of BOTH the resistance and the reactance and why you will learn soon to calculate the impedance of a capacitor as Zc = 1/wC at some value of frequency and then essentially draw a right triangle to determine the impedance (phasors) which has a magnitude and a phase shift.
Zc = 1/wC
where w is the frequency of interest in radians/second.
You might ask, why not just take the voltage across the capacitor and divide it by the current determined using Ic = C dV/dt like we did earlier. Well if you wanted instantaneous resistance you could do exactly that and you would be correct (though I'm not sure it would mean much depending on your goals). This is because it isn't the voltage which is determining the amount of current so why would you care (actually there are very rare cases where it is of use (see finite element analysis)). Rather it is the change in voltage. The voltage source can be at a value of 1, 10, 50 or even 100 KV and still have identical capacitor currents as long as the slope of the voltages at those points are equal. And what of the problem of having the sine wave voltage at it's peak extremes (where the slope and current are zero). Since real resistance is Voltage/Amps this would give you an infinite result. Instead we normally care about the reactance of a capacitor which is similar to resistance but 90 degrees out-of-phase. Impedance is the combined effect of BOTH the resistance and the reactance and why you will learn soon to calculate the impedance of a capacitor as Zc = 1/wC at some value of frequency and then essentially draw a right triangle to determine the impedance (phasors) which has a magnitude and a phase shift.
I agree. As an aside, V/I is not Ohm's law. We can discuss that later.
I disagree. Caps are a lot more complicated because they store energy while resistors do not. Resistors reduce current because they dissipate electrical energy. Caps reduce current because they produce a back voltage. Each reduces current by entirely different methods.
The basic definition of capacitance is the amount of charge that can be deposited on one plate of the cap per volt of difference between plates, C = Q/V . Capacity is the ability of the capacitor to store energy and change charge on both plates. As V increases, so does Q. This means that C will be constant for a given cap.
No tricks involved. Direct calculation is all that is needed. Recall Q = C V. Differentiating both sides with respect to t gives dQ/dt = I = C dV/t = C d(5 sqrt(2) sin(w t))/dt = C w 5 sqrt(2) cos(w t) = 0.47E-6 2 pi 100 5 sqrt(2) cos(2 pi 100) = 2.09 E-3 cos(628.32t) . So I = 2.09E-3 cos(628.32 t) , where 628.32 is radians/sec . Keep in mind that that no charge passes through the dielectric of the cap.;
Differential and integral calculus uses radians.
Ratch
Nevertheless, for dQ/dt = I the "I" would be physically measured as current flowing through the capacitor.
LOL. Never go into teaching.
Do you really mean charge flowing through the cap? Doesn't happen. If it did, the cap would be leaky and defective. What you will observe is the charge accumulating and depleting on the plates of the cap. That is much different than current existing through the cap. I thought we settled that earlier. What is it about flow of charge you don't understand?
Perhaps this thought will help. If current existed through the cap, charge would not accumulate and deplete on its plates, and the cap could not be energized to a voltage. Just like a resistor cannot be energized.
Ratch
You say that without reading the 5th post of this thread? https://www.electronicspoint.com/threads/lpf-and-hpf.274055/#post-1652226 . Are you averring that the professors who wrote those physics books should not be teaching either. Peruse, cogitate, and learn. Come back to the side of light, the side of right, and exit the dark side.
Ratch
Do you honestly think I didn't know that already. In fact I think not only did I realize this, I suspect he did as well. Are you just trying to spout out all you know in an effort to confuse or perhaps bolster your ego? Teaching is about understanding where your students are and taking them to the next step. Not proving to them how much you know and in the process only serve to confuse matters more for them. It's immaterial to the point and doesn't serve to clarify his question.
Not one of your "clarifications" to me wasn't previously understood by myself and not one served to promote his understanding. Leave out the superfluous until they understand the big picture, "What does the equation mean and how do you use it?". How would pointing out that current (I) is equal to dQ/dt resulting in dQ/dt divided by dV/dt resulting in dQ/dV have provided clarification to someone who is just learning what rms means let alone a solid foundation in differential calculus? Are you trying to frustrate them and drive them off from what is a fascinating and, yes, easy to understand science? Should we tell them capacitors are very complicated and not easily understood then confuse them just to prove it? His equation was in the form of either C = Ic / dV/dt or Ic = C * dV/dt.
By the way, I considered starting with C=Q/V then deriving his equation and rejected it, knowing he hadn't had DE and perhaps not even Trigonometry. Also why I didn't correct my subtle miss statement about the slope of sine being 1 at the origin. It isn't, its equal to 2*pi*amplitude = 6.28 for a 1 Hz signal because the sine source is actually sin(2*pi*t). Think that would have helped him too? If he had recognized this only then would I have tried to clarify.
Didn't I sufficiently explain what I meant? Perhaps a diagram will provide clarification. The current 'I' in the equation I=C⋅dV/dt is measured as the current flowing through the capacitor. Would there be any other possible way to measure 'I'?
What do you think?
You are confused. Look at post #30 of this thread and observe to whom my reply was addressed. It certainly was not to you. I do assume you know that charge does not flow through the dielectric of a cap, but Laplace doesn't seem to understand that. He has challenged me several times about this point in this and past threads.
I do not consider myself a teacher or those who ask questions in this thread my students. I am more of a knowledge base who may impart some knowledge to those who ask.
In you reply above, you refer to "he" and "his question". To whom does the pronoun refer, Laplace or the OP?
The OP asked where to find dv and dt. I gave him a succinct answer, not a rambling dialog or a DE course. It must have answered his question because I did see any follow up. By the way, surely you must agree that caps are more than a tiny bit more complicated than resistors. Telling someone otherwise does not help them.
If you ask ten different people the OP's question, you would probably get ten different answers. The OP has to decide which answer his likes best.
Ratch
I think you are measuring the charge flow released by the dielectric next to the right plate of the capacitor. No charge is passing through the dielectric. If the charge were passing through the dielectric, you would be able to sustain the charge flow value. You can't and won't do so unless you constantly increase the voltage value of your voltage source.
Ratch
And that would be wrong. One needs to think of this in terms of a complete circuit. The current being measured is the 'I' circulating in the loop. Maxwell provided a correction to Ampere's Law in order to explain why this is so; but that is more complicated than is necessary to understand how a capacitor behaves in a circuit. Nothing more is necessary for circuit analysis than the differential equation relating the capacitor voltage to the current flowing through the capacitor.
I have exhausted my ability to explain the operation of a capacitor to you. Perhaps this analog description will help you understand. https://commons.wikimedia.org/wiki/File:Water_cap_and_diode.png . Notice that water goes into the tank and out of the tank, but does not go all the way through the tank. Also the amount of water in/out is limited by the ability of the diaphragm to flex. If you cannot understand this, then I cannot help you further.
Ratch
You seem to misunderstand the situation here. I am trying to help YOU understand how we do circuit analysis with capacitors.
I have been doing circuit analysis since the year one. I don't think you have anything to teach me in regard to this subject.
Ratch
My apologies. I misunderstood to whom it was directed.
As for current flowing through the dielectric I might be able to clarify. Current is defined as the net flow rate of electrons through a component. It might surprise some to find out when you are measuring the current flowing through a wire the electrons do not really flow through the wire. Much like when I turn on a hose the water immediately begins flowing out the end even though the water flowing out of the spigot has yet to traverse the entire hose. It is the force which propagates through to the outlet not necessarily a specific droplet of water. Some of the water molecules actually traverse back into the spigot. It is the net flow of water we care about.
Similarly when we look at a capacitor with current "flowing through it" we are looking at the system as a whole. We don't normally track and aren't normally concerned with the movement of a single electron. The electrons drift from one collision to the next in a haphazard manner. Some even flow back towards the source. Also some positive ions flow towards the source and even towards the load. The ion flow is much lower due to their larger mass and the crystalline structure of the metal. In fact almost half of the free-electrons in a wire flow in a direction opposite the flow of current. If 1 amp of current is flowing through a wire this amounts to 1 Coulomb of charge per second tending to flow more towards the negative terminal of a battery. Yet the number of free electrons moving in the wire is many millions of Coulombs. In other words if you have 5 million Coulombs of electrons flowing in the wire only a small percentage above 50 percent tend to flow in the direction of current. It is this small imbalance of electron flow we care about on the macroscopic level.
It is entirely correct that if you were to follow the movement of a particular electron it is unlikely to flow through the dielectric (some small amount do however). If the voltage rating of the capacitor is exceeded you will get enough electrons tunneling through the dielectric to destroy the component due to heating. Under normal operating conditions the electrons tend to build up at the plates until the "back-voltage" matches the source voltage.
If you charge the plates of a capacitor then disconnect the source from the capacitor charge will remain on the plates. However, this charge will not remain indefinitely. It will slowly drain over time. Not all of which occurs through the gases surrounding the terminals. Even in a vacuum, some charge will slowly leak through the capacitor dielectric. So a real capacitor is modeled as an ideal capacitor with a leakage resistor in parallel.
For the most part the question as to whether electrons are flowing through the dielectric is one of semantics. It doesn't matter in terms of what we are interested in (normally). We are concerned with the system as a whole. The times we are concerned with dielectric current mainly involve understanding capacitors at the atomic level. In order to understand capacitors it does help to pretend they are ideal capacitors which have no leakage initially. Especially since it is the build-up of charge at the plates and the electric field which results which is so critical to the functioning of a capacitor. Once the physics are understood at the atomic level we must abandon the ideal model in some situations if leakage current is of concern.
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