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Capacitors and AC 

When we apply ac to a capacitor as shown to the right, we know that the capacitor will draw current to oppose any change in voltage across itself. But that doesn't tell us how much opposition the capacitor will offer, or how much current it will draw. So how can we determine just how much current will flow through C?
We find the answer by going back to the original equation for capacitive current, i_{c}, which we introduced when we looked at RC time constants with an applied dc voltage. This equation uses differential calculus, and is written as:
i_{C} = C  dv_{C} 
dt 
Now we are applying an ac voltage to the capacitor. Therefore, v_{c} is a sine wave of some frequency, not a fixed dc voltage. Technically:
In this type of equation, the Greek letter omega (ω) represents the frequency in radians per second, where ω = 2πf. v_{p} is the amplitude of the ac generator or other source. So how do we find the derivative of v_{p}sin(ωt) to determine i_{C}?
Since these pages are not intended to be a rigorous treatment of mathematics (especially calculus), we will not go into a process of evolving the derivative of a sine function. Instead, we will simply fall back on the following general expression from a book of math tables:
d  sin(u) =  du  cos(u) 
dx  dx 
In this expression, "x" is the generalized independent variable. For our specific case, this will be "t," for time. The variable "u" is the generalized expression or function of "x" which is used as the argument of the sine function. Making these substitutions, we get:
i_{C} =  C  d  v_{p}  sin(ωt) 
dt  
=  v_{p}C  dωt  cos(ωt)  
dt  
=  v_{p}C ω cos(ωt)  
=  ωC v_{p} cos(ωt) 
The factor ωC, or 2πfC, amounts to a "constant of proportionality" that relates the voltage and current in the capacitor. Note that it depends on both the value of the capacitance and the frequency of the sine wave. As either factor is increased, the capacitor current will increase for the same applied voltage. Note that this is exactly the opposite behavior from a resistance. Can we make use of this factor in a similar way?
The derived equation above for the alternating current in a capacitor tells us several important things. One of these is that the when the applied ac voltage is a sine wave, as shown in red in the graph to the right, the resulting current is actually shifted in phase by 90° — it is a cosine wave, as shown in blue in the graph. The current actually leads the applied voltage by ¼ cycle.
This actually fits what we know about the capacitor, which is that it will draw current in its attempt to oppose any change in voltage across its terminals. Thus, the capacitor reacts to the applied ac voltage by drawing current ahead of the applied voltage changes.
As to that factor of ωC (or 2πfC), if we invert it and use the factor 1/ωC or 1/2πfC, it will behave like the capacitive equivalent of resistance. We can't properly call it resistance, of course, but because the capacitor does react to the application of an ac voltage, we can properly call it a reactance. This is typically designated with the letter X, and capacitive reactance is designated X_{C}. Mathematically:
Capacitive reactance is measured in ohms, just like resistance, and works like resistance in many ways. However, its value depends on frequency as well as on the value of the capacitance. If we plot a graph of X_{C} versus the product ωC using logarithmic scales, we get the graph shown to the right. This graph extends indefinitely in both directions, to cover any value of C and ω. It is not possible to get an X_{C} of zero with any finite frequency, other than by setting C = 0.
In a purely capacitive circuit, we can use X_{C} for the various capacitors just as if they were resistors. Ohm's Law still applies to such circuits. However, as we will see on another page a little later on, we cannot simply add values of X_{C} and R. That phase shift introduced by the capacitor adds a bit of a complication that must be dealt with. We'll see how when we reach that page.


 
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