
Use your knowledge of capacitor action to explain these observations. 
It seems that in Circuit A the current is being blocked, while in Circuit B the current is being allowed to pass.
We see this because the capacitor is being continually charged and discharged. No current is flowing through the capacitor. It cannot because there is an insulating layer. However we will treat the capacitor as if it were allowing a current to flow.
If we replace the 1000 mF capacitor with one of 100 mF, the charging and discharging currents are less. Therefore the bulb will glow less brightly. We can make the bulb brighter again by increasing the frequency.
From this we can conclude:
Large capacitors offer “less opposition” to AC.
Increasing the frequency increases the current in a circuit containing a capacitor since the same charge flows on and off the plates in a shorter time.
The reactance of a capacitor is the ratio of the voltage to the current. The symbol for reactance is X_{C} and the units are ohms (W). It is the “effective” resistance of the capacitor at a particular frequency. The reactance is linked to the capacitance, C, and the frequency, f, in the following relationship:
X_{C}
= __1__
2pfC
where:
f is the frequency in Hertz
C is the capacitance in farads.
The capacitance must be in farads. Watch out for this bear trap.
We can also define the reactance in terms of the rms voltage and the rms current of a sinusoidal waveform:
X_{C}
= V_{rms}
I_{rms}
Why do we use the term reactance to describe the ratio of voltage to current, in preference to resistance? 
Notice that in calculations we treat reactance like resistance as in Ohm’s Law.
Worked Example
A 400 mF capacitor is connected to a 56 Hz supply. If the supply voltage is 14 V rms, what is the rms current through the capacitor?
Use X_{C} = __1__ to work out the capacitor reactance. 2pfC
X_{C} = __1__ = __________1_____________ = 7.11 W 2pfC 2 ´ p ´ 56 Hz ´ 400 ´ 10^{6} F
Current = 14 V ¸ 7.11 W = 1.97 A

What is the reactance of a 1000 mF capacitor at a frequency of 1000 Hz? 
We can plot a graph of how the reactance varies as frequency. As X_{C} varies as 1/f, the graph will be a hyperbola.
To get a straight line, we would need to plot X_{C} against 1/f, the reciprocal of the frequency, which is T, the period. The gradient will be 2pC.
The essential difference between a reactive and a resistive circuit is that:
Power is dissipated in a resistive component
No power is dissipated in a reactive component.
In a resistive circuit, energy is dissipated as heat.
In a reactive circuit with a capacitor, the energy is used to build up the electric field.
A capacitor in series or parallel with a resistor can be used to make a filter circuit that allows us to select frequencies. A filter circuit consisting of a capacitor in series with a resistor can be made to act as a voltage divider, in the same way as two series resistors form a potential divider.
A series circuit with a capacitor C and a resistor R is connected to an AC supply of rms voltage V_{s} and frequency f.
We know that for a
series circuit, the current is the same all the way round.
For any frequency we know that:
X_{C} = __1__ and that V_{C} = I X_{C}
2pfC
We also know that V_{R} = IR
However, if we add the voltages across the resistor and the capacitor, they DO NOT add up to the supply voltage. For example, if the voltages across the capacitor and the resistor were 3 V and 4 V respectively, we would find that the supply voltage was not 7 V, but 5 V.
Why does this happen? The answer is that in a capacitor, the current and charge are not in phase. Phase relationships are discussed in my electrical engineering notes, so we won't consider them any further here.
Worked
Example
A series circuit consisting of a 0.1 mF capacitor and a 2000 W resistor is connected to a 2.0 V supply set at a frequency of 1000 Hz.
(a) What is the reactance of the capacitor? (b) What is the impedance of the circuit? (c) Work out the current. (d) Work out the voltage across and the resistor. (e) Work out the voltage across the capacitor. 
(a) First work out the reactance of the capacitor.
X_{C} = __1__ = ____________1____________ = 1590 W 2pfC 2 × p × 1000 Hz × 0.1 × 10^{6} F
(b) Now work out
the impedance. Z^{2} = R^{2} + X_{C}^{2} = (2200)^{2} + (1590)^{2} = 7.36 × 10^{6} W^{2} => Z = (7.36 × 10^{6})^{1/2} = 2714 W
(
Note: I have had to use the notation X^{1/2} to denote the square
root)
(c) The current
can be worked out by I = V/Z V_{C} = IX_{C} = 7.4 × 10^{4} A × 1590 W = 1.77 V 
There are occasions where a filter circuit is useful:
To boost low frequency sounds when music is played at low level. The ear does not detect low frequency sounds very well.
Gets rid of high frequency noise such as a tape hiss
Compensates for imperfection in sound sources.
The RC circuit is a passive filter circuit. It cuts treble or bass frequencies, but cannot boost them. To boost the treble or bass, we need an amplifier in the circuit; this is an active circuit.
What is the difference between an active and a passive circuit? 
The RC circuit is like a potential divider. Formula:
This circuit acts as a voltage balance:
If R_{1} is smaller than R_{2}, then the voltage across R_{2} (V_{out}) will be larger than the voltage across R_{1}, and vice versa.
The voltages across both resistors add up to the supply voltage.
We can alter the voltage balance by changing either of the resistances.
We can lay out the RC circuit in exactly the same way.
V = IR
X_{C}
=
1_
2pfC
V_{C}
= IX_{C}
At low frequencies, the reactance of the capacitor is high. This means that the voltage across the capacitor is high, since the current is the same throughout the circuit.
The voltage across the resistor is low at low frequencies.
Explain what happens to the voltage across the capacitor when the frequency is high. 
There is a frequency at which
the voltage across the capacitor is the same as the voltage across the resistor.
This is called the break frequency. Therefore the reactance and the resistance are the
same:
We can measure the voltage across the capacitor at different frequencies and plot the log of the voltage against the log of the frequency.
A logarithm is a number expressed as a power of 10. So 10 000 = 10^{4}. Therefore log_{10} 10000 = 4. Logarithmic scales allow for a much larger range of values to be displayed than a linear scale. 
The plot is almost level until the break frequency. There is a change in the voltage balance, but remember that the effect will be less marked as the voltages add up as vectors. The effect is made even less by the logarithmic scale.
The plot passes under the frequency axis at 1 volt, since log_{10} 1 = 0. Voltages of less than 1 volt will show up as negative values on the log_{10} V axis.
A filter circuit has a 10 mF capacitor in series with a 1 kilohm resistor. What is the break frequency? 
If
we place the input to an amplifier at V_{out}, the amplifier would amplify
preferentially the lower frequencies, since the voltage produced by higher
frequencies would be very small.
If we turn the RC circuit upside down, we get a passive bass cut filter.
If we increase the frequency, the reactance of the capacitor goes down. Therefore the voltage across the capacitor must go down as well. So the voltage balance shifts and the voltage across the resistor goes up.
V_{out} will be high when the frequency is high. A graph can be plotted of log_{10} voltage against log_{10} frequency:
Explain why the bass cut filter behaves in this way. 
Active filters are discussed in the next part of the Tutorial.