Download Capacitors and AC/Complex Numbers notes

Devices and Applications
Ctec 201.
Capacitors and AC
Supplement
Prepared by Mike Crompton. (Rev. 15 July 2003)
Capacitors and AC
When an AC voltage is applied to a capacitor, the current flow in the circuit will follow
the supply voltage but the voltage across the capacitor will be delayed. The delay is equal
to 90 degrees of the sine wave. That is, the capacitor voltage is always 90 degrees behind
the current no matter what the freq. If the freq is high, the voltage across the plates may
not reach maximum before the supply changes direction and starts to decrease. This is
significant in that you recall that the current in the circuit decreases as the voltage across
the plates increases. This means the voltage across the plates acts like Resistance. The
greater Vc the less current flows. The smaller Vc the greater the current. If the capacitor
does not have time to charge to the supply voltage, (the freq is too high) the less the
“Resistance” of the capacitor. The slower the rate of change (low freq) the greater Vc and
the greater the Resistance. Remember that with a DC the resistance increased to infinity
i.e. No current when the capacitor was fully charged. This means that the “Resistance” of
a capacitor is frequency sensitive, dependant on the ‘reaction’ time of the capacitor and is
thus called ‘Capacitive reactance’. Capacitive reactance, sometimes known as AC
Resistance uses the letters XC and is measured in Ohms. Since the capacitors reaction
depends on it’s size and the frequency of the applied voltage, the formula for Xc is:
XC = 1 / (2 FC)
Where XC is in Ohms, F in Hertz and C in Farads.
e.g. What is the capacitive reactance of a 10F capacitor at a freq of 45 kHz and 4.5 Hz
@ 45kHz XC = 1/ (2 *  * 45,000 * 0.00001) = 1 / 2.826 = 0.3537
@ 4.5kHz XC = 1/ (2 *  * 4.5 * 0.00001) = 1 / 0.0002826 = 3537
Note that a reduction by a factor of 10,000 in frequency produces an increase in XC by a
similar 10,000 factor.
What is the Xc of a 100pF capacitor at 45 KHz and 4.5 Hz.
@ 45kHz XC = 1/ (2 *  * 45,000 * 100 * 10-12) = 1 / (2.826 * 10-5) = 35.37 k
@ 4.5Hz XC = 1/ (2 *  * 4.5 *100 * 10-12) = 1 / (2.826 * 10-9) = 35.37 M
Note that a reduction by a factor of 100,000 in capacitance (10F to 100F) from the
example above produces an increase in XC by a similar 100,000 factor.
The capacitive reactance behaves like resistance and could be used in an Ohms law
formula in place of resistance to determine circuit current, or voltage across the capacitor
(Vc). It can also be combined with resistance to determine total opposition to current
flow known as impedance, because it impedes current flow. The letter Z is used to denote
impedance.
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Summary of Vectors & J Operator
But .......... remember that Xc is in fact a measure of how much a capacitors voltage will
reduce current flow and that capacitor voltage is delayed by 90 degrees compared to
circuit current. So the phase of Xc is going to be different to the phase of normal
resistance (R) (which has no difference or phase change between it’s voltage and
current). You cannot therefore just add Xc and R, you must add them ‘Vectorially’.
People often ask, “how do you add vectorially” to which we reply “very precisely”.
A vector is a line whose length indicates an amount, quantity or number and whose
position indicates direction, phase or offset from a point. In order to draw a vector you
need reference axes to allow you to plot the vector’s length and direction.
e.g. A compass has cardinal points, 4 of which are
N, S, E, & W. (fig 1). Assuming each small line on
each of the four axes (N, S, E, W) represents 1 mile
we can plot the Resultant Vector for a distance of 3
miles along the N axis and 2 miles along the E axis.
(3N & 2E)
A second example could be a circle, which of
course has 360 degrees. A vector could be 1” long
at an angle of 45 degrees (fig 2)
These two methods of describing a particular point
(end of vector) within the graph space are called
‘Rectangular’ (fig. 1) and “Polar” (fig. 2)
In electronics we can use either or both by
combining and modifying the two graphs. Since we
are interested in adding two vectors (or subtracting,
multiplying or dividing) we need numerical values
on both axes. This presents a problem. Take the
horizontal axis for example. We can label from the
centre to the right from 0 to +6 (or any other
quantity). and from the centre to the left from 0 to –
6 (fig 3)
O
90
45
45
O
180
Fig. 2
O
O
0/360
270 O
Vector
O
2.5cm at 45
0
-6
-5
-4
-3
-2
-1
+1
+2
+3
+4
+5
+6
Fig. 3
3
O
But.... When we go to add the vertical axis we encounter the problem of how we will
indicate a value above the horizontal axis and below the horizontal axis, we have already
used the + & -??? Solution.... Call the vertical axis “J” (‘i’ if you are a mathematician) so
anything above the horizontal axis would be +J and below, -J (See fig 4)
+j5
Examining point X on our Fig. 4 graph it can
be seen that it’s position could be described as:
+j4
Point 'X'
–4 + J3 (or –4 + 3J)
+j3
+j2
Point Y is +4 – J5
Should we wish to add the two points (X & Y)
together we just algebraically add them.
+j1
-4
-5
-3
-2
-1
+1
+2
+3
+4
+5
-j1
-j2
-4
+4
0
+J3
-J5
-J2
-j3
Point 'Z'
Point 'Y'
-j4
Fig. 4
-j5
Which is point Z on our graph
This is the “Rectangular” method of representing a point in the graph space using a
Complex Number (A number using both + & - and the J factor) and how two points can
be added (or subtracted). Addition or subtraction of complex numbers is most easily
accomplished if the numbers are in “Rectangular” format.
90
Fig 5 shows our graphs from Fig. 2 and Fig. 4
combined.
0
143.1
0
+j5
+j4
Point 'X'
+j3
Point X (same as above) can now be described
as:
+j2
5
180
+j1
0
0
-4 + J3 (Rectangular)
-5
-4
-3
-2
-1
+1
+2
+3
-j1
or as:
-j2
5 143.1 (Polar)
which translates to 5 units long at 143.1 degrees
-j3
-j4
Fig. 5
-j5
270 0
4
+4
0
+5
The graph type used in electronics is a combined
graph similar to Fig 5 with one main difference.
See Fig 6 on the following page.
+j8
90
0
POINT 'A'
0
+2 +j7 or 7.28 +74
+j6
7.28
+j4
180
-8
+j2
0
-6
-4
74 0
+2
-2
-j2
0
+6
+4
-51.4
6.4
As can be seen we have changed the 360 scale to
two 180 scales. A 0 to +180 scale going CCW
from 0 and remaining above the horizontal axis,
and 0 to –180 going CW from 0 and remaining
below the horizontal axis. This means that any
point above the horizontal axis will have a +J
element and a +ve angle. Any point below the
horizontal will have a – J element and a -ve angle.
Looking at Point Y from Fig 4 (and duplicated on
Fig 6), it still has the same Rectangular
components i.e. +4 – J5 but it’s Polar components
are 6.4 -51.4 instead of 6.4 308.6 as it would
be with a 360 scale.
0
+8
0
-j4
POINT 'Y'
-j6
Fig. 6
+4 -j5 or 6.4 -51.4
0
0
-90
-j8
In electronics (particularly AC Circuits) we often have to add, subtract, multiply or divide
complex numbers. “Rectangular” format is best for addition and subtraction, “Polar” for
multiplication and division.
+j4
Converting from “Rectangular” to “Polar” and vice versa.
Point B on Fig 7 corresponds to the rectangular coordinates +4
+J3 and the lines from those points on the axes explain why
the word ‘Rectangular’ is used. It should also be obvious that
the line from corner to corner also represents the ‘Polar” vector
for the same coordinates and bisects the rectangle into two
right angled triangles each with one side equal to 4 and one 3
(J3). The hypotenuse of both triangles is common to both and
its length will be the length of the Polar Vector. The angle 
(theta) will be the angle of the Polar Vector.
Point 'B'
+j3
+j2
+j1
-4
-3
-2
O
+1
-1
+2
+3
-j1
-j2
Fig. 7
-j3
-j4
The ‘Rectangle’ from Fig 7 above is enlarged in Fig 8 and the traditional names for the
sides of a triangle (opposite, adjacent and hypotenuse) have been added. The length of the
hypotenuse can obviously be found by Pythagorean theorem
C2 = A2 + B2
or
C =  A2 + B2
Where C is the Hypotenuse, A the Adjacent and B the Opposite.
Thus the calculation is:
Fig. 8
C =  42 + 32 = 25 = 5 (The length of the vector)
H
yp
O
Opp
Adj
4
5
3
+4
To calculate the angle we use trigonometry (SIN = O/H, COS = A/H, TAN = O/A):
Since Tan  = Opposite/Adjacent then  = Inv Tan Opposite/Adjacent
 = Inv Tan ¾
= Inv Tan 0.75
= 36.87
Therefore “Rectangular” +4 +J3 = “Polar” 5 36.87
To convert from Polar to Rectangular Trig will
again be used, and we will convert our polar vector
back to rectangular.
Fig 9 shows the Polar Vector with a vertical line
from the 5 36.87 point down to the horizontal
axis, forming a right angled triangle where the
opposite side will = J and the adjacent side = to our
real number.
By trig, the adjacent side (real number) is:
Cos  = Adj / Hyp
Point 'B'
5 36.87 0
5
O
Opp
(j)
0
36.87
Adj
Fig. 9
= Cos 36.87 = Adj / 5
So Adj = Cos 36.87 * 5 = 0.799 * 5 = 3.99 (or 4 if rounding is taken into account)
Meaning or first rectangular number is +4.
By trig, the Opposite side (or J) is:
Sin  = Opp / Hyp
= Sin 36.87 = Opp / 5
So Opp = Sin 36.87 * 5 = 0.6 * 5 = 3
Meaning our second rectangular number is +3J and our conversion is complete:
Polar 5 36.87 is equal to +4 +3J.
To relate the sides of the triangle to electronics, the Hypotenuse = Z, the Adjacent = R
and the Opposite = XC. Later it will be shown that they can also be Hypotenuse =
VSUPPLY, Adjacent = VR and Opposite = VC.
We will now look at resistance and capacitors in the same series circuit.
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R.C. Series Circuits
We left capacitors at the point of considering the situation when there is capacitive
reactance (Xc) and resistance in the same series circuit. Recall that the charge across the
capacitor plates was responsible for the “AC resistance” (Xc) and that charge (Vc) was
delayed by 90 with respect to the circuit current.
Fig 1 shows a simple series circuit with a 1k
resistor and a 0.1F capacitor. Let us first
calculate Xc (AC resistance of Capacitor)
Xc = 1 / (2FC)
= 1 / (6.28 * 1500 * 0.1 * 10-6)
Xc = 1062 and will lag R by 90
I total (IT) will be the same in all parts of the
circuit (series circuit), and can be calculated by
dividing VR by R.
IT = 10V / 1000 = 0.01A or 10mA
Vc using Ohms law will be IT * Xc (equivalent to V = I * R)
Vc = 0.01A * 1062 = 10.62V and will lag VR (and IT) by 90
We now have all our circuit values EXCEPT VSUPPLY & RTOTAL
How do we calculate VSUPPLY & RTOTAL ?
In a series circuit with two resistors, RTOTAL = R1
+ R2, but one of our resistors is Xc and is
derived from Vc which is lagging I by 90. For
the resistor, VR & I are in phase. Xc could more
correctly be written as –J1062 and this gives us a
clue. Recall our complex number graph (Fig 2).
If we change the scale from 1 to 1000 and add
our R & Xc values we can vectorially add Xc &
R and call the resultant Z (Impedance). Or:
+j2000
90
0
+j1000
-2000
180
-1000
+1000
+2000
0
Z
0
R = 1000
-j1000
Z = 1000 -J1062 (Rectangular)
To convert Z to Polar:
Xc = -j1062
Fig. 2
0
-j2000
-90
Z = 10002 + 10622 = 1457.7
7
0
The angle () is Tan-1 -Xc / R = Tan-1 = -1062 / 1000 = -46.7
Therefore Z = 1457.7 -46.7 Polar
VSUPPLY can now be determined in either of 2 ways:
1. Ohm’s law VT = IT * RT (Where RT is actually Z)
VT = 0.01A +/- 0 * 1457.7 -46.7 (Algebraically add the angles)
VSUPPLY = 14.587V -46.7
2. By vectorial addition of VR & VC , which would be the identical process used to find
Z, substituting VR for R & VC for XC
By graph VSUPPLY rectangular = +10 –J10·62
Convert to Polar, Z = 102 + 10.622 = 14.58V
The angle () is Tan-1 Vc / VR = Tan-1 = -10.62 / 10 = -46.7
VSUPPLY = 14.58V -46.7
The angle associated with VSUPPLY is the angle between VAPPL and circuit current (IT)
and is sometimes referred to as the ‘Circuit Angle’.
Note: J has been assigned the numerical value of -1 which is of course a fictitious
number as –ve numbers cannot be square rooted. However, in theory J2 is +1, and 1
times any number is that number and therefore J does not interfere with most
mathematical processes.
If the circuit contains 2 (or more) resistors and 2 (or more) capacitors in any order,
reduce them to their equivalent values before doing the vectorial addition. Remember
that capacitors in series behave like resistors in parallel (and capacitors in parallel behave
like resistors in series) for determining CTOTAL.
Series CTOTAL = 1/ CTOTAL = 1/C1 + 1/C2 + 1/C3 etc.
Parallel CTOTAL = C1 + C2 + C3 etc.
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