Download Inductors and AC

Devices and Applications
Ctec 201.
Inductors and AC
Supplement
Prepared by Mike Crompton. (Rev. 15 July 2003)
Inductors and AC
If an AC voltage/current is applied to an inductive circuit the current is changing all the
time and therefore an induced voltage is being produced all the time. The greater the
frequency of the AC, the greater the induced voltage and the greater the opposition to
current flow (Resistance). This means that the AC resistance of a coil/inductor is
frequency sensitive, as is that of a capacitor. The difference being that the AC resistance
of a coil is directly proportional to its inductance (L) and to the frequency whereas in a
capacitor it is inversely proportional to C and frequency.
The correct term for AC resistance of a coil is Inductive Reactance (XL) and is calculated
by the formula: XL = 2 FL Where F is in Hz, L is in Henrys, XL is in Ohms
Series R.L. Circuits
The behavior of a simple series R.L. circuit as shown in Fig 1 is very similar to an R.C.
circuit. The calculations have to take into account the delayed current and resulting phase
changes produced by this delay.
Looking at Fig. 1 below right, let us first calculate the AC resistance of the coil (XL).
XL = 2 FL = 6.28 x 1500 x 0·1
XL = 942
Bearing in mind that this 942 is opposition to
current flow produced by the induced voltage
across the coil (i.e. VL) and the voltage leads the
current by 90, XL is in fact leading R by 90. XL
as a vector will actually be +J942, and to find Z
(the total opposition to current flow (XL + R) we
can resort to our vector diagram Fig 2
In rectangular form our Z is:
Z = +1000
+J942
Convert to polar:
+j2000
90
0
Z = 10002 + 9422 Tan-1 942/1000
+1000 +j942
+j1000
Note that the angle is + because J is +
Z
Z = 1373.8 +43.29
-2000
180
0
-1000
0
0
+1000
By using the voltage across R (VR) and Ohm’s
law we can determine the circuit current.
-j1000
ITOTAL = VR / R = 10/1000 = 0.01A
Fig. 2
0
-j2000
-90
2
+2000
It is very important to remember that IL has been delayed, and this is a series circuit so
ITOTAL is also delayed (current is the same in series circuit). The amount of delay
compared to VSUPPLY will become obvious when VSUPPLY is calculated.
Now calculate VL using Ohm’s law.
VL = ITOTAL * XL
= 0.01 * 942 = 9.42 Volts (or +J9.42V)
VSUPPLY = VR + VL
= +10 + J9.42
Convert to polar:
VSUPPLY = 102 + 9.422 Tan-1 9.42/10
= 13.74V +43.29
This means VSUPPLY is 13.74 Volts and is 43.29 ahead of ITOTAL or that ITOTAL lags
VSUPPLY by 43.29 .
A quick double check using Ohm’s law:
VSUPPLY = ITOTAL * Z
= 0.01 * 1373.8
= 13.74 Volts
Note that ITOTAL is used as our 0 reference vector (VR & R would also be at 0) the
circuit phase angle when inductance and resistance are in series, will always be a + angle.
When there is more than one inductor or resistor in series, find REQUIV and LEQUIV before
proceeding. For LEQUIV series inductors behave like resistors in series. LTOTAL = L1+L2
etc. Parallel inductors behave like resistors in parallel.
There is one other problem associated with RL circuits, and that is the problem of the
actual resistance created by the copper wire used to wind the coil.
You will recall (I hope) that a perfect coil has 0 resistance to D.C. but since a coil is
just a series of turns of copper wire, there is DC resistance proportional to the number
(and size) of turns and inversely proportional to the wire diameter. This resistance is
“unwanted” but cannot be ignored and is referred to as the “internal resistance” (R INT), it
is not frequency sensitive as is XL, but it does produce a voltage drop which appears as a
part of VL. Furthermore RINT does not produce a phase change (as does XL) and can give
a distorted calculation of circuit phase angle. The greater R INT the greater the error in
phase angle. In fact, the “quality” (Q) of a coil is the ratio of XL: R or XL/R.
This means that any part of VL that is the result of RINT should be added to the voltage
drop across the actual resistance (R) of the circuit, thus giving the correct calculation of
the phase angle . The example, along with Fig 3, on the following page, clarifies this
(hopefully).
3
The following values are measured/given.
L = 117mH
VR = 10V as measured with DVM
RINT = 310 as measured with DVM
VL = 8V as measured with DVM
 = 30º as measured with scope
VSUPPLY = 15V as measured with DVM
From the calculations on the following page,
there must be an error in these figures
If VR is 10v & VL is 8V then VSUPPLY should be:VSUPPLY = 102 + 82 Tan-1 8/10
= 12.8V +38.7º
(Measured as 15V +30º) ???
OR ..... If VSUPPLY = 15V and VR = 10V then VL should be:
15² = 10² + VL²
VL = 125
then VL² = 15² - 10²
= 11.18V
VL² = 125
(Measured value = 8V) ???
To find our error first use Ohms law to calculate ITOTAL :
ITOTAL = VR / R
= 10 / 1000
= 0.01A
Now calculate the voltage drop across RINT :
VR INT = ITOTAL * RINT
= 0.01 * 310 = 3.1V
Fig. 4
10
2
2
8 = 3.1 +
VXL2
VXL = 54.39
then
VXL2
2
= 8 - 3.1
2
= 54.39
6
8V
We can calculate a value for VXL from our diagram:
8
4
2
VXL
?
VL
Since VL was measured as 8V it is obviously the vectorial sum of
VRINT + VXL. Remember the voltage across RINT is at 0 and
across XL at +90. A simple vector diagram (Fig 4) shows this.
O
= 7.375V
2
4
6
VR 3.1V
4
Now we can do a true vector plot of VR + VRINT and VXL that should confirm our
measured values of  and VSUPPLY (See Fig 5 below).
To confirm the plotted values we can calculate VSUPPLY and 
VSUPPLY =  (VR + VRINT)2 + VXL2
Tan-1 VXL / VR + VRINT
8
= (10 + 3.1)2 + 7.3752
VSUPPLY = 15.03V
Tan-1 7.375 / 13.1
+29.4º
6
VXL 7.375V
4
So our measured values of 15V +30º compare
well with our calculated and plotted values.
pp
Vsu
15V
2
Further more, knowing VXL and ITOTAL we can
calculate XL using Ohm’s law.
XL = VXL / ITOTAL
Fig. 5
+29.4
VRint 3.1V
0
VR 10V
O
2
4
6
8
10
= 7.375 / .01 =737.5 +90º
And using the formula for XL the value of L can also be calculated:
XL = 2 FL then
L = XL / (2 F)
L = 737.5 / (6.28 * 1000)
= 0.1174H or 117.4mH
(Given value = 117mH)
A similar condition can exist with an R.C. circuit. Remember that the DC resistance
(RINT) of a capacitor is supposed to be infinity, so a relatively low value of RINT is due
mainly to “leakage”. Electrons leaking through the insulating material between the plates.
In most modern capacitors, the insulation is excellent and leakage is negligible. Most
electrolytic capacitors however do have leakage, and a quick check of calculated  versus
measured  will show if the leakage is significant. i.e. Measure VR , VC and the phase
angle . Then calculate  and compare the measured and calculated values, any major
difference will indicate leakage.
The major differences between inductors and capacitors are detailed on the next page.
5
12
14
XL delays the current through the inductor and results in the current being 90 (lagging)
behind the voltage VL. This is the opposite to a capacitor where current leads VC by 90.
An easy way to remember ‘what leads or lags what, in what circuit’ is to use the word:
CIVIL
Break the word into 2 parts;
CIV, in a Capacitor (C), I leads V.
VIL, V leads I in an inductor (L).
The fundamental differences between capacitors and inductors is a follows (Assuming a
pure/perfect coil and capacitor):
Capacitor
Parameter
Inductor
Resistance to DC
Infinity
0
Tends to resist changes in
Voltage
Current
Voltage/Current phase relationship
I leads V by 90
V leads I by 90
AC resistance (Reactance)
XC
XL
AC resistance (Reactance)
Inv. Proportional to C
Dir. Proportional to L
AC resistance (Reactance)
Inv. Proportional to F
Dir. Proportional to F
AC resistance (Reactance)
1 / (2FC)
2FL
6