IAN LANG ELECTRONICS

L

Calculating Capacitive Impedance

Traditionally the formula for capacitive impedance is given as Z= (R + X c) which is putting in the resistive element too and which is introducing needless complexity, so let’s arrange it so we don’t have to work as hard. Putting anything to the power of 0.5 is exactly the same as saying “square root of” and if it’s the square root of the right- hand side then it must be the square of the left. So we can say:

Z = R + X c which makes it a pythagorean equation and less likely we’ll get it wrong. Whichever way we do it we still need to find Xc first . Our example circuit looks like this:

2

2

2

2

2

0.5

When we calculated Xc before we used the formula Xc=1/(2 f C) . In our example we are given no example frequencies. Let’s have two of them so we can compare the effects. For the first, we shall have 50 Hz, and for emphasis the second can be 10Hz.

As before, we need to get our units straight. Let C be in Farads:

200/1000000 = 0.002F,

let R be in ohms:

2 x 1000000 = 2,000,000 ohm

So, let's do Xc at 50 Hz:

start with fC which is 50 X 0.0002 = 0.01

multiply that answer by 2 = 0.0628

now put that as a reciprocal 1 / 00628 = 15.9 ohms of capacitive reactance (Xc).

At 10 Hz:

10 X 0.0002 = 0.002

multiply that by 2 = 0.01256

and now the reciprocal 1 / 0.01256 = 79.6 ohms of capacitive reactance (Xc).

This illustrates an important point: Xc goes down as frequency goes up.

Our impedance at 50 Hz then is:

Z = 2,000,000 + 15.91 and squaring everything on the right hand side:

Z = 4,000,000,000,000 + 253.13 and adding them up:

Z = 4000,000,000,253.1281 if that’s Z2 then finding the square root gives us Z:

Z = 2,000,000.000063284 ohms of total impedance.

and thus the circuit contains hardly any reactance at all in comparison. The power factor of such a circuit would be as close to 1 as to make scant difference

If we were to work the 10Hz frequency in the same way we’d get an answer of 2,000,000.001583 ohms which is still almost entirely resistive. This is because the resistor is such a big value. Nevertheless an important point has been illustrated : unlike resistance impedance is dependent upon the frequency of the supply. And for capacitors, as frequency goes up, reactance goes down. As reactance goes down, so does impedance.

The above is given as question 5 in BTEC assesment 2. The below is question 10, which concerns itself with doing the same thing for inductors.

2

2

2

2

2

In this case we have no capacitors but an inductor, at 300/1000 Henries (300mH) and so we need to be getting a figure for that: 300/1000 = 0.3

We do the same for the resistor:

2,700,000 ohms.

We need to find the inductive reactance before the impedance.

The formula we need is:

X = 2 f L

Right then let's get busy. We'll do 50 Hz and 10Hz again, so at 50 Hz then our inductive reactance is:

6.28 X 50 X 0.3 (2pi X f X L)

and 50 X 0.3 = 15

and 6.28 X 15 = 94.2 ohms

At 10 Hz

6.28 X 10 X 0.3

and 10 X 0.3 =3

and 6.28 X 3 = 18.8 ohms

In this case the greater the frequency, the greater the reactance which is the opposite of the behaviour of the capacitor.

The square of the total impedance in LR (inductive resistive) circuits is given by:

Z = R + X

putting in some numbers:

Z = 2700000 + 94.2 (at 50 Hz) and doing the sums:

Z = 7290000000000+ 887364 = 7290000008873.64and if that's the square, finding the root gives the impedance Z:

Z= 2700000.00164327

At 50 Hz the circuit is almost entirely resistive and this is because the resistor is so big.

So far we have looked at capacitive and inductive reactance, and the opposition to AC that comes with either one or the other and with resistance. In an RLC circuit, all three properties are present and the calculations are slightly different. Let us look at our example circuit:

We have an inductor of 300mH, a capacitor of 300mF and a 2.7 Megohm resistor. The first thing we need to do is to get our subdivisions and multiples into the base units. So:

300mH = 300/1000 H = 0.3H

300mF = 300/1000000 = 0.0003F

2.7 Megohm =2,700,000 ohm

We are not given any frequencies and so let us use 50Hz as a low frequency and 1MHz as a high for comparative values to see what happens. 1Mhz is of course 1,000,000 Hz

The formula we need is the somewhat arcane looking Z= R +((XL-XC) ) This is undoubtedly easier to remember when put as a pythagorean equation and so transposing it we get:

Z = R + (XL-XC) and moving that square root over the other side makes working the right hand side a little easier.

It will do us no harm to go over the formulae for capacitive and inductive reactance again.

So at 50Hz:

Xc= 1/(2pi fC) = 1/ (6.283)(50)(0.0003)=1/0.094245 = 10.61W

XL= 2pi fC = (6.283)(50) (0.3) =94.25W

and at 1,000,000 Hz (or 1 MHz)

Xc= 1/(2pi fC) = 1/ (6.283)(1,000,000)(0.0003)= 1/1884.9 = 0.00053 ohm or 0.53 mohm

XL= 2pfC = (6.283)(1,000,000) (0.3) = 1884900 ohmor 1.9 M

An important point is illustrated again here; reactance is dependant on frequency. As frequency increases, capacitive reactance Xc decreases, and inductive reactance XL increases. Observe now how this affects impedance:

for 50Hz:

Z = R + (XL-XC) ........Z = R + (94.25-10.61) ........Z = R + (83.64) ......Z = R + 6995.6496......

Z = 7290000000000+ 6995.6496......Z = 7290000006995.6496......... Z=2700000.001295 ohm

At this low a frequency the reactance in this highly resistive circuit is insignificant, and opposition to electron flow is almost purely resistive. This circuit will have a power factor close to 1.

for 1,000,000 Hz

Z = R + (XL-XC) ......Z = R + (1884900-0.00053) .......Z = R + 3552848010000..........

Z = 7290000000000 + 3552848010000..........Z = 10842848010000........Z=3292848W or 3.3Mohm

there will be a power factor of about 0.8 in this circuit, and the current will lag the voltage by about 35 degrees.

The sum of this lesson is that impedance is made up not only of resistance but also of reactance, and unlike resistance, impedance depends on the frequency of the supply.

14. Frequency of Resonant Circuits.

The frequency for a resonant circuit is given by the equation

f=1/(2pi LC)

Where L is in Henries, C is in Farads and putting anything to the power of 0.5 is the same as saying square root of. Once again we have to sort out our subdivisions first. We are given 300mF and 800mH, and from what we did above we know 300mF=0.0003F and 800/1000 gives 0.8H.

This calculation is tricky, so we shall "bite-size" it . I find it easier to start by multiplying LC first and working backwards, so:

f=1/(2pi LC) and putting in numbers for LC

f=1/(2pi 0.8 x 0.0003) and multiplying

f=1/2pi 0.00024) and doing the square root

f=1/2pi 1.549) and multiplying by 2pi

f=1/0.0973 and doing the reciprocal gives us the answer:

f= 10.277Hz

2

2

2

L

2

2

2

2

2

2

0.5

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

0.5

0.5

0.5

0.5