# Mathematics for the RAE

### Typical calculations

We will now apply these rules to the solution of problems likely to be met in radio work as a lead in to some typical numerical multiple choice questions.

Answers to three significant figures as given by a slide-rule or four-figure logarithm tables are satisfactory for most radio purposes and the eight figures given by the electronic calculator should certainly be rounded off.

The most important aspect is to remember that the units met with are most likely to be the practical ones such as microfarads, picofarads, milliamperes, millihenrys etc. These must be converted into the basic units of farads, amperes and henrys before substituting them into the appropriate formula. This involves multiplying or dividing by 1000 (103), 1,000,000 (106) and so on. Therefore the important thing is to get the decimal point in the right place or the right number of noughts in the answer. The commonest conversions are as follows

There are 106 microfarads in one farad

hence 8µF = 8 x 10-6 farads

There are 1012 picofarads in one farad

hence 22pF = 22 x 10-12 farads

(The use of 'nano' or 10-9 is now fairly common; there are 10-9 nanofarads in 1 farad so 1nF = 1 x 10-9 farads, but such a capacitor may well be marked '1000pF'.) Similarly other conversions are

50µH = 50 x 10-6 henrys

3mH = 3 x 10-3 henrys

45mA = 45 x 10-3 amperes

10µV = 10 x 10-6 volts

#### Problem 1

What value of resistor is required to drop 150V when the current flowing through it is 25mA?

This involves Ohm's Law which can be expressed in symbols in three ways:

where R is in ohms, V in volts and I in amperes. Clearly the first, R = V/I, is needed. First of all, we must express the current (25mA) in amperes.

Substituting values for V and I

(we are dividing by 25/1000, ie multiplying by 1000/25 )

25 'goes into' 150 six times, so

####

#### Problem 2

What power is being dissipated by the resistor in Problem 1 ?

The power dissipated in the resistor is power (watts) = V (volts) x I (amps). By Ohm's Law, power can be expressed in two other forms.

Because we know V, I and R we can use any of the above relationships, say

Two 'noughts' on the top and bottom can be cancelled, leaving

Cancelling 15 into 60 leaves

The other two forms will, of course, give the same answer - try them!

####

#### Problem 3

Resistors of 12Ω, 15Ω, and 20Ω are in parallel. What is the effective resistance?

60 is the lowest common denominator of 12, 15 and 20, so

This is a simple equation in R, and the first step in solving it is to 'cross-multiply'. It may be shown that the denominator of one side multiplied by the numerator of the other side is equal to the numerator of the first side multiplied by the denominator of the other side, thus

Hence, dividing each side by 12

####

#### Problem 4

Capacitors of 330pF, 680pF and 0.001mF are in parallel. What is the effective capacitance?

The first step is to express all the capacitors in the same units which can be either picofarads or microfarads

(there are 1,000,000µF in 1pF) and hence

Effective capacitance is therefore

####

#### Problem 5

What is the reactance of a 30H smoothing choke at a frequency of 100Hz?

We take π to be 3.14 so

####

#### Problem 6

What is the reactance of a 100pF capacitor at a frequency of 20MHz?

(XC is in ohms, when f is in hertz, L in henrys and C in farads.)

(It is much more convenient here to use the index notation)

Hence

Note that we have kept 1/2π intact because 1/2π = 0.16, thus

####

#### Problem 7

What is the impedance (Z) of an inductance which has a resistance (R) of 4Ω and a reactance (Z) of 3Ω

####

#### Problem 8

At what frequency do a capacitor of 100pF and an inductance of 100mF resonate?

At resonance

Hence

(f is in hertz, L is in henrys, C is in farads)

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