A number of sample calculations involving resistance, capacitance, inductance, reactance and resonance follow.
The mathematics involved are explained in Appendix 2. The calculations required in the RAE are generally more simple than these and often the answers can be obtained by inspection or approximation.
By Ohm's Law
By Ohm's Law
Therefore, by cross-multiplying
Alternatively, the value of two resistors in parallel can be found by dividing the product of their values by the sum.
Effective capacitance
Before addition can be effected, the values must first of all be expressed either in picofarads or in microfarads. Since there are 1,000,000pF to 1μF,
Therefore the effective capacitance is
Alternatively,
and
so that the effective capacitance is
The 10 and 20μH coils in series are equivalent to (10 + 20) = 30μH
The 30 and 40μH coils in series are equivalent to (30 + 40) = 70μH
These two equivalent inductances of 30μH and 70μH respectively are in parallel and therefore equivalent to one single inductance of
(a) input power = W = DC volts x DC amperes
= 20 x 3 = 60W
(b) output power = W = (load current)2 x load resistance
= 0.9x0.9x50 = 40.5W output power
(c) efficiency
or
(a)
(b) At eight times the frequency, X is eight times as great, ie 37,600Ω.
Note how the frequency in kilohertz must be multiplied by 1000 to bring it to hertz, and how the inductance in micro-henrys must be divided by 1,000,000 to bring it to henrys.
(a)
where
f is the frequency and C the capacitance in farads.
(b) At eight times the frequency, X is one-eighth of the above value, ie 200Ω.
(a) At DC the reactance is zero and only the resistance opposes the passage of current. By Ohm's Law
(b) At AC Ohm's Law may still be used, provided Z (the impedance) is used in place of R (the resistance).
(a) A capacitor blocks the passage of direct current, therefore the current is zero amperes.
(b) Ohm's Law still holds, provided Z (the impedance) is used in place of R (the resistance).
For resonance
hence
farads
microfarads
and inserting the given values
For resonance
hence
From the resonance formula
the inductance is
Expressing the frequency and the capacitance in the basic units (f = 400x103Hz and C = 500x10-12F)
=
Taking π2 = 10
=
The antenna and coil together resonate at a frequency determined by the capacitance and the sum of the antenna effective inductance and the loading coil inductance.
Here the relevant values of inductance and capacitance expressed in the basic units are
Therefore
(a) What value of inductance in series is required to tune the circuit to
resonance?
(b) At resonance, what is the current in the circuit?
(a) For the calculation of the inductance, the resistance can be ignored, since it has no effect on the resonant frequency, which is given by
rearranging
Expressing the frequency and the capacitance in the proper units (f=5x106/π Hz and C=100x10-12F)
(b) At resonance, the inductive and capacitive reactances cancel out and the circuit has a purely resistive impedance of 10Ω. The current I through the circuit at resonance can then be calculated directly from Ohm's Law: I = V/R. Since V = 10V and R = 10Ω, the current at resonance is
Bandwidth is approximately 666Hz.