Fundamentals of Electricity & Radio Communications sample paper-2
By
The Radio Society of Sri Lanka
The copy right of the information contained in this sample exam are with the Radio Society of Sri Lanka (RSSL).
These questions are mainly targeting the Radio Amateur exam for Novice and General class. The recommended time duration is 1 hour. For more information please contact Mr. Jayasiri Wijeratne (4S7VJ) 4s7vj@rssl.lk
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Question 1 |
An AGC circuit in a receiver usually controls the
A | RF and IF stages |
B | power supply |
C | audio stage |
D | mixer stage |
Question 2 |
Three 15 picofarad capacitors are wired in parallel. The value of the combination is
A | 18 pF |
B | 45 pF |
C | 12pF |
D | 5 pF |
Question 2 Explanation:
For parallel combination of capacitors use C = C₁ + C₂ + C₃, and for series combination use 1/C = 1/C₁ + 1/C₂ + 1/C₃
Question 3 |
A | Drain of a junction FET |
B | Gate of a junction FET |
C | Collector of pnp transistor |
D | Collector of npn transistor |
Question 3 Explanation:
The figure shown a PNP transistor. 1 = emitter, 2 = base, 3 = collector. (If the arrow head in opposite direction it is NPN.)
Question 4 |
The function of a shunt in an ammeter is to
A | increase the meter’s resistance |
B | by pass the current |
C | decrease the voltage drop |
D | increase the current in the coil |
Question 4 Explanation:
When you connect a low resistance (few µΩ) parallel with ( as shunt) a micro ammeter or
milli ammeter, it will be converted to an ameter. More than 99% of current bypass through the shunt. If you connect a large resistor (several kΩ ) in series it will be converted to a Voltmeter.
Question 5 |
Two resistors are connected in parallel and are connected across a 40 V battery. If each resistor is 1000 Ohms, the total battery current is
A | 80 A |
B | 80 mA |
C | 40 mA |
D | 40 A |
Question 5 Explanation:
V = I x R, therefore I = V / R, V = 40 , R = 1000/2 = 500 therefore I = 40 / 500 = 0.08 A = 80 mA
Question 6 |
A receiver with high selectivity has a
A | wide bandwidth |
B | narrow tuning range |
C | wide tuning range |
D | narrow bandwidth |
Question 6 Explanation:
When increase the selectivty , the bandwidth is decreases. The result is high selectivity has a narrow bandwith and low selectivity has a wide bandwith. Tuning range is not depend on the bandwidth.
Question 7 |
A varactor diode acts like a variable
A | capacitance |
B | voltage regulator |
C | resistance |
D | inductance |
Question 7 Explanation:
The capacitance of a varactor diode is dependant on the voltage applied to the diode.
Therefore it acts like a variable capacitor.
Question 8 |
Radio wave energy on frequencies below 4 MHz. during daylight hours is almost completely absorbed by ionospheric
A | E layer |
B | C layer |
C | D layer |
D | F layer |
Question 8 Explanation:
The D layer is formed soon after sunrise and disappears at sunset and absorbs low frequencies, roughly below 4 MHz therefor during times of darkness low frequencies go through and are not absorbed by the D-layer and is refracted by the F layer for long distant propagation.
Question 9 |
An inductor and a capacitor are connected in parallel. At the resonance frequency the resulting impedance is
A | totally inductive |
B | maximum |
C | totally reactive |
D | minimum |
Question 9 Explanation:
This is act as a parallel resonance LRC circuit.
at resonance impedance = maximum . Almost infinity. Practically this is use as traps in multi band antenna system.
Question 10 |
The effect of adding a series inductance to an antenna is to
A | have no change |
B | decrease the resonant frequency |
C | have little effect |
D | increase the resonant frequency |
Question 10 Explanation:
Consider the theory of loaded dipole.
Question 11 |
A moving coil meter by itself only responds to
A | Power |
B | AC |
C | Electric field |
D | DC |
Question 12 |
The polarization of a radio wave is defined by the direction of
A | H field |
B | Propagation |
C | receiving antenna |
D | E field |
Question 13 |
Power factor is given by
A | cos φ |
B | sin (2πf) |
C | cos (2πf) |
D | apparent power |
Question 13 Explanation:
Power Factor defined as (true power) / (Apparent power) ; If φ is the phase angle between voltage and current in an inductor then, power factor = cos φ
Question 14 |
Changes in received signal strength when sky wave propagation is used are called
A | diffraction loss |
B | sunspot |
C | modulation loss |
D | fading |
Question 14 Explanation:
In sky wave propagation signals can arrive at the receiving antenna in more than one angle of refraction. The phase difference of the multiple signals create fading. Diversity reception using two rexeivers can overcome this to a great extent.
Question 15 |
A power amplifier requires 30 mA at 300 V. The DC input power is
A | 300 W |
B | 9000 W |
C | 6 W |
D | 9 W |
Question 15 Explanation:
P = V x I, V = 300Volt, I = 30mA = 0.03 A, P = 300 x 0.03 = 9 W
Question 16 |
The following unit is used to perform rectifying operation
A | vacuum tube |
B | resistor |
C | a fuse |
D | full wave diode bridge |
Question 16 Explanation:
:- A single diode is used for half-wave rectifiers. There are two rectifier systems for Full wave
Rectifiers. 1. Diode bridge or bridge rectifier made out of 4 diodes. 2, Two diodes and a center tapped transformer is used. Rectifier vacuum diode tubes were used by old timers.
Question 17 |
An attenuator network has 10 V rms applied to it’s input with 1 V rms measured at its output. The attenuation of the network is
A | 40 dB |
B | 10 dB |
C | 6 dB |
D | 20 dB |
Question 17 Explanation:
Formula for Attenuation or gain for Voltage is 20 log(V₁ / V₂) = 20 log(10/1) = 20 dB; formula for Power :- 10 log ( P₁ / P₂)
Question 18 |
The designed output impedance of the antenna socket of most modern transmitter is normally
A | 75 Ohm |
B | 100 Ohm |
C | 50 Ohm |
D | 25 Ohm |
Question 19 |
A half-wave dipole antenna is normally fed at the point of
A | maximum resistance |
B | resonance |
C | maximum voltage |
D | maximum current |
Question 20 |
The portion of HF radiation which is directly affected by the surface of the earth is called
A | Inverted wave |
B | ground wave |
C | local field wave |
D | ionospheric wave |
Question 20 Explanation:
In HF the sky wave is refracted off the ionosphere. There is also radiation along the ground, which is attinuated as by the ground, this is more during the day while at night it could travel quite a distance
Question 21 |
A current of 200 mA flows through a lamp of 25 Ohm resistance. The voltage across the lamp is
A | 5 V |
B | 50 V |
C | 500 V |
D | 750 V |
Question 21 Explanation:
V = I x R, I = 200mA = 0.2 A, R = 25 Ohm therefore V = 0.2 x 25 = 5 V
Question 22 |
A frequency range of the “70 centimeter” band is
A | 435 to 438 MHz |
B | 430 to 450 MHz |
C | 430 to 460 MHz |
D | 430 to 440 MHz |
Question 23 |
The effective capacitance between A and B in the circuit shown
A | 0.5 C |
B | 1 C |
C | 2 C |
D | 0.1 C |
Question 23 Explanation:
Equivalent capacitance of two, 1C parallel capacitors is 2C. That 2C and the other 2C in the circuit (left upper) are in series. The effectiv capacitance of these two is equal to 1C because they are in series. This 1C and the other 1C in the circuit (lower side 1C) are in parallel and it is equal to 2C.
Question 24 |
The unit of the power is the
A | Ampere |
B | Volt |
C | Watt |
D | Ohm |
Question 24 Explanation:
Also use µW, mW, kW and MW for Power, Ohm is the unit of Resistance, Ampere is the unit of current. Volt is the unit of Potential difference or voltage.
Question 25 |
Which of the following layers tend to combine into a single layer at night ?
A | F1, F2 |
B | E, F1 |
C | D, F |
D | E, F2 |
Question 26 |
During the daytime the ionosphere is composed of
A | D, E, F₁, F₂ layers |
B | D, F₂ layers only |
C | D, F₁, layers only |
D | D, E layers only |
Question 26 Explanation:
During day time D, E, F₁ and F₂ layers are present. At night D and E dissapears and F₁ and F₂ merged as one single F layer.
Question 27 |
The unit of E field strength is
A | Ampere/meter |
B | Volts/meter |
C | Henry |
D | Ampere hour |
Question 28 |
The following meter could be used to measure the power supply current drawn by a small transistorized receiver.
A | A d.c. milli ammeter |
B | An RF ammeter |
C | A power meter |
D | An electrostatic voltmeter |
Question 28 Explanation:
:- Small transistorized receiver draws a d.c. current of about 100mA. Therefore you can use a d.c. milli-ammeter .For smaller currents we can use a micro-ammeter and for larger currents we can use an ammeter. For a.c. current we have to use an a.c. ammeter or Clip-on meter. Power meter use to measure the power in Watts. Electro static volt meter is use to measure d.c. high voltages.
Question 29 |
The following two quantities should be multiplied together to find power
A | resistance and capacitance |
B | inductance and capacitance |
C | voltage and current
|
D | voltage and inductance |
Question 29 Explanation:
P = V I and you must remember these two also. P = V² / R and P = I² R
Question 30 |
Class AB amplifier with sinusoidal wave input signal, the output current flows for
A | more than a half cycle |
B | less than a half cycle |
C | half cycle |
D | full cycle |
Question 30 Explanation:
Class-A output is full cycle and Class-B out put is a half cycle. Therefore Class-AB output is
more than a half cycle.
Question 31 |
The “S meter” on a receiver
A | indicates where the squelch control should be set |
B | indicates the standing wave ratio |
C | indicates relative incoming signal strength |
D | indicates the state of battery voltage |
Question 32 |
One megahertz is equal to
A | 0.0001 Hz |
B | 100kHz |
C | 1000 kHz |
D | 10 Hz |
Question 33 |
Power Factor of a pure inductor is
A | 1/√2 |
B | zero |
C | 1 |
D | √3/2 |
Question 33 Explanation:
If there are resistors and inductors in a circuit, we can define the Power Factor = true
power / Apperent power . This is always between 0 and 1. For a pure resistance, PF =1, for a pure inductance, PF = 0
Question 34 |
A 25 Ω resistor dissipates 4 W of power. The voltage across the resistor is
A | 100 V |
B | 10 V |
C | 12.5 V |
D | 25 V |
Question 34 Explanation:
P = V² / R , therefore V² = P R = 4 x 25 = 100, therefore V = 10
Question 35 |
Incoming signal to a superhet receiver is of 3540 kHz. Local oscillator produce a signal of 3995 kHz. The IF is tuned to
A | 455 kHz |
B | 3995 kHz |
C | 7435 kHz |
D | 3540 kHz |
Question 35 Explanation:
In super Heterodyne System, Local Oscillator signal is always higher than the incoming signal and the difference between them is equal to IF. For normal broadcast receivers IF is 455 kHz.
Question 36 |
A 10mA current is measured in a 100 Ohm resistor. The voltage across the resistor will be
A | 1 V |
B | 0.001 V |
C | 0.1 V |
D | 0.01 V |
Question 36 Explanation:
V = I x R, I = 10mA = 0.01 A, R = 100 Ohm, therefore V = 0.01 x 100 = 1 V
Question 37 |
A half-wave antenna resonant at 7100 kHz is approximately this long
A | 40 meters |
B | 80 meters |
C | 20 meters |
D | 160 meters |
Question 37 Explanation:
wave length (m) x frequency (MHz) = 300 given frequency = 7100 kHz = 7.1 MHz, therefore wave length = 300 / 7.1 = 42 m, half wave length = 20m (approx)
Question 38 |
The current through an inductor
A | can change suddenly |
B | is never zero |
C | is always zero |
D | cannot change suddenly |
Question 39 |
In a forward bias pn junction, the electrons
A | remain in the p region |
B | flow from p to n |
C | remain in n region |
D | flow from n to p |
Question 39 Explanation:
In forward bias P-N junction, current flows from P to N and electrons flow from N to P. In
the diode symbol arrow head shows the direction of current.
Question 40 |
The reactance of an inductor, increases as the
A | applied voltage increases |
B | frequency increases |
C | frequency decreases |
D | applied voltage decreases |
Question 40 Explanation:
Consider the formula X = 2 π f L, When "f" or "L" increases "X" is increasing.
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