## Fundamentals of Electricity & Radio Communications sample paper-1

*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 |

Which of the following could be attached to a moving coil meter in an attempt to measure power ?

A | a thermocouple |

B | a thermostat |

C | a thermistor |

D | a resistor |

Question 1 Explanation:

1. If you convert a moving coil meter to a voltmeter and measure the voltage across the resistor and calculate the power using P = V² / R. 2. If you convert a moving coil meter to an ammeter, measure the current through the resistor and calculate the power with using P = I² R 3. If you don't know the resistance, measure the voltage across the resistor and current through the resistor and calculate the power using, P = V I. Thermister is a special kind of resistor, it's resistance varies with the temperature. Thermocouple generates an electric current , when it is heating up. Thermostat is a switch operating according to the temperature.

Question 2 |

The Reactance of a 15 H smoothing choke at a frequency of 50 Hz is approximately

A | 750 Ω |

B | 1500 Ω |

C | 3000 Ω |

D | 4700 Ω |

Question 2 Explanation:

Apply X = 2 π f L = 2 x 3.14 x 50 x15 = 4700 Ω (approximately:- 2x3x50x15 = 4500)

Question 3 |

Transformers operates on

A | both on ac. and dc |

B | a dc. supply only |

C | an ac. supply only |

D | all the above are correct |

Question 3 Explanation:

If the current through primary winding varies, a voltage is induced on the secondary. This action continues, only while an A.C. supply connected to the primary.

Question 4 |

When constant D. C. voltage applied to capacitor, it is acts as

A | Zero resistor |

B | A finite resistor |

C | Current source |

D | A infinite resistor |

Question 4 Explanation:

Electric charges do not pass through a capacitor. Therefore it acts as a infinite resistor.

Question 5 |

A varactor diode acts like

A | a variable inductor |

B | a variable regulator |

C | a variable capacitor |

D | a variable resistor |

Question 5 Explanation:

Variable capacitor is a tuning condenser or a varactor diode; Variable resistor is a potentiometer or variac or reostat; There is no meaning on variable regulator because regulator is a constant supply.

Question 6 |

The symbol in the figure is that of a

A | Switching diode |

B | varactor diode |

C | Light Emitting Diode (LED) |

D | Zener Diode |

Question 7 |

The effective resistance between A and B in the circuit shown is

A | 225 Ω |

B | 55Ω |

C | 115Ω |

D | 60Ω |

Question 7 Explanation:

Two resistors of 110 Ohm are in parallel. It is equivalent to 55 Ohms. This 55 Ohm is connected in series with 5 Ohms. This is equivalent to 60 Ohms.

Question 8 |

What is the power consumed by a transmitter taking 1.5 A at 12V ?

A | 1.5W |

B | 12W |

C | 8W |

D | 18W |

Question 8 Explanation:

apply the formula of, P = V I, V = 12V, I = 1.5 A, therefore, P = 12 x 1.5 = 18 Watt

Question 9 |

Resistors of 100Ω and 150 Ω are connected in parallel the effective value is

A | 100 Ω |

B | 60 Ω |

C | 150 Ω |

D | 250 Ω |

Question 9 Explanation:

For two parallel resistors, Equivalent resistance is R = R₁ R₂ / (R₁ + R₂), therefore R = 100 x 150 / ( 100+150) = 100 x 150 / 250 = 60 Ω

Question 10 |

The effective capacitance between A and B is

A | 100 pF |

B | 24 pF |

C | 55 pF |

D | 45 pF |

Question 10 Explanation:

parallel connection of 33pF and 27pF is equivalent to 60 pF ( 33+27 =60), same way other two equivalent to 40 pF. Then these 60pF and 40 pF connected as series. Their equivalent value is 60x40 / (60+40) =24pF.

Question 11 |

The conductivity of a current carrying conductor can be increased by

A | none of the above |

B | adding a resistor |

C | increasing its length |

D | increasing its diameter |

Question 11 Explanation:

Consider R = ρ l / A formula, R = resistance, ρ = resistivity of the material, l = length of
conductor, A = cross section of the conductor. The conductivity is increased by the decrease of the resistance, R. For decrease R; ρ and l should decrease or A should increase.

Question 12 |

To minimise unwanted radiation of sub-harmonics and harmonics, a VHF transmitter should be followed by

A | a low pass filter |

B | a high pass filter |

C | a band pass filter |

D | a notch filter |

Question 12 Explanation:

Low pass filter is filtering only higher frequencies and it allow to pass low frequencies. Band pass filter allow to pass all frequencies within the specified frequency range. All lower frequencies and all higher frequencies get cut off. High pass filter stops all lower frequencies and allows to passing of higher values only. Notch filter cuts only one perticular frequency.

Question 13 |

To check that a crystal is working on its correct overtone the simplest piece of equipment is

A | a Voltmeter |

B | a dip oscillator |

C | an ammeter |

D | an absorption wave meter |

Question 13 Explanation:

Voltmeter and Ammeter can't do anything on this. An absorption wave meter can use only if there is enough power output. But dip meter (dip oscillator or GDO) can measure the frequency in the oscillator circuit. It operates by extracting a very small amount of energy from the oscillating circuit to be measured.

Question 14 |

The typical accuracy of a dip oscillator might be

A | 10% |

B | 1% |

C | 0.001% |

D | 0.05% |

Question 14 Explanation:

From a dip meter, frequency is measured with an analog dial. Error is not more than ±10% . It operates by extracting a very small amount of energy from the oscillating circuit to be measured.

Question 15 |

Which one of the following frequencies is in the VHF range ?

A | 5MHz. |

B | 25 MHz. |

C | 250 MHz. |

D | 950 MHz. |

Question 15 Explanation:

VLF (very low frequency) = 10 to 30 kHz. ; LF (low frequency or long wave) = 30 to 300 kHz. ; MF (medium frequency or medium wave) = 300 to 3000 kHz. ; HF (high frequency or short wave) = 3 to 30 MHz. ; VHF (very high frequency) = 30 to 300 MHz. ; UHF (ultra high frequency) = 300 to 3000 MHz. ; SHF (super high frequency) = 3 to 30 GHz. ; EHF ( extra high frequency or micro wave) = 30 to 300 GHz.

Question 16 |

A transformer has a coil 8Ω resistance and a reactance of 6Ω The impedance is

A | 8Ω |

B | 10Ω |

C | 6Ω |

D | 12Ω |

Question 16 Explanation:

Use the formula, Z² = X² + R², Therefore Z² = 6² + 8² = 36 + 64 = 100, therefore Z = √100 = 10 Ohm.

Question 17 |

As the frequency rises the reactance of a capacitor

A | stays constant |

B | none of these |

C | decreases |

D | increases |

Question 17 Explanation:

Apply the following formula for the reactance of capacitor, X = 1 / 2 π f C, When "f" increasing "X" is decreasing.

Question 18 |

The typical accuracy of a moving coil meter is

A | 3% |

B | 0.03% |

C | 10% |

D | 0.3% |

Question 18 Explanation:

The question says accuracy, but it would have been better to say the error of the moving coil meter is about 3%. That means accuracy is 97%.

Question 19 |

The circuit is shown in the figure for

A | voltage multiplication |

B | full wave rectification |

C | half wave rectification |

D | reverse bias protection |

Question 19 Explanation:

As shown in the diagram, when a diode connected in series with an a.c. supply it’s a half wave rectification. If you connect a bridge rectifier with the a.c. supply it is Fullwave rectification. If a diode connected with a d.c. supply as forward bias, it is a reverse bias protection. That means, if you connect a d.c. supply with wrong polarity by mistake, protect your equipment. There is a special system to connect few diodes and capcitors with an a.c. supply you can multiply the voltage accordingly.

Question 20 |

The input power of a transmitter stage running at 24V, 2.5A is

A | 150 W |

B | 600 W |

C | 60 W |

D | 300 W |

Question 20 Explanation:

Apply P = V I formula. Therefore, P = 24 x 2.5 = 60W

Question 21 |

A moving coil meter depends on which of the following in order to operate

A | the interaction of two permanent magnetic fields |

B | an electric only |

C | interaction of an electric and magnetic field |

D | interaction of a permanent and electromagnetic field |

Question 21 Explanation:

The moving coil generates an electro-magnet ic field, while a current flow through the coil. This coil installed on a permanent magnetic field. This coil is turning (moving) due to the interaction between those fields.

Question 22 |

The maximum current that may be safely passed through a 10000Ω resistor rated at 25 W is

A | 1 A |

B | 0.005 A |

C | 0.05 A |

D | 0.5 A |

Question 22 Explanation:

Use the formula, P = I² R , P = 25W, R = 10000Ohm, Therefore I² = P / R = 25 / 10000.
Therefore I = √( 25 / 10000) = 5 / 100 = 0.05 A or 50 mA

Question 23 |

For a silicon transistor the base emitter voltage for biasing must be above

A | 0.8V |

B | 0.7 V |

C | 1.0 V |

D | 0.3 V |

Question 23 Explanation:

For silicon transistors, base-emitter bias voltage must be above 0.7 V. For germanium transistors, it is above 0.3 V. Same principle apply for diodes also.

Question 24 |

The most widely used conducting materials are

A | copper and gold |

B | copper and aluminum |

C | copper and silver |

D | gold and silver |

Question 24 Explanation:

The best conductor is Silver, next copper, next gold , next aluminium. But silver and gold are too expensive for common use. That is why copper and aluminium are commonly used.

Question 25 |

An SWR meter is inserted into a perfectly matched transmitter/antenna system the value shown should Indicate

A | 1:1 SWR |

B | 10 W reflected power |

C | 1:0 SWR |

D | 0:1 SWR |

Question 25 Explanation:

If SWR equal to 1 (or 1:1) the antenna system is perfectly matched. If it is totally mismatched, SWR = infinity, If it is less than 2, it is practically acceptable. If it is 3, only 50% of power will radiate. Always SWR exist between 1 and infinity.

Question 26 |

There are two basic forms of transistors, these are

A | NPP and PNN |

B | PNP and NNP |

C | PNP and NPN |

D | PPN and NNP |

Question 27 |

A good dummy load is constructed from

A | light bulb |

B | a column |

C | non-reactive resistors |

D | wire-wound resistors |

Question 27 Explanation:

A good dummy load must have a pure resistance, without having a reactance. Only then will it have a 1:1 SWR. Wire wound resistors or electric bulbs have some reactance. But in practically, testing an amateur TX having about 50 Watts we can use a filament type bulb (100 W 230 V) as a dummy load.

Question 28 |

A light bulb is rated at 12V , 3W. The current drawn when used on a 12V source is

A | 36 A |

B | 750 mA |

C | 4 A |

D | 250 mA |

Question 28 Explanation:

Use the formula of P = V I, P = 3W, V = 12V, therefore I = P/V = 3/12 = 0.25 A = 250 mA

Question 29 |

The current at the center of a given λ/2 antenna is found to be 0.5 A. If this antenna has a radiation resistance of 70 Ω find the radiated power.

A | 70 W |

B | none of these |

C | 17.5 W |

D | 50 W |

Question 29 Explanation:

Apply P = I² R formula, I = 0.5 A, R = 70Ω Therefore P = 0.5 x 0.5 x 70 = 17.5 W

Question 30 |

The output signal of a balanced modulator is

A | FM |

B | AM |

C | DSB |

D | SSB |

Question 30 Explanation:

SSB transmitter produces a DSB (Double Side Band) signal from the balanced modulator and passes through the side band filter for USB or LSB signal and the non selected side band gets filtered out.

Question 31 |

Volt is a measuring unit of

A | Resistance |

B | Current |

C | Electrical potential |

D | Power |

Question 31 Explanation:

The unit of Power is Watt. The unit of potential or voltage is Volt. The unit of current is Ampere. The unit of resistance is Ohm.

Question 32 |

The tolerance of a resistor is given as 10%. If the nominal value is 4700Ω, then its value must b be lie Between

A | 4656 and 5170 Ω |

B | 4653 and 4747 Ω |

C | 4230 and 5170 Ω |

D | 4230 and 4747 Ω |

Question 32 Explanation:

If the tolerance is 10% , The value is 4700±10%. 10% of 4700 is 470. Therefore 4700±10% is equals to 4700±470. Therefore 4700+ 470 = 5170 and 4700 - 470 = 4230.

Question 33 |

In the smoothing circuit of a power supply capacitors of 8 μF, 4 μF and 2 μF are connected in parallel, the effective capacitance is

A | 8 μF |

B | 8/7 μF |

C | 4 μF |

D | 14 μF |

Question 33 Explanation:

Formula for the equivalent value of parallel capacitors is C = C₁ + C₂ + C₃ = 8+4+2 =14 μF

Question 34 |

3GHz is equivalent to

A | 3000 MHz |

B | 300 MHz |

C | 30 MHz |

D | 30000 MHz |

Question 34 Explanation:

1 GHz = 1000 MHz

Question 35 |

What is the power consumed by a transmitter taking 0.5A at 24V ?

A | 2.4W |

B | 4.8W |

C | 18W |

D | 12W |

Question 35 Explanation:

Apply P = V I, V = 24 Volt, I = 0.5 A, therefore P = 24 x 0.5 = 12 W

Question 36 |

Which of the instrument below has highest accuracy ?

A | A digital frequency counter |

B | An oscilloscope |

C | A heterodyne wave meter |

D | An absorption wave meter |

Question 36 Explanation:

A good frequency counter is accurate as ± 1 Hz. When using an Oscilloscope we have to measure the wave length according to scale on the screen an calculate the frequency. The error will be about 10%. An absorption wave meter has an analog dial, it operates by extracting a small amount of energy from the oscillating circuit to be measured. Because of that the error will be more ( about 10%) .

Question 37 |

Which of the following represents 100% amplitude modulation ?

A | (a) |

B | (b) |

C | (d) |

D | (c) |

Question 38 |

The unit of resistance is

A | Ohm |

B | Ampere |

C | Farad |

D | Watt |

Question 38 Explanation:

The unit of resistance is Ohm. The unit of current is Ampere. The unit of capacitance is Farad. The unit of power is Watt.

Question 39 |

For ideal amplitude modulation the modulating index must be

A | greater than one |

B | smaller than one |

C | zero |

D | unity |

Question 39 Explanation:

If modulating index = 1, it is 100% modulated. If it slightly increased, it is over modulated. Then signal becomes distorted. Therefore always modulating index < 1

Question 40 |

Ground wave communication is most effective in the frequency range of

A | 300 kHz to 3 MHz |

B | 3 MHz to 30 MHz |

C | above 300 MHz |

D | 30 MHz to 300 MHz |

Question 40 Explanation:

Ground wave propagation occurs on MF (300kHz to 3MHz) and HF (3 to 30MHz) only. Above 30 MHz, propagate mostly with line of sight.

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