## Fundamentals of Electricity & Radio Communications sample paper-3

*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 amateur radio transmitter/ antenna system has an ERP of 100 W. If the antenna gain is 10 dB, transmitter out put power is

A | 1000 W |

B | 100 W |

C | 10 W |

D | 1 W |

Question 1 Explanation:

ERP = Effective Radiated Power, apply the formula,
Power gain (dB) = 10 Log( P₁ / P₂), Power gain = 10 dB,
P₁ = 100 W and find P₂ , therefore 10 = 10 Log (100 / P₂),
therefore 1 = Log (100 / P₂), Log 10 = Log (100 / P₂), 10 = 100 / P₂ therefore P₂ = 10

Question 2 |

The term “PTT” means

A | Phase testing terminal |

B | Press to talk |

C | Phased transmission transponder |

D | Piezo-electric transducer transistor |

Question 3 |

For best reception, the S/N ratio should be

A | high |

B | zero |

C | none of these |

D | low |

Question 3 Explanation:

If the signal level is high and noise level low, that means S/N is high and you can have a good reception, If S/N is low the reception will be poor.

Question 4 |

A | 150Ω |

B | 250Ω |

C | 260Ω |

D | 160Ω |

Question 4 Explanation:

First you must know the resistor color code:- 0 = black, 1 = brown, 2 = red, 3 = orange, 4 = yellow, 5 = green, 6 = blue, 7 = violet, 8 = gray, 9 = white; In this question, 1st color band is red represent 2 and the 2nd green represents 5 and 3rd brown represents 1 and it’s the multiplyer. It is equal to 10¹ . So the value is 250. If it is Orange, blue, green , the value is 3600000 Ω or 3.6 MΩ.

Question 5 |

The frequency range 300 kHz to 30 Mhz; includes

A | LF, MF ranges |

B | MF, HF ranges |

C | UHF, VHF ranges |

D | HF, VHF ranges |

Question 5 Explanation:

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 6 |

In a class B amplifier with a sinusoidal input signal the output current flows for a

A | full cycle |

B | Half cycle |

C | ¾ of cycle |

D | Quarter cycle |

Question 6 Explanation:

Class A is Full cycle, class B is half cycle, class C is Quater or less than half cycle. In between there are some little bit different classes like, class AB, AB1 ect.

Question 7 |

The reflection coefficient of an open circuited transmission line is

A | -1 |

B | +1 |

C | zero |

D | infinity |

Question 7 Explanation:

Reflection coefficient = sqr(Ref.pwr / Forw. Pwr) = Ref. Voltage/Forw. Voltage, ;Perfectly matched antenna has no reflected power, so Reflected coefficent = 0, Total power is rflect in perfectly mismatch antenna , so Reflected coefficent = 1, Always reflected coefficient having a value between 0 and 1. It will not take a value of -1 or infinity.

Question 8 |

The total inductance between A and B in the circuit shown is

A | 1L |

B | 10L |

C | 2L |

D | 5L |

Question 8 Explanation:

As same as Resistances. Parallel two inductors having 4L (lower side) is equivalent to 2L, This 2L and other 2L are parallel. It is equivalent to 1L. This 1L is in series with other 3L and L. It is equivalent to 5L.

Question 9 |

A dc voltmeter can be used to measure

A | RMS value |

B | Power factor |

C | Power |

D | Polarity |

Question 9 Explanation:

If you connect a dc meter with correct polarity, indicator is moving correct way ( clockwise). If it is on wrong polarity indicator slightly move wrong way (anti clockwise) and getting stuck. Power or power factor or RMS value canot measure with a dc voltmeter.

Question 10 |

The total resistance between A and B is the circuit shown is,

A | 5R |

B | 3R |

C | R |

D | 2R |

Question 10 Explanation:

three resistore of 3R are connected in parallel and equivalent to R, two resistors of 2R are also connected in parallel and equivalent to R. these two resistors of R are in series having a value of 2R.

Question 11 |

Power advantage of SSB over AM is

A | 3 : 1 |

B | 4 : 1 |

C | 4 : 3 |

D | 3 : 4 |

Question 11 Explanation:

When 100W of SSB signal and 400 W AM signal are comming from the same station can receive approximately with equal strength.

Question 12 |

A beat frequency oscillator (BFO) is used in the demodulation of

A | SSB signal |

B | PM signal |

C | FM signals |

D | AM signals |

Question 12 Explanation:

BFO use to demodulate SSB and CW signals.

Question 13 |

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

A | 5Ω |

B | 30Ω |

C | 10Ω |

D | 47Ω |

Question 13 Explanation:

Use the formula 1/R = 1 / R₁ + 1 / R₂ + 1 / R₃, ; 1 / R = 1/12+1/15+1/20 = 5/60+ 4/60+ 3/60 = 12/60 = 1/5, therefore R = 5

Question 14 |

Capacitors of 8μF, 4μF and 2μF are connected in Parallel. What is the effective
Capacitance?

A | 14 F |

B | 1.14μF |

C | 1.5μF |

D | 14μF |

Question 14 Explanation:

Apply C = C₁ + C₂ + C₃ for parallel capacitors, therefore C = 8 + 4 + 2 = 14

Question 15 |

Two inductors of 10 μH & 20 μH are connected in series; two others of 30 μH &
40μH are also connected in series. What is the equivalent inductance if these series
combinations are connected in parallel?

A | 23.8μH |

B | 21μH |

C | 100μH |

D | 20μH |

Question 15 Explanation:

Series combination of 10 μH and 20 μH is equivalent to 30 μH. Same way other two equivalent to 70 μH. Parrallel combination of 30 μH and 70 μH is equivalent to 30 x 70 / ( 30+70) = 2100/100 = 21 μH.

Question 16 |

At what frequency do a capacitor of 100pF & an inductance of 100μH resonance?

A | 1.6 kHz |

B | 1.6 MHz |

C | 3.2 MHz |

D | 3.2 kHz |

Question 16 Explanation:

Apply the formula, f = 1 / {2π √(LC)} therefore
f = 1 / {2π √(100x10ˉ⁶ x 100x10-12)}
= 1 / 2π x 10-7
= (10/2π) x 10-6 Hz
= (10/2π) MHz = 1.6 MHz

Question 17 |

The current flow through the 27 Ω resistor has a value of

A | 60 mA |

B | 27 mA |

C | 33 mA |

D | 100 mA |

Question 17 Explanation:

27Ω and 33 Ω are in series. It is equivalent to 60 Ω. The current flow through this 60Ω is equal to the current flow through the 27Ω resistor.
Apply V = I R to that 60 Ω, V=6, R=60, therefore 6 = I x 60 , therefore I = 6/60 = 0.1 A = 100 mA

Question 18 |

A loudspeaker speech coil has an impedance of 4Ω. If the voltage across it is 4 V, then the power in the speech coil is

A | 4W |

B | 9W |

C | 6W |

D | 3W |

Question 18 Explanation:

An AC current passing through the speech coil.
Therefore the power is equal to VI or I² Z or V² / Z , V = 4Volt, Z = 4Ω,.
Therefore Power = 4² / 4 = 4 Watt

Question 19 |

Ohm is a unit of

A | Capacitance |

B | Resistance |

C | Inductance |

D | Frequency |

Question 19 Explanation:

Ohm is the unit of resistance, Henry is the unit of Inductance, Farad is the unit of capacitance, Hertz is the unit of frequency.

Question 20 |

The peak value of the 230V, 50Hz mains supply is

A | 230V |

B | 230 / √2 V |

C | 230√2 V |

D | 2 x 230√2 V |

Question 20 Explanation:

Peak value = √2 x RMS value or RMS value = (1/√2) x Peak value.

Question 21 |

If the length of a current carrying conductor is doubled, the resistance will become

A | one fourth |

B | same |

C | double |

D | half |

Question 21 Explanation:

Apply the formula R = ρ l / A, for a conducter. “ ρ “ is the resistivity of the material, “ l “ is the length and “ A “ is the aera of the cross section. According to the formula R is doubled, if length doubled or A become half. “ ρ “ is constant for a particular material.

Question 22 |

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

A | 6 Ω |

B | 8 Ω |

C | 2 Ω |

D | 4 Ω |

Question 22 Explanation:

Two series resistors of 4Ω equivelent to 8Ω. This one and other 8 Ω resistor connected in parallel and equivalent to 4Ω. This 4Ω combined in series with two resistors of 2Ω. It is equivalent to 8Ω.

Question 23 |

A | 2 A |

B | 3/2 A |

C | 1 A |

D | 1/2 A |

Question 23 Explanation:

First calculate the equivalent resistance between A and B. Two resistors of 2 Ω (Right hand and upper side) are equivalent to 4Ω. This one is parallel with othe 4Ω in the circuit. This parallel combination equivalent to 2Ω and it is connected in parellel with other 2Ω (middle one) and equivalent to 1Ω. This one in series with remain 2Ω and the final value is 3Ω. Then apply Ohm’s law to this 3Ω resistor. V = 3 Volt, R = 3Ω. Therefore I = V / R = 3/3 = 1

Question 24 |

The two basic forms of transistors are

A | PPP and NNN |

B | PNP and NPN |

C | PPN and NNP |

D | PNN and NPP |

Question 24 Explanation:

There are no semiconductors like PPN, NNP, PNN, NPP, PPP, NNN.

Question 25 |

What is the total capacitance between the points A and B?

A | C |

B | ⅔C |

C | 4C |

D | 5C |

Question 25 Explanation:

Two capacitors equal to C are in parallel and they are equivalent to 2C. This 2C connected in series with other two capacitore having a value of 2C and it is equivalent to 2C/3.

Question 26 |

The total inductance between A and B in the circuit shown is

A | H |

B | 2H/3 |

C | 2 H |

D | 3H/2 |

Question 26 Explanation:

Two inductors having 2H are in parallel. They are equivalent to H. This is in series with other H and equivalenet to 2H.

Question 27 |

A half wave antenna is resonant at 100MHz. It’s approximate length will be

A | 4.5m |

B | 3m |

C | 6m |

D | 1.5m |

Question 27 Explanation:

apply the formula f (MHz) x wave length (m) = 300 , (speed of light = f λ).
Therefore 100 λ =300, λ = 300/100 = 3, λ/2 = 3/2 = 1.5 m

Question 28 |

A 400Ω resistor dissipates a power of 0.25W. The current flowing across the resistor is

A | 0.025A |

B | 0.25A |

C | 0.028A |

D | 0.0013A |

Question 28 Explanation:

Apply P = I² R , P = 0.25 W, R = 400Ω.
Therefore 0.25 = I² x 400, therefore I² = 0.25 / 400 = 25 / 40000
And I = 5/200 A = 25 mA

Question 29 |

A 3dB power gain is equivalent to an increase of gain by

A | 10 times |

B | 3 times |

C | 2 times |

D | 30 times |

Question 29 Explanation:

Apply the formula power gain (dB) = 10 Log(P₁ / P₂)
Therefore 3 = 10 Log (P₁ / P₂) ,
therefore 0.3 = Log (P₁ / P₂) ,
Log 2 = 0.3010, approximatly Log 2 = 0. 3 , therefore Log 2 = Log(P₁ / P₂) ,
Therefore 2 = (P₁ / P₂) , double the power.

Question 30 |

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

A | 60 W |

B | 24 W |

C | 48 W |

D | 72 W |

Question 30 Explanation:

Apply P = VI therefore P = 24 x 2.5 = 60

Question 31 |

The radiation resistance of a folded dipole antenna is

A | 75 Ω |

B | 100 Ω |

C | 50 Ω |

D | 300 Ω |

Question 31 Explanation:

Naturally it is 300Ω, Simple dipole is 75Ω

Question 32 |

The wavelength of a signal at 60 MHz in free space is

A | 5m |

B | 0.5m |

C | 50m |

D | 10m |

Question 32 Explanation:

apply the formula f (MHz) x wave length (m) = 300 , (speed of light = f λ).
Therefore 60λ =300 , λ = 300/60 = 5 m.

Question 33 |

As the frequency rises, the reactance of an inductor

A | decreases |

B | none of these |

C | stays constant |

D | increases |

Question 33 Explanation:

Apply the formula X = 2 π f L, If f increases X also increase.

Question 34 |

Which one is incorrect from following statments

A | Power factor of a pure inductor is zero |

B | Power factor = actual power / apperent power |

C | Power Factor of a pure resistor is 1 |

D | Power factor of a capacitor is infinity |

Question 34 Explanation:

Power factor lying between 0 and 1. For a capacitor PF = 0

Question 35 |

The power dissipation of the 1Ω resistor of the circuit shown is

A | 250 mW |

B | 2 W |

C | 500 mW |

D | 1 W |

Question 35 Explanation:

The effective value three 3Ω resistors is 1Ω, because they are in parallel. This 1Ω is connected in series with the other 1Ω resistor in the circuit and getting the equivalent value of 2Ω.
Therefore the power dissipation is = V² / R = 2² / 2 = 2 W

Question 36 |

The symbol shown indicates a

A | npn transistor |

B | diode |

C | pnp transistor |

D | field effect transistor |

Question 37 |

The conductivity of a current carrying conductor can be decreased by

A | reducing its length |

B | reducing its temperature |

C | none of the above |

D | reducing its diameter |

Question 37 Explanation:

apply the formula R = ρ l / A, ρ is the resistivity of the material, it is a constant. If the conductivity decrease, Resistance will be increased. When you consider this formula, R increased, if “ l “ increase or “A” decrease. “A” decrease mean reducing the diameter.

Question 38 |

The output signal of a balanced modulator is

A | SSB |

B | FM |

C | DSB |

D | AM |

Question 38 Explanation:

Balance modulater giving a DSB signal. When it is passing through the side band filter , output will be USB or LSB accordingly.

Question 39 |

When 2V e.m.f. applied across an 1F. capacitor, the energy stored in the capacitor is

A | 1 J |

B | 2 J |

C | 4 J |

D | 8 J |

Question 39 Explanation:

Use the formula E = ½ C V², C = 1 F, V = 2 volts , therefore E = ½ x 1 x 2² = 2 Jule

Question 40 |

A coil has a resistance of 3Ω and an inductive reactance of 4 Ω. The impedance of the coil is

A | 5 Ω |

B | 21 Ω |

C | 3 Ω |

D | 7 Ω |

Question 40 Explanation:

Use the formula Z² = R² + X² , R = 3 Ω, X = 4 Ω,
therefore Z² = 3² + 4² = 9 + 16 = 25 ,
therefore Z = 5 Ω

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