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An antenna has a radiation resistance of 72 Ω a loss resistance of 8 Ω and a power gain of 16. Find efficiency and directivity.
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An antenna has a radiation resistance of 72 Ω a loss resistance of 8 Ω and a power gain of 16. Find efficiency and directivity.

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Description :

Define the following terms related to antennas; (i) Antenna resistance (ii) Directivity (iii) Antenna gain (iv) Power density

Answer :

Antenna Resistance - The resistance of an antenna has two components: 1. Its radiation resistance due to conversion of power into electromagnetic waves 2. The resistance due to actual losses in the ... the transmitter power divided by the surface area of a sphere (4πR2) at that distance.

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Description :

Define the following terms: (i) Polarization (ii) Antenna gain (iii)Antenna resistance (iv)Directivity

Answer :

i) Polarization:- It is defined as the direction of electric field vector in the EM wave radiated by the transmitting antenna. ii) Antenna Gain:- Antenna gain is defined as the ratio of ... in only one direction in which the radiation is maximum. That is directivity = Max. directive gain

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Description :

Describe with respect to antenna (i) radiation pattern (ii) directive gain (iii) power gam (iv) polarization

Answer :

(i) Radiation pattern:-A graph or diagram which tells us about the manner in which an antenna radiates more power in different directions is known as the radiation patteren of antenna.  ( ... as the direction of the electric vector in the electromagnetic wave radiated by the transmitting antenna. 

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Description :

When a dipole antenna of 1/8 length has an equivalent total resistance of 1.5 Watt then the efficiency of the antenna is   (A) 0.89159% (B) 8.9159% (C) 89.159% (D) 891.59%

Answer :

When a dipole antenna of 1/8 length has an equivalent total resistance of 1.5 Watt then the efficiency of the antenna is 89.159%

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Description :

Explain loop antenna with neat sketch. Draw radiation pattern. State its advantages and applications.

Answer :

Loop antenna:-The single turn coil carrying RF current through it having length less than the wavelength.   Advantages:- 1. highly directive 2. Small size Applications:- 1. For direction finding 2. In portable receivers 3. In navigation 

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Description :

Draw the structure of horn antenna and its radiation pattern. List its any two applications.

Answer :

The structure of horn antenna Radiation pattern of Horn Antenna Application:- i) Used at microwave frequency. ii) Used in satellite tracking 

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Description :

Draw radiation pattern for following antenna i) Yagi-Uda antenna ii) Loop antenna iii) Dish antenna iv) Horn antenna

Answer :

Type of antenna Radiation Patteren  Yagi-Uda antenna Loop antenna Dish antenna Horn antenna 

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Description :

Explain half dipole antenna ( Resonant antenna ) with its radiation patteren. 

Answer :

Half wave dipole antenna Explanation:  1. It is a resonant antenna 2. It is exact half wavelength (λ /2) long & open circuited at one end. 3. The dipole antennas have ... pattern is bidirectional.  The radiation pattern of half wave dipole antenna is - The radiation pattern

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Description :

Draw Yagiuda antenna with its radiation pattern.

Answer :

Construction of Yagi-Uda Antenna Radiation Pattern of Yagi-Uda Antenna

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Description :

Draw the radiation pattern for Dipole antenna: (i) Half wave dipole (ii) Folded dipole.

Answer :

The radiation pattern for Half wave dipole antenna The radiation pattern for Folded dipole antenna.

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6 views 1 answer
Description :

Dish antenna is parabolic in shape and has meshy structure. Give reasons.

Answer :

A Practical reflector employing the properties of the parabola will be a three dimensional bowl-shaped surface, obtained by revolving the parabola about the axis AB. The resulting geometric ... (directrix) is constant. These geometric properties yield an excellent microwave or light reflector. 

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Description :

The gain of a parabolic dish antenna of 1 m diameter -and 80% efficiency at 9.5 GHz is: a. 39 dB b. 78 dB c. 40 dB d. 45 dB

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Description :

Define and explain the term beam width related to antenna with a sketch.

Answer :

Definition: The beam width of an antenna is described as the angles created by comparing the half power point (3dB) on the main radiation lobe to its maximum power point. As an example the beam width ... max voltage at center of lobe (these point are known as half power points.) Sketch-  

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Draw the structure and state applications of: i) Ferrite loop (rod) antenna ii) Horn antenna

Answer :

Horn antenna: Application:- i) Used at microwave frequency. ii) Used in satellite tracking.  Ferrite loop antenna: Application:- In Am radio receiver to receive MW and SW band signals. In FM radio receiver

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Description :

Draw construction of Yagi-Uda antenna and explain.

Answer :

Explanation A Yagi-Uda antenna, commonly known as a Yagi antenna, is a directional antenna consisting of multiple parallel elements in a line, usually half-wave dipoles made of ... receiver with a transmission line and additional parasitic elements called reflector and one or more directors.

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14 views 1 answer
Description :

Explain how modulation reduces height of antenna and avoid mixing of signals.

Answer :

Modulation reduces antenna height: For the transmission of radio signals, the antenna height must be multiple of λ/4 ,where λ is the wavelength . λ = c /f where c : is the ... will occupy different slots in the frequency domain (different channels). Thus, modulation avoids mixing of signals.

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Description :

The parameters of Transmission line are R = 50 Ω/ km, L= 1mH/km ,C = 0.1µf/km, G = 2µV/km. calculate characteristic impedance.

Answer :

The parameters of Transmission line are R = 50 Ω/ km, L= 1mH/km ,C = 0.1µf/km, G = 2µV/km. calculate characteristic impedance.

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Description :

The ratio of the monthly salaries of A and B is in the ratio 15 : 16 and that of B and C is in the ratio 17 : 18. Find the monthly income of C if the total of their monthly salary is Rs 1,87,450. A) Rs 66,240 B) Rs 72,100 C) Rs 62,200 D) Rs 65,800 E) Rs 60,300

Answer :

Answer: A A/B = 15/16 and B/C = 17*18 So A : B : C = 15*17 : 16*17 : 16*18 = 255 : 272 : 288 So C’s salary = [288/(255+272+288)] * 1,87,450= Rs 66,240

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Description :

An a.c series circuit has resistance of 10ohm, inductance of 0.1H and capacitance of 10µf, voltage applied to circuit is 200V. find (i) Resonant frequency (ii) Current at resonance (iii) Power at resonance 

Answer :

Solution: For RLC series circuit

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Description :

A superheterodyne radio receiver with an IF of 455KHZ is turned to 1000KHZ. Find: (i) Image frequency (ii) Local oscillator frequency

Answer :

Given Intermediate Frequency fi=455KHz Signal frequency =fs=1000KHz Local oscillator frequency fo=fs+fi  Fo=1000KHz+455KHz  =1455KHz Image frequency is the input frequency which produces the same intermediate frequency fsi=fs+2fi  =1000KHz+2*455KHz  =1910KHz 

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Description :

If resistance is 8 Ohm and current is 16 ampere then what will be the voltage and power?

Answer :

128 volts ,2048 watts

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Write one application of following antenna i. Rectangular antenna ii. Dish antenna iii. Yagi-Uda antenna iv. Horn antenna 

Answer :

i. Rectangular antenna is used in direction finding in portable recievers. ii. Dish antenna is used to transmit and receive signal from satellite. iii. Yagi-Uda antenna is used in HF and VHF range as a TV receiving antenna. iv. Horn antenna is used in satellite tracking.

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For an earth station transmitter, with an antenna output power of 40dBW 10,000W, a back-off loss of 3dB, a total branching and feeder loss of 3dB, a total branching and feeder loss of 3dB and transmit antenna gain of 4dB, the effective isotropic radiated power (EIRP will be  a) 38dBW b) 40 dBW c 36 dBW d) 47 dBW

Answer :

For an earth station transmitter, with an antenna output power of 40dBW 10,000W, a back-off loss of 3dB, a total branching and feeder loss of 3dB, a total branching and feeder loss of 3dB and transmit antenna gain of 4dB, the effective isotropic radiated power (EIRP will be 38dBW

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How many kg of dal at Rs.8.40 per kg be mixed with32 kg of dall at Rs10.20 per kg to get a mixture worth Rs.9.60 per kg? a) 23 b) 49 c) 43 d) 72 e) 16

Answer :

E By the rule of allegation ,we have Quantity of 1st kind dall : Quantity of 2nd kind dall = 60 : 120  => Quantity of 1st kind dall:32 = 1:2  Quantity of 1st kind dall =32 *1/2 = 16

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Description :

 Find the speed and average speed of a bus which leaves salem at 4 p.m. and reaches madurai in the same day at 8 p.m. The distance between the two stations is 216 km and the total time for stoppage is 1 hour between these stations. a) 54,72 b) 62,84 c) 72,54 d) 82,64

Answer :

C Total time taken = 4 hours; Time of stoppage = 1 hour, that is, actual time taken = 4 hours - 1 hours = 3 hours Speed = Distance/Time = 216/3 = 72 km/hr Average speed = Total Distance/ Total Time = 216/4 = 54 km/hr

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Description :

To find current in a resistance connected in a network, Thevenin's theorem is used VTH = 20 V and RTH = 5 Ω. The current through the resistance: (1) is 4 A (2) is 4 A or less (3) is less than 4 A (4) May be 4 A or less or more than 4 A

Answer :

To find current in a resistance connected in a network, Thevenin's theorem is used VTH = 20 V and RTH = 5 Ω. The current through the resistance: is less than 4 A 

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Description :

Each branch of Y-connected load has resistance of 10 Ω. The resistance of each branch of an equivalent ∆-connected load will be   (a) 30 Ω (b) 100 Ω (c) 110 Ω (d) None of these

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Description :

A vendor sells a radio at Rs. 1680 at a gain of 40% and another for Rs. 1920 at the loss of 8%. Find his total gain percent

Answer :

So, C.P. of 1st radio= (100/140∗1680)=1200 C.P. of 2nd radio= (100/92∗1920)=2086.95 Total C.P. = 3286.95 Total S.P. = 3600 Gain = 3600 – 3286.95 = 313.05 Gain% = 9.5%

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Description :

A substance is purchased for Rs. 300. If one fourth of the substance is sold at a loss of 10% and the remaining at a gain of 5%, Find out the overall gain or loss percentage.

Answer :

Price Received by selling one fourth of the substance at a loss of 10% = (1/4) * 300 * (90/100) = Rs. 67.5 Price Received by remaining substance at a gain of 5% = (3/4) * 300 * (105/100) = Rs.236. ... 75 Profit = 303.75 - 300 = 3.75 Profit%=(Gain/Cost∗100)% =(3.75/300∗100)% =1.25%

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Description :

Dharan bought a donkey and a carriage for Rs.3000. He sold the donkey at a gain of 20% and the carriage at a loss of 10%, thereby gaining 2% on the whole. Find the cost of the donkey? 

Answer :

Let the cost price of the donkey be Rs ‘X’  Then cost price of the carriage = Rs(3000-X)  20% of X – 10% of (3000 – X) = 2% of 3000  X/5 – (3000-X)/10 = 60  2X – 3000 + X = 600  X = 1200  Hence cost price of the donkey = Rs.1200

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Description :

After selling a fan at 6% gain and a fridge at 9% gain, a shopkeeper gains Rs 5100. But if he sells the fan at 9% gain and the fridge at 6% loss, he gains Rs 1800 on the whole transaction. Find the original price of the fan.

Answer :

let the price of fan be X and fridge be Y 6x/100+9y/100 = 5100 6x+9y=510000------(1) then 9x/100-6y/100=1800 9x-6y=180000-------(2) Calculate both, the we will get x=Rs40000

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a) For a transmission line, Find SWR and reflection coefficient R if, i. There is no reflected voltage. ii. Reflected voltage and incident voltage is equal. iii. If reflected voltage=20V and incident voltage=10V. iv. If reflected voltage=10V and incident voltage =20V.

Answer :

reflection coefficient R=Vr/Vi i. There is no reflected voltage. i.e,Vr=0 R=0 SWR= 1+R/1-R=1 ii. Reflected voltage and incident voltage is equal. Vr=Vi; R=1 SWR= 1+R/1-R=1+1/1-1=infinity iii. If reflected ... and incident voltage =20V. Vr=10 and Vi=20 R=10/20=0.5 SWR= 1+R/1-R=1+.5/1-.5=3 

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A super heterodyne AM receiver is tuned to a station operating at 1200 KHz .Find local oscillator frequency and image frequency.

Answer :

A super heterodyne AM receiver is tuned to a station operating at 1200 KHz Intermediate frequency is 455KHz. IF frequency=f0-fs Local oscillator frequency is f0=IF +fs=455K+1200K=1655kHz The image frequency which gives the same IF is f0+2*IF=2110KHz

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A telephone cable has following primary constants per loop kilometer ,R=30Ω, L=20mH,C=0.06µF,G=0.If the applied signal has an angular frequency of 5000 rad/sec.., Determine (i) Characteristics impedence (ii) Attenuation constant

Answer :

A telephone cable has following primary constants per loop kilometer ,R=30Ω, L=20mH,C=0.06µF,G=0.If the applied signal has an angular frequency of 5000 rad/sec.., Determine (i) Characteristics impedence (ii) Attenuation constant

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Description :

An SSB transmitter produces a peak-to-peak voltage of 200 V across a 100 Ω antenna load. What is the Peak Envelope Power? A) 49.98 W B) 79.98 W C) 99.96 W D) 199.93 W

Answer :

An SSB transmitter produces a peak-to-peak voltage of 200 V across a 100 Ω antenna load. What is the Peak Envelope Power? A) 49.98 W B) 79.98 W C) 99.96 W D) 199.93 W

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Description :

An 8-pole, 50Hz, three-phase induction motor is running at 705rpm and has a rotor copper loss of 5kW. Its rotor input is (A) 5.06 kW (B) 0.3 kW (C) 100 kW (D) 83.33 kW

Answer :

An 8-pole, 50Hz, three-phase induction motor is running at 705rpm and has a rotor copper loss of 5kW. Its rotor input is 83.33 kW

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A half wavelength Dipole antenna has _________ Radiation Pattern  a. Hemispherical b. Omni-Directional c. Isotropic d. End fire

Answer :

A half wavelength Dipole antenna has Omni-Directional Radiation Pattern 

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Description :

An amplifier without feedback has a voltage gain of 50, input resistance 1kΩ and output resistance of 2.5kΩ. The input resistance of the current-shunt negative feedback amplifier using the above amplifier with a feedback factor of 0.2 would be: (1) 1/11 kΩ (2) 1/5 kΩ (3) 5 kΩ (4) 11 kΩ

Answer :

An amplifier without feedback has a voltage gain of 50, input resistance 1kΩ and output resistance of 2.5kΩ. The input resistance of the current-shunt negative feedback amplifier using the above amplifier with a feedback factor of 0.2 would be: 1/11 kΩ

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An amplifier which has almost same gain and input resistance of a CE amplifier but of higher bandwidth is  A) Cascode amplifier B) Cascade amplifier C) Darligton amplfier D) CC amplifier

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Description :

A load of 200 ohm is used to match 300 ohm transmission line to achieve SWR=1. Find out the required characteristic impedance of a quarter of a quarter wave transformer connected directly to the load.

Answer :

Solution :

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Description :

In a FM system, the maximum deviation is 75KHz. Find bandwidth for modulating frequency i. fm=500Hz ii. fm=5KHz iii. fm=10KHz Draw conclusion for bandwidth of FM from answer.

Answer :

Given deviation∆=75kHz i) fm=500Hz bandwidth B.W=2(∆+fm)  =2(75k+500)=151kHz  ii). fm=5KHz bandwidth B.W=2(∆+fm)  =2(75k+5k)=160KHz iii) fm=10KHz bandwidth B.W=2(∆+fm)  =2(75k+10k)=170KHz As the modulating frequency increases bandwidth also increases.

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Description :

Answer :

Req= (1/1200 +1/560 + 1/2.1)^-1        = 2.1 ohms RT= 2.1 + 7.8      = 9.9 ohms IT = V/R = 24/9.9                = 2.4 A Veq= R I =  2.1 2 ... 7.8         = 12.096 w                 = 44.928 w _________________________________________ By: Ellis-Junior Kamatjipose [email protected]

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Description :

The mean of 8 article was found to be 15. On rechecking, it was found that two article were wrongly taken as 11 and 9 instead of 16 and 14 respectively. Find the correct mean. A) 17.25 B) 13.65 C) 16.54 D) 16.25

Answer :

D) Calculated mean of 8 articles = 15 Incorrect sum of these 8 articles = (15*8) = 120. Correct sum of these 8 articles = (incorrect sum) - (sum of incorrect articles) + (sum of actual articles) = [120 ... (30)] = 130 Therefore, correct mean = 130/8 = 16.25 Hence, the correct mean is 16.25.

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Description :

Calculate the resistance of shunt required to make a milliammeter which gives maximum deflection for a current of 15 mA and which has a resistance of 5 Ω; read up to 10 Amp.

Answer :

Answer :

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Description :

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Answer :

A moving coil meter has a resistance of 3 Ω and gives full scale deflection with 30 mA. To measure voltage up to 300 V, the external resistance to be connected in series with the instrument is 9997 Ω

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Description :

DC ammeter has a resistance of 0.1 Ω and its current range is 0-100 A. If the range is to be extended to 0-300 A, what shunt resistance need to be connected?  A) 0.025 Ω B) 0.05 Ω C) 0.015 Ω D) 0.03 Ω

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Description :

Consider a 32 - bit microprocessor, with a 16 - bit external data bus, driven by an 8 MHz input clock. Assume that the microprocessor has a bus cycle whose minimum duration equals four input clock cycles. What is the maximum data transfer rate for this microprocessor? (A) 8x106 bytes/sec (B) 4x106 bytes/sec (C) 16x106 bytes/sec (D) 4x109 bytes/sec 

Answer :

(B) 4x106 bytes/sec

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