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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.

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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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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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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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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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

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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 

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Answer :

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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 :

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 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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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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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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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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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.
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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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?
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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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Description :

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Answer :

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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 :

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Description :

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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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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Answer :

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Answer :

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

Answer :

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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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Description :

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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asked Feb 24 by anonymous
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.
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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asked Feb 24 by anonymous
A supply of 24 V DC is applied across a circuit which consist of three resistors connected in parallel R1= 1.2KΩ, R2= 560 Ω, R3= 2.1Ω and a fourth resistor of 7.8 Ω connected in series with the parallel group. Determine the current, voltage and power across each resistor
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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asked Aug 5, 2019 by A supply of 24 V DC is applied across a
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
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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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 :

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 (A) 9997 Ω (B) 10000 Ω (C) 997 Ω (D) 10003 Ω

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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asked Apr 28, 2018 by anonymous
A moving coli ammeter has a fixed shunt of 0.02 Ω with a coil circuit resistance of R=1 kΩ and needs potential...
Description :

A moving coli ammeter has a fixed shunt of 0.02 Ω with a coil circuit resistance of R=1 kΩ and needs potential of 0.5 V across it for full scale deflection. Calculate the value of shunt to give full scale deflection when the total current is 10 A. (1) 0.05 Ω (2) 0.005 Ω (3) 0.5 Ω (4) 0.0005 Ω

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asked Apr 11, 2018 by anonymous
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...
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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asked Apr 9, 2018 by anonymous
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
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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asked Feb 8 by anonymous
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