Electronics and Communication Engineering - Electromagnetic Field Theory - Discussion

5. 

A broadside array operating at 100 cm wavelength consist of 4 half-wave dipoles spaced 50 cm apart. Each element carries radio frequency current in the same phase and of magnitude 0.5 A. The radiated power will be

[A]. 196 W
[B]. 73 W
[C]. 36.5 W
[D]. 18.25 W

Answer: Option A

Explanation:

P = n . I2 . Rr, where Rr = 80p2 Ω.

n = 4.


Pratik said: (Aug 26, 2014)  
What will be "dl" here?

Alex said: (Apr 10, 2015)  
Could you please put the complete equation?

Usha said: (Oct 10, 2015)  
There dl=50.

Blessy said: (Sep 19, 2016)  
What is the value of p in the Rr equation?

Vivek said: (Nov 9, 2016)  
n = 4; i = 0.5; Here p is pi(3.14....);dl = 50cm.

P = n * I^(2) * Rr = >;P = Rr;

Rr = 80 * (9.8) * (1/4) = 196 W;(pi^(2) approximately 9.8).

Pranab said: (Mar 4, 2017)  
But half dipole Rr=73 ohm.

Chandraraj said: (Jul 18, 2018)  
The correct one is option C.

Krishna Chaitanya said: (Feb 24, 2020)  
P= n I^2 Rr.

Where n=4, I=0.5=1/2A, Rr=73 ohms.
So P= (4) (1/2) ^2 (73) = 73W.

CORRECT OPTION IS B.

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