Electronics and Communication Engineering - Electromagnetic Field Theory - Discussion
Discussion Forum : Electromagnetic Field Theory - Section 1 (Q.No. 5)
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
Answer: Option
Explanation:
P = n . I2 . Rr, where Rr = 80p2 
Ω.
n = 4.
Discussion:
9 comments Page 1 of 1.
JMS Orps said:
2 years ago
Rr = 80.
n = 4.
=80π^2 (50 cm/100cm)^2.
= 196 W.
NOTE: 80 π is constant.
n = 4.
=80π^2 (50 cm/100cm)^2.
= 196 W.
NOTE: 80 π is constant.
Krishna Chaitanya said:
6 years ago
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.
Where n=4, I=0.5=1/2A, Rr=73 ohms.
So P= (4) (1/2) ^2 (73) = 73W.
CORRECT OPTION IS B.
Chandraraj said:
7 years ago
The correct one is option C.
Pranab said:
9 years ago
But half dipole Rr=73 ohm.
Vivek said:
9 years ago
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).
P = n * I^(2) * Rr = >;P = Rr;
Rr = 80 * (9.8) * (1/4) = 196 W;(pi^(2) approximately 9.8).
Blessy said:
9 years ago
What is the value of p in the Rr equation?
Usha said:
10 years ago
There dl=50.
Alex said:
1 decade ago
Could you please put the complete equation?
Pratik said:
1 decade ago
What will be "dl" here?
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