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
196 W
73 W
36.5 W
18.25 W
Answer: Option
Explanation:

P = n . I2 . Rr, where Rr = 80p2 Ω.

n = 4.

Discussion:
9 comments Page 1 of 1.

JMS Orps said:   10 months ago
Rr = 80.
n = 4.
=80π^2 (50 cm/100cm)^2.
= 196 W.

NOTE: 80 π is constant.

Krishna Chaitanya said:   4 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.

Chandraraj said:   6 years ago
The correct one is option C.

Pranab said:   7 years ago
But half dipole Rr=73 ohm.

Vivek said:   7 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).

Blessy said:   8 years ago
What is the value of p in the Rr equation?

Usha said:   9 years ago
There dl=50.

Alex said:   9 years ago
Could you please put the complete equation?

Pratik said:   10 years ago
What will be "dl" here?

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