# 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* . I^{2} . R_{r}, where R_{r} = 80p^{2} 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.

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.

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).

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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