Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 7 (Q.No. 12)
12.
In a class C operation VCC = 40 V, RL = 50 Ω. The maximum load power can be
Discussion:
5 comments Page 1 of 1.
Gopalsamy said:
2 years ago
The formula is VCC^2/2*RL.
Zulu Chungtia said:
6 years ago
Vpp^2/8*Rl.
80*80/8*50.
= 16.
80*80/8*50.
= 16.
Esaki Diode said:
7 years ago
Answer is correct.
class C = 50%.
P = (V^2/R)*50%
P = ( 40^2/50)*0.5,
= 16.
class C = 50%.
P = (V^2/R)*50%
P = ( 40^2/50)*0.5,
= 16.
Tamzilan said:
8 years ago
I think B is the answer.
P = V^2 / R.
= 40*40/50.
= 32.
P = V^2 / R.
= 40*40/50.
= 32.
Mukund said:
8 years ago
Can any one explain how?
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers