What is the necessary a.c. input power from the transformer secondary used in a half wave rectifier to deliver 500 W of d.c. power to the load?
A. 1232 W
B. 848 W
C. 616 W
D. 308 W
Answer: Option A
This Question Belongs to Electronics And Communications Engineering >> Electronic Devices And Circuits
Join The Discussion
Comments ( 5 )
The forbidden energy gap between the valence band and conduction band will be least in case of
A. Metals
B. Semiconductors
C. Insulators
D. All of the above
For a NPN bipolar transistor, what is the main stream of current in the base region?
A. Drift of holes
B. Diffusion of holes
C. Drift of electrons
D. Diffusion of electrons
A. Both A and R are true and R is correct explanation of A
B. Both A and R are true but R is not a correct explanation of A
C. A is true but R is false
D. A is false but R is true
For a P-N diode, the number of minority carriers crossing the junction depends on
A. Forward bias voltage
B. Potential barrier
C. Rate of thermal generation of electron hole pairs
D. None of the above
40% efficiency of rectification does not mean that 60% of power is lost in the rectifier circuit. In fact, a crystal diode consumes little power due to its small internal resistance. The 100 W a.c. power is contained as 50 watts in positive half-cycles and 50 watts in negative half-cycles. The 50 watts in the negative half-cycles are not supplied at all. Only 50 watts in the positive half-cycles are converted into 40 watts
so input power *40% = output of halfwave rectifier
input power =500/(40/100)
=5000/4=1250
Given,
Dc load power = 500 W,
We need to find ac input power from the transformer.
As we know, the formula for the efficiency of half wave rectifier is as below,
efficiency = dc load power/ac input power.
We know that the efficiency of half-wave rectifier =40.6% or 0.406.
Therefore,
0.406=500/ac input power.
Ac input power = 500/0.406.
Ac input power = 1231.52.
How
40%= 500/Ac input power
Efficiency of half wave rectifire= DC load power/AC load power
40.5%