UNIT#10 ELECTRONICS Flashcards
The applied input a.c. power to a half-wave rectifier is 100 watts. The D.C output power obtained is 40 watts. What is the rectification of efficiency?
40%
Efficiency= D.C Output Power/ A.C. input power x 100
The ripple factor indicates the number of ripples in the:
D.C Output
For a half wave or full rectifier the Peak Inverse Voltage of the rectifier is always:
Smaller than the input voltage
The device which converts A.C into D.C is called:
Rectifier
The types of rectifications are:
2
Rectification is possible by:
Diode
The semiconductor diode can be used as a rectifier because:
It has low resistance to the current flow when forward biased & high resistance when reverse biased
The simplest type of rectification known as half-wave rectification is obtained by:
Suppressing half wave of A.C supply by using diode
The output of half wave rectifier is suitable only:
Charging batteries
During the interval 0➡T/2 the forward biased diode offers:
Very small resistance
In a half wave rectifier, the frequency of the input is N, and the frequency and form of the output will be:
➡N
➡Pulsating
The most common device used as the filter is:
Capacitor
The method by which only half of the A.C cycle is converted into the direct current is called:
Half wave rectification
If the time period of input T in the full wave bridge rectifier circuit, then the time period of the pulsating current will be:
T/2
In a bridge rectifier how many diodes conduct during each half cycle of input A.C
2
If a full wave rectifier circuit is operating from 50Hz mains, the fundamental frequency in the ripple will be;
100Hz
In a full-wave rectifier, the frequency of the output wave is double of the input wave
To reduce ripples in the output of the bridge rectifier we should use:
A filter circuit
In full wave rectification, the output D.C. voltage across the load is obtained for:
The complete cycle of input A.C
The output voltage of the rectifier is not smooth. It can be made smooth by using a circuit known as;
Filter Circuit
The basic reason why a full wave rectifier has twice the efficiency of a half-wave rectifier is that:
It utilizes both half cycles of the input
If an A.C voltage of rms value of 10V is applied as input of half wave rectifier, then the rms voltage value of D.C output will be:
9.3V
General potential barrier of Si=0.7V
If a half-wave rectifier is used to convert 50Hz A.C into D.C then the number of pulses present in rectifier voltage is:
50
For half wave, f=f
The output from a full wave rectifier is:
➡Uni-Directional
➡Pulsating DC
The transformer is used in a rectifier to:
Step-Down Voltage
Which has less ripple factor:
Full wave Rectifier
Ripple factor of full wave rectifier?
1.21
The work done by the electric field on a particle when it is moved from one point on an equipotential surface to another point on the same equipotential surface is;
Always zero
If input free is 50Hz for full wave rectifier, then ripple frequency will be:
100Hz
If the input voltage of input De if 10V for full wave rectifier output De voltage will be;
10
The bridge rectifier is used more than a full rectifier because:
➡Perfomrs on less secondary voltage
➡No centre tap transformer required
➡Power loss is minimum
The half wave rectifier has an input of 10V its de component is:
10/π
Half wave rectifier contains, how many diodes
1
Ripple factor of half wave rectifier?
1.21
The efficiency of a full wave rectifier?
81.2%
In peak rectifier Vp=100 and R=10KΩ f=60Hz … find the factor for which diode is conduction:
4.02%
The bridge rectifier is an alternative for:
Full wave rectifier
A rectifier is a:
Linear Device
The ripple factor indicates the number of ripples in the:
D.C output
The ripple factor is:
Ripple factor = √(Vrms/Vdc)^2-1
Rectifier circuits are:
SIngle-Phase