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Ac to dc
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Answer : NO IT ONLY CONVERT FROM AC TO DC BECAUSE DIODE WHICH ARE USED IN IT, ARE UNIDIRECTIONAL.

Answer : IT CHANGES NEGATIVE CYCLE IN TO POSITIVE CYCLE FROM AC CURRENT.

Answer : Rectifier is use in power supply to convert Alternating current (AC) to Direct Current (DC) .

Answer : Full wave bridge rectifier is used in mobile chqrger.

Description : Single-phase mid-point controlled rectifier with RL load

Answer : Single-phase mid-point controlled rectifier with RL load : The circuit configuration of single-phase midpoint controlled rectifier is shown in the figure. During positive half-cycle of ... some part of next positive half-cycle of voltage with reversed polarity can appear across load.

Description : Single-phase Midpoint controlled rectifier with Resistive load

Answer : Single-phase Midpoint controlled rectifier with Resistive load:  1) During positive half cycle of AC supply, a is positive with respect to b , this makes T1 forward biased and T2 is reverse ... voltage reverses the polarity and T2 is turned off. The operation is as shown in waveforms.

Description : When the firing angle a of a single-phase, fully controlled rectifier feeding constant direct current into a load is 30°, the displacement power factor of the rectifier is

Answer : When the firing angle a of a single-phase, fully controlled rectifier feeding constant direct current into a load is 30°, the displacement power factor of the rectifier is √3/2

Answer : the efficiency of centre tapped full wave rectifier and bridge rectifier is same that is 80.2% but the because of the use of Central trap the Transformer utilisation factor It's become low.hence we can say the bridge rectifier is more efficient

Answer : rectifier is a device which converts AC voltage to pulsating DC voltage using one or more diode diode conducts only in unit direction that is forward bias condition practically it does not conduct ... types one is with using two diodes and Centre tap Transformer and another is bridge wave rectifier

Description :  An AC supply of 230 V is applied to a half-wave rectifier circuit through a transformer having primary to secondary turn’s ratio 12:1. The peak inverse voltage is A) 8.62 V B) 12 V C) 19.17 V D) 27.11 V 

Answer :  An AC supply of 230 V is applied to a half-wave rectifier circuit through a transformer having primary to secondary turn’s ratio 12:1. The peak inverse voltage is A) 8.62 V B) 12 V C) 19.17 V D) 27.11 V 

Description : List out advantages and disadvantages of bridge rectifier.

Answer : Advantages :  1. Bridge rectifier can be used in applications allowing floating output terminals i.e. not output terminal is grounded.  2. The need of center-tapped transformer is eliminated.  ... The value of the diodes used should be precise , else there will be an error in rectification

Description : Define PIV, TUF, ripple factor, efficiency of rectifier.

Answer : PIV: Peak Inverse Voltage (PIV) is defined as the maximum negative voltage which appears across non-conducting reverse biased diode. TUF : Transformer Utilization Factor (TUF) is defined as the ratio of DC output ... Efficiency of rectifier: η = DC output power/ AC input power = PLdc / Pac

Description : operation of half wave rectifier type AC Voltmeter.

Answer :  Half-wave Rectifier Voltmeter The d'Arsonval meter movement only responds to the average or dc value of the current through the moving coil. In order to measure alternating ... ) input voltage signal alternating signal needs to be rectified first by using diode rectifier to produce 

Description : need of rectifier and filter. 

Answer : Need of rectifier: Every electronics circuit needs a D.C. power source for its operation. The rectifier circuit is used to convert A.C. supply to unidirectional pulses .  Need of filter: A filter ... only the D.C. components to reach the load. So filter is used to provide ripple free output.

Description : Explain full wave bridged rectifier with the help of circuit diagram and input output waveform.

Answer : 1. In positive half cycle (0 to Π): The end A of the secondary winding becomes positive and end B negative. This makes diode D1 and D4 forward biased while diode D2 and D3 are reverse biased. These two diodes will ... D2 and D3 when it is conducting is as follows.  B - D2 - RL - D3 - A 

Description : In a half-wave rectifier, if an a.c supply is 60 Hz. Then what is the a.c ripple at output? (A) 30 Hz (B) 60 Hz (C) 20 Hz (D) 15 Hz

Answer : In a half-wave rectifier, if an a.c supply is 60 Hz. Then what is the a.c ripple at output? (A) 30 Hz (B) 60 Hz (C) 20 Hz (D) 15 Hz

Answer : The rectifier instrument is not free from frequency error.

Answer : The scale of a rectifier instrument is linear.

Answer : Rectifier is a device which convert AC to DC.

Description : State classification of Phase controlled rectifiers.

Answer : Classification of phase controlled rectifiers:

Description : How is a thyristor used to convert AC to DC?

Description : Can transformer convert AC to DC?

Answer : No, Transformer cannot convert AC to DC.

Description : Which device use to convert DC to AC?

Answer : Rectifier

Description : A freewheeling diode in phase -controlled rectifiers  (a) enables inverter operation (b) is responsible for additional reactive power (c) improves the line power factor (d) is responsible for additional harmonics

Answer : A freewheeling diode in phase -controlled rectifiers improves the line power factor

Description : Electronic Voltmeters which use rectifiers employ negative feedback. This is done -  a) To increase the overall gain b) To improve stability c) To overcome non -linearity of the diodes d) None of the above

Answer : Electronic Voltmeters which use rectifiers employ negative feedback. This is done - To overcome non -linearity of the diodes

Description : A compiler is used to convert the following to object code which can be executed (A) High-level language (B) Low-level language (C) Assembly language (D) Natural language

Answer : (A) High-level language

Description : What is the name of the device used to convert alternating current into direct current ? (1) Ammeter (2) Galvanometer (3) Rectifier (4) Transformer

Answer : Rectifier

Description : The device used to convert solar energy into electricity is (1) Photovoltaic cell (2) Daniell cell (3) Electrochemical cell (4) Galvanic cell

Answer : Photovoltaic cell

Description : A dynamo is used to convert (1) mechanical energy into electrical energy (2) electrical energy into mechanical energy (3) electrical energy into magnetic energy (4) magnetic energy into mechanical energy

Answer : mechanical energy into electrical energy

Description :  A bimetallic strip is used to convert a temperature change - a) into mechanical displacement b) into electrical displacement c) into neither mechanical nor electrical displacement d) all of the above

Answer :  A bimetallic strip is used to convert a temperature change - into mechanical displacement

Description : Which of the following are the parts of a DC generator that convert the AC in the armature to DC in the external circuit. Are these parts the: w) field poles x) slip rings and the brushes y) commutator and the brushes z) armature and the core

Answer : ANSWER: Y -- COMMUTATOR AND THE BRUSHES

Description : "Goals are dreams we convert to plans and take action to fulfill." - Zig Ziglar  

Description : To convert a galvanometer to a voltmeter, you should add a: w) high resistance in series x) high resistance in parallel y) low resistance in series z) low resistance in parallel

Answer : ANSWER: W -- HIGH RESISTANCE IN SERIES

Description : If we convert ∃u ∀v ∀x ∃y (P(f(u),v, x, y) → Q(u,v,y)) to ∀v ∀x (P(f(a),v, x, g(v,x)) → Q(a,v,g(v,x))) This process is known as (A) Simplification (B) Unification (C) Skolemization (D) Resolution

Answer : (C) Skolemization

Description : Biofertilizers convert nitrogen to _______ . (1) nitrates (2) ammonia (3) nitrogenase (4) amino acids

Answer : ammonia

Description : Transformer is a device to convert (1) D.C. to A.C. (2) Low voltage D.C. into high voltage D.C. (3) Low voltage A.C. into high voltage A.C. (4) Mechanical energy into Electrical energy 

Answer : Low voltage A.C. into high voltage A.C.

Description : The function of ball bearings in a wheel is : (1) to increase friction (2) to convert kinetic friction into rolling friction (3) to convert static friction into kinetic friction (4) just for convenience

Answer : to convert kinetic friction into rolling friction

Description : Explain the suitable example to convert a practical current source into equivalent voltage source.

Answer : The open- circuit voltage across terminals A and B is Voc = drop across R  =5x2=10V Hence, voltage source has a voltage of 10V and the same resistance of 2Ω

Description : A PMMC instrument has FSD of 100 μA and a coil resistance of 1 kΩ. To convert the instrument into an ammeter with full scale deflection of 100 mA, the required shunt resistance is (A) 1 Ω (B) 1.001 Ω (C) 0.5 Ω (D) 10 Ω

Answer : A PMMC instrument has FSD of 100 μA and a coil resistance of 1 kΩ. To convert the instrument into an ammeter with full scale deflection of 100 mA, the required shunt resistance is 1.001 Ω

Description : To convert a S-R flip-flop to D flip-flop (a) D input is given to S and D input to clock (b) D input is given to S and D to R (c) D input is given to clock and D to S input (d) D input to S and clock to R

Description : Convert hexadecimal value 16 to decimal: - a) 2210 b) 1610 c) 1010 d) 2010

Answer : Convert hexadecimal value 16 to decimal: - 2210

Description : Which rectifier can convert AC to DC? A. Half wave rectifier B. Full wave rectifier C. Bridge wave rectifier D. All of these 

Answer : Which rectifier can convert AC to DC? A. Half wave rectifier B. Full wave rectifier C. Bridge wave rectifier D. All of these 

Answer : The binary number 1011010 to hexadecimal is 5A.

Description : It is desired to convert a 0-1000A meter movement, with an internal resistance of 100 ohms, into a 0-100mA meter. The required value of shunt resistance is about 

Answer : It is desired to convert a 0-1000A meter movement, with an internal resistance of 100 ohms, into a 0-100mA meter. The required value of shunt resistance is about 1 ohms.

Description : Can you convert analog signal to digital?

Description : Explain various performance parameter used to evaluate the quality of inverters.

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