Transformers-Construction

• Main elements of the transformer as below:

         1) Magnetic circuit: limbs, yokes
         2) Electrical circuit:primary secondary and tertiary windings, formers, 
             insulation and bracing devices.
         3) Terminals, tappings and tapping switches, terminal insulators 
         4) Tank oil, cooling devices, conservators, dryers 

• Special alloy steel of high resistance and low hysteresis loss is used as transformer core, CRGO steel.
• Cut the transformer sheets as far as possible along the grain which is the direction in which the material has a higher permeability
• For large three phase transformer, the core is five-limbed core than , which needs a cross section in the yokes less than that required in usual three limbed core but core losses increase
• 
• on account of the easier insulation facilities, the low voltage winding is placed nearer to the core in case of core type and on the outside positions in the case of shell type transformers. 
• Cylindrical concentric helix windings, commonly employed for core-type transformers.
• Cross over coils are suitable for currents not exceeding about 20A. They are used for HV windings in comparatively small transformers, and comprise wires of small circular section with double cotton covering
• Disc coils are made up of a number of flat sections, generally with rectangular wire
• In Disc coils every turn is contact with oil hence better cooling
• Sandwich windings , commonly employed for shell-type transformers, allow of easy control over reactance.
• Bushings are the porcelain insulators used to provide the insulation between conductor end terminals and whole part of the T/F. 
• Conservators are required to take up the expansion and contraction of the oil with changes of temperature in service without allowing the oil to come in contact with air. 
• Breather extracts the moisture from the air. It consists of calcium chloride or silica gel.
• Transformer oil serves the double purpose of cooling and insulating.
• Sludging is the slow formation of semi-solid hydrocarbons, sometimes of an acidic nature, which are deposited on windings and tank walls. The formation of sludge is due to heat and oxidation. 
• Sludge formation is more in the presence of bright copper surfaces

Objective Questions:

1) Sludge formation in transformer oil is due to which one of the following?

 a. Ingress of dust particles and moisture in the oil.
 b. Appearance of small fragments of paper, varnish, cotton and other organic materials in the oil
 c. Chemical reaction of transformer oil with the insulating materials
 d. Oxidation of transformer oil

Ans: d

Explanation: Refer above theory

2) Cores of large power transformers are made from which one of the following? 
a. Hot-rolled steel
b. Cold-rolled non-grain oriented steel
c. Cold-rolled grain oriented steel
d. Ferrite

Ans: c
Explanation: to increase the relative permeability

3)The function of oil in a transformer is to provide

a. Insulation and cooling
b. Protection against lightning
c. Protection against short circuit
d. Lubrication

Ans: a

Explanation: Refer above theory

4)Consider the following statements relating to the constructional features of a large power transformer:

1. The conservator is used to maintain the level of oil in the transformer tank 2. The bushing is used to protect transformer insulation against lightning over-voltages
3. The Buchholz relay is an over current relay
4. Silica gel is used to absorb moisture.
Which of these statements are correct?
a. 1, 2, 3 and 4
b. 2 and 3
c. 1 and 4
d. 1, 2 and 4 

Ans: c

Explanation: Refer above theory

5) In  large power transformer, a "conservator" drum is provided above the tank and connected to it by a short pipe. The conservator drum is linked to external air through a breather. What is the purpose pf providing the conservator?

a) To store reserve oil to make up oil losses due to leakage.

b) To prevent explosion due to rise in oil pressure inside the tank during a fault.

c) To accommodate change in oil level during the "load cycle" of the transformer load

d) To exert additional pressure by the conservator oil on the inside the main tank to prevent disintegration of oil at high temperature. 

Ans: c

Explanation: Refer above theory

6)Match List I (Type of Coil) with List II (Use of Coil) and select the correct answer using the code given below the lists:

List I
A. Sandwitch coils
B. Disc coils
C. Cross-over coils
D. Spiral type

List II
1. Low voltage coils for currents above 100 A
2. High voltage windings of small transformers
3. Cooling oil is in contact with each turn of the winding
4. Shell-type transformer core

Codes;
   A B C D
a. 2 3 4 1
b. 4 1 2 3
c. 2 1 4 3
d. 4 3 2 1

Ans: b

Explanation: Refer above theory

7) Assertion(A):For obtaning improved magnetic properties, the transformer magnetic core is assembled using cold rolled silicon steel sheets.

   Reason(R):  The laminations for the core could be cut out of the cold rolled

  silicon steel sheets, cutting either in the direction of rolling or transverse

 therof, without affecting the magnetic properties in any way. 

a) Both A and R are individually true and R is the correct explanation of A
b) Both A and R are individually true but  R is not the correct explanation of A  
c) A is true but R is false
d) A is false but R is true

Ans: c

Explanation: Refer above theory

8) Assertion(A):Transformer is not used in a D.C. line

   Reason(R): Losses in the Dc circuit are not negligible  

a) Both A and R are individually true and R is the correct explanation of A
b) Both A and R are individually true but  R is not the correct explanation of A  
c) A is true but R is false
d) A is false but R is true

Ans: b

Explanation: Both are true. R is not reason to A. because when T/F operates on Dc line it burns. 

9) Match List I  with List II  and select the correct answer using the code given below the lists:

List I
A. Silica gel
B. Porcelain
C. Mercury
D. Fins

List II
1. Bushing
2. Buccholz relay
3. tank
4. Breather

Codes;
    A B C D
a. 2 1 4 3 
b. 4 3 2 1
c. 2 3 4 1 
d. 4 1 2 3

Ans: d

Explanation: Refer above theory

10) When are eddy current losses in a transformer reduced?   
a. if laminations are thick
b. if the number of turns in primary winding is reduced
c. if the number of turns in secondary winding is reduced
d. if laminations are thin

Ans: d
Explanation: Pe=ke*f^2*B^2*t^2 

Transformers basics1

 

• The physical basis of transformer (T/F) is mutual induction between two circuits linked by a common magnetic field. It is a static device.

• Mutual inductance is the same irrespective of which circuit is primary and which circuit is secondary.

• The voltage applied to the primary is almost completely concerned in opposing the induced e.m.f.

• If the primary voltage is constant, the mutual flux remains approximately constant regardless of the load connected across the secondary coil.
• Important tasks performed by T/F are 

            1) Changing voltage and current levels in power system.     

            2) Matching source and load impedance for maximum power transfer 

            3) Electrical isolation

• In core type, to avoid leakage flux, it is usual to have half the primary and half of the secondary winding side-by-side or concentrically on each limb; not primary on one limb and secondary on other limb.

• The three phase core type, the principle that the sum of the fluxes in each phase in a given direction along the cores is zero.

• In core type inspection of coils and core is easy.

• The shell type is more robust mechanically.

• In shell type core, the cooling is good.

• The e.m.f induced in transformer given by: E = 4.44fΦmT volts

• Volt/turn is same in primary winding and secondary winding provided only that it links the same flux in both winding.

• Both primary and secondary e.m.f.’s are in phase.

• The e.m.f’s lag by 90 degree in time on the flux.

• The applied voltage V1 opposes E1, while E2 provides the secondary output voltage V2. 

• E1/E2 = I2/I1 = T1/T2

• A small magnetizing current is needed to maintain the magnetic circuit or core in magnetized state, when the secondary is open
• The m.m.f. of the primary on no load is of the order of 5 per cent of its m.m.f. on full load.

• The no load current, Io has two components, magnetizing component, I_m and loss component, I_r.

• Leakage between primary and secondary could be eliminated if the winding could be made to occupy the same space.
• Reductions in leakage flux can achieve by sectionalizing and interleaving the primary and secondary coils.

• An equivalent circuit is useful for calculations of regulation, efficiency, parallel operation, etc.
• I2R loss and per-unit reactance voltage in primary and secondary is same.

• Core loss due to the pulsation of the magnetic flux in the iron producing eddy current and hysteresis loss.
• I2R loss due to heating of the conductors by the passage of current.
• Stray loss due to stay magnetic field causing eddy currents in the conductors or in surrounding metal (tank).
• Dielectric loss in the insulating materials (oil and solid insulation of HT T/F).
• Efficiency of transformer is given by:

• Where S is full load kVA, x is per-unit load, P_i is iron loss, P_c is full load copper loss.
• For maximum efficiency occurs when the variable loss is equal to the constant loss, i.e. x² P_c= P_i.
• Maximum efficiency occurs below full load.
• Maximum efficiency point independent of power factor.
• The regulation of a transformer refers to the change of secondary terminal voltage between no-load and load conditions: it is usually quoted as a percent value for full load at given power factor.

• Per unit regulation:

• Maximum regulation occurs when

• Zero regulation occurs when

i.e. leading p.f.

• On account of the easier insulation facilities, the low-voltage winding is placed nearer to the core in the case of core type and on the outside positions in the case of shell type transformers.
• Under no-load conditions, the “hum” developed by energized power transformer originates in the core, where the lamination tend to vibrate by magnetic forces
• The essential factors in noise production are magnetostriction, mechanical vibrations by the lamination.
• In an Ideal transformer, there are no voltage drops in resistance or leakage reactance, the MMF required to maintain the main flux is small and there are no core losses.
• Magnetizing MMF is a function of the length, the net cross sectional area and the permeability of the iron path. 

Objective Questions:

1).If the applied voltage of a certain transformer is increased by 50% and the frequency is reduced to 50% (assuming) that the magnetic circuit remains unsaturated), the maximum core /flux density will

a. Change to three, times the original value
b. Change to 1.5 times the original value
c. Change to 0.5 times the original value
d. Remain the same as the original value

Ans: d

Explanation: B∝ V/f

2). In a transformer, zero voltage regulation at full load is

     a. not possible
     b. possible at leading power factor load
     c. possible at lagging power factor load
     d. possible at unity power factor load

Ans: b
Explanation: 

• Zero regulation occurs when

i.e. leading pf.
3). The full load copper loss and iron loss of a transformer are 6400W and 500W respectively. The above copper loss and iron loss at half load will be

    a) 3200 W and 250 W respectively
    b) 3200 W and 500 W respectively
    c) 1600 W and 125 W respectively
    d) 1600 W and 500 W respectively


Ans: d
Explanation: Refer above theory 
                  Pcu=x^2*Pcu.fl=(0.5)^2*6400=1600W
                  Pi=500W  ( independent of on the load)

4). A 4KVA, 400V/200V single phase transformer has resistance of 0.02 pu and reactance of 0.06 pu. The resistance and reactance referred to high voltage side are

    a) 0.2 Ω and 0.6 Ω
    b) 0.8 Ω and 2.4 Ω
    c) 0.08 Ω and 0.24 Ω
    d) 1 Ω and 3 Ω


Ans: c
Explanation: R2+jX2=0.02+j0.06
                  a=V2/V1=200/400=0.5
                  R2'+jX2'=(R2+jX2)/(a^2)=0.08+j0.24

5). A single-phase transformer rated for 220/440 V, 50 Hz operates at no load at 220 V, 40Hz.  This frequency operation at rated voltage results in which one of the following?
a. Increases of both eddy-current and hysteresis losses
b.  Reduction of both eddy-current and hysteresis losses
c. Reduction of hysteresis loss and increase in eddy-current loss
d. Increase of hysteresis loss and no change in the eddy-current loss


Ans: d
Explanation: Ph=kh*f*B^x;   B∝ V/f    hysteresis loss increases with decrease in frequency. 
                  Pe=ke*f^2*B^2*t^2 , B∝ V/f  eddy current loss depend on the square of the applied voltage and independent of the frequency. 


6). What is the load at which maximum efficiency occurs in case of a 100 kVA transformer with iron loss of 1 kW and full–load copper loss of 2 kW?
a. 100 kVA
b. 70.7 kVA
c. 50.5 kVA
d. 25.2 kVA


Ans: b
Explanation: Pcu*x^2=Pi;   x=sqrt(Pi/Pcu)= sqrt(1/2);
                  load KVA=100*x=100*sqrt(1/2)=70.7 KVA


7). Cores of large power transformers are made from which one of the following? 
a. Hot-rolled steel
b. Cold-rolled non-grain oriented steel
c. Cold-rolled grain oriented steel
d. Ferrite


Ans: c
Explanation: to increase the relative permeability


8). A transformer has a percentage resistance of 2% and percentage reactance of 4%. What are its regulations at power factor 0.8 leading, respectively?
a. 4% and - 0.8%
b. 3.2% and – 1.6%
c. 1.6% and – 3.2 %
d. 4.8% and – 0.6%


Ans: a
Explanation: Per unit regulation:



                                 %regulation, lagging = 2*0.8+4*0.6= 4%
                      %regulation, leading = 2*0.8-4*0.6= -0.8%

9) Assertion(A): Both the efficiency and regulation of a 3 winding ideal

 transformer are 100%

Reason(R): The flux leakage and the magnetic reluctance of the magnetic core 

in an ideal transformer is zero.moreover losses are absent in ideal 

transformers

a) Both A and R are individually true and R is the correct explanation of A
b) Both A and R are individually true but  R is not the correct explanation of A  
c) A is true but R is false
d) A is false but R is true


Ans: d
Explanation: Ideal transformer has 100% efficiency and 0% regulation

10) When are eddy current losses in a transformer reduced?
      


a. if laminations are thick
b. if the number of turns in primary winding is reduced
c. if the number of turns in secondary winding is reduced
d. if laminations are thin


Ans: d
Explanation: Pe=ke*f^2*B^2*t^2  

11) on which of the following factors does hysteresis loss depend?
   1. flux density
   2. frequency
   3. thickness of lamination
   4. time
   
select the correct answer using the codes given below. 
  a) 2 and 3        a) 1 and 2       a) 3 and 4         a) 1 and 4


Ans: d
Explanation:  Ph=kh*f*B^x;

12) If P1 and P2 are the iron and copper losses of a transformer at full load, and the maximum efficiency of the transformer is at 75% of the full load, then what is the ratio of P1 and P2?
a. 9/16
b. 10/16
c. 3/4
d. 3/16


Ans: a
Explanation: Pcu*x^2=Pi;  x=0.75=3/4;  P1/P2=Pi/Pcu=x^2= (3/4)^2=9/16;

13). If the iron core of a transformer is replaced by an air core, then the hysteresis losses in the transformer will 
a. increase 
b. decrease
c. remain unchanged
d. become zero


Ans: d
Explanation:  air core transformer is free from hysteresis loss

14). The equivalent circuit of a transformer has the leakage reactance X1, X '2 and the magnetizing reactance Xm. What is the relationship between their magnitudes?
a. X1 >> X'2 >> Xm
b. X1 << X'2 << Xm
c. X1≈ X2 >> Xm
d. X1≈ X'2 << Xm


Ans: d
Explanation: Xm is parallel circuit in  equivalent  circuit 


16). If the voltage applied to a transformer primary is increased by keeping the V/f ration fixed, then the magnetizing current and the core loss will, respectively, 
a. decrease and remain the same
b. increase and decrease
c. remains the same and remains the same
d. remains the same and increase


Ans: d
Explanation: Im ∝ V/f;   Im unchanged and core loss depends on the frequency so increases. 

17) The use of higher flux density in the transformer design
a. Reduces the weight per kVA
b. Increases the weight per kVA
c. Has no relation with the weight of transformer
d. Increases the weight per kW


Ans: a
Explanation:   B=Φ/A . for same Φ, A is less. so weight will reduce 

18) The function of oil in a transformer is to provide
a. Insulation and cooling
b. Protection against lightning
c. Protection against short circuit
d. Lubrication  


Ans: a
Explanation: Insulation and cooling

19)A single-phase transformer when supplied from 220 V, 50 Hz has eddy current loss of 50 W. If the transformer is connected to a voltage of 330V, 50 Hz, the eddy current loss will be
a. 168.75 W
b. 112.5 W
c. 75W
d. 50W


Ans: b
Explanation: Pe=ke*f^2*B^2*t^2 , B∝ V/f  eddy current loss depend on the square of the applied voltage and independent of the frequency.

         Pe2/Pe1=(V2/V1)^2= (330/220)^2=2.25;

          Pe2=50*2.25=112.5 W

20) Match List I with List II and select the correct answer:
List I
A. Shell type with wound core
B. Core type with core of laminated sheets
C. Shell type with laminated core
D. Core type with wound core
List II



   A B C D
a. 4 3 1 2
b. 4 2 1 3
c. 1 2 4 3
d. 1 3 4 2

Ans: b
Explanation: Refer above theory

Diode-Circuits

Diode Convention:

• Positive terminal of the diode is called Anode
• Negative terminal of the diode is called Cathode.
• If a negative voltage applied to the diode , no current flows and diode behaves like open circuit. This mode called Reverse bias.
• If a positive voltage applied to the diode , current flows and diode behaves like short circuit. This mode called Forward bias.

Ideal Diode

Practical Diode Characteristics:

• Reverse saturation current is intedependent of the applied reverse bias.
• Cutin voltage: In forward bias, voltage below the Cutin voltage, diode forward current is very small.
• Cutin voltage for Germanium =0.2V, Silicon=0.7V.
• Reverse saturation current is 1000 times more in Germanium when compared to Silicon diode. 
• At higher currents diode behaves like a resistor than a diode and current increases linearly rather than exponentially with applied voltage.  

• The reverse saturation current approximately doubles for every 10 degree C temperature rise for both Si and Ge materials. 
• The p-n junction voltage decreases by about 2.5mV/degree C with rise in temperature.

Diode Circuits: 

       Simple Rectifier:

Basic Rectifier Circuit

Voltage relations

Transfer Characteristics

Half-Wave Rectifier:

• The diode only conducts when it is forward biased, therefore only half of the AC cycle passes through the diode to the output.
• The DC output voltage is 0.318Vm  =
  (Vm/pi) 
   , where Vm = the peak AC voltage

• The diode is only forward biased for one-half of the AC cycle, it is also reverse biased for one-half cycle.
• It is important that the reverse breakdown voltage rating of the diode be high enough to withstand the peak, reverse-biasing AC voltage.

 

Full-Wave Rectifier- Centertap Transformer Rectifier:

•Two diodes

•Center-tapped transformer

Full wave Rectifier

Circuit Operation

Full wave Rectifier Transfer Characteristics

Full-Wave Rectifier- Bridge Rectifier: 

• 4 diodes

Bridge Rectifier

Circuit Operation

Transfer Characteristics

For Full wave rectifier, the  fundamental frequency of the ripple voltage is twice that of the AC supply 

frequency (120Hz) where for the half-wave rectifier it is exactly equal to the supply frequency (60Hz).

half- wave Rectifier with Smoothing Capacitor: 

Full- wave Rectifier with Smoothing Capacitor: 

Diode Clipper Circuits:

 Diode Clamping circuits:

Voltage Doubler circuit:

• This half-wave voltage doubler’s output can be calculated by: Vout = VC2 = 2Vm,  where Vm = peak secondary voltage of the transformer.  

•Positive Half-Cycle

•           D1 conducts
•           D2 is switched off
•           Capacitor C1 charges to Vm

•Negative Half-Cycle

•         D1 is switched off
•          D2 conducts
•         Capacitor C2 charges to Vm

                                                    Vout = VC2 = 2Vm

Zener Diode:

• The Zener is a diode operated in reverse bias at the Zener Voltage (Vz).

    When  Vi ³ VZ

                   –The Zener is on

                   –Voltage across the Zener is VZ

       –Zener current: IZ = IR – IRL

       –The Zener Power: PZ = VZIZ

        When Vi < VZ

–The Zener is off

–The Zener acts as an open circuit





Objective questions:


1) A 700mW maximum power dissipation diode at 25degree C has 5mW/degree C de-rating factor. if the forward voltage drop remains constant at 0.7 V, the maximum forward current at 65 degree C is
     
         a) 700mA   b) 714mA    c) 1A  d) 1mA

Ans: b) 714 mA.

Explanation:

           P at 25= 700mW;
         
           dP/dT= -5mW/degree C

            P at 65 = P at 25- dP/dT * (65-25)= 700-5*40=500mW;
         
            I=P/V=500/0.7=714 mA.

2)
         
       

For the circuit shown above, using ideal diode, the values of voltage and current are respectively,

a) -3 V and 0.6 mA    b) 3 V and zero   c) 3V and 0.6mA  c) -3V and zero

Ans: a) -3 V and 0.6 mA

Explanation:

              I= (3-(-3))/10k=  0.6mA;  V=-3 V;  diode is in forward bias; 

3) A half wave rectifier has an input voltage of 240V(rms). if the step down transformer has a turns ratio of 8:1 . What is the peak load voltage? ignore diode drop.

a) 27.5 V   b) 86.5 V   c) 30.0 V   d) 42.5 V

Ans: d) 42.5 V

V1rms= 240V, V2rms= 240/8=30 V;

V2peak= 30*sqrt(2)= 42.5 V;

4) Assertion(A): The reverse saturation current approximately doubles for every 10 degree C temperature rise for both Si and Ge materials.
    Reason(R): At room temperature,the p-n junction voltage decreases by about 2.5mV/degree C with rise in temperature.
a) Both A and R are individually true and R is the correct explanation of A
b)  Both A and R are individually true but  R is not the correct explanation of A  
c) A is true but R is false
d) A is false but R is true

Ans: b) Both A and R are individually true but  R is not the correct explanation of A
 Explanation: Refer above topic
5) Assertion(A): Cut in voltage for Germanium diode is greater than that for Silicon diode.
    Reason(R): Germanium diode has a higher reverse saturation current than Silicon diode.

Ans) d)   A is false but R is true
Explanation: Refer above topic

6) When a junction diode is used as a half wave rectifier with purely resistive load and sinusoidal input voltage, what is the value of diode conduction angle(where Φ  is the ignition angle corresponding to the cut in volatge) ?
     a)π                    b)π-Φ                      c)π-2Φ             d)Slightly greater than π 
Ans: c)π-2Φ 

7) Consider the following statements for a p-n junction diode:
1. it is an active component
2. Depletion layer width decrease with forward biasing
3. In the reverse biasing case, saturation current increases with increasing temperature 
which of the statements given above are correct?
a) 1,2 and 3    b) 1 and 2     c) 2 and 3   d) 1 and 3
Ans: a) 1,2 and 3.

8) Which one of the following statements is not correct?
a) Reverse saturation current in a BJT approximately doubles for every 10 degree C rise in temperature.
b) The reverse resistance of a junction diode increases with increase in temperature
c) Reverse saturation current of a silicon diode is much smaller than that of a germanium diode.
d) The cut-in voltage of silicon diode is larger than that of germanium
Ans: b
Explanation: Refer above topic

9) When a positive DC voltage is applied to the n-side relative to p-side , a diode is said to be given a
a) forward bias  b) reverse bias   c) zero bias  d) neutral bias
Ans: b) reverse bias 
Explanation: Refer above topic

10) A combination of two diodes connected in parallel when compared to a single diode can withstand 
a) twice he value of peak inverse voltage
b) twice the value of maximum forward current
c) a larger leakage current
d) twice the value of cut in voltage
Ans: b) twice the value of maximum forward current

11) The AC resistance of a forward biased p-n junction diode operating at a bias voltage V and carrying current I is
a) zero  b) a constant value independent of V and I           c) V/I             d) ΔV/ΔI
ans: d) ΔV/ΔI

12) 

Ans: a

13) A rectifier with fundamental ripple frequency equal to twice the mains frequency, has a ripple factor of 0.482 and power conversion efficiency equal to 81.2%. The rectifier is

1. bridge rectifier

2. full wave (non-bridge rectifier

3. half wave rectifier 

which of these are correct?

a) 2 and 3         b) 2 only        c) 1 and 2        d) 1,2 and 3

Ans: c) 1 and 2

14) Consider the following statements:

When compared with a bridge rectifier, a centre-tapped full wave rectifier

1. has larger transformer utilization factor

2. can be used for floating output terminals i.e., no input terminal is grounded

3. needs two diodes instead of four

4. needs diodes of lower PIV rating

Which of the above statements are correct?

a) 1 and 2     b) 1,2,3 and 4    c) 3 only   d) 3 and 4

Ans: c) 3 only

15) 

The output Vdc from the above circuit is

a) 12√2        b) 12/π    c) 24/π    d) 12/√2

Ans:b) 12/π 

16)

Ans: c) 2Vm

 Explanation: Voltage doubler circuit

17) Silicon diodes are less suited for low voltage rectifier operation because

a) it cannot withstand high temperatures

b) its reverse saturation current is low

c) its cut in voltage is high

d) its breakdown voltage is high

Ans: c) its cut in voltage is high

Explanation: Refer above topic

18)

100 ohm




in the given circuit D1 is an ideal germanium diode and D2 is a silicon diode having its cut in voltage 0.7 V , forward resistance as 20 ohm and reverse saturation current as 10 mA. What are the values of I and V for this circuit respectively
.
a) 60 mA and zero   b) 50 mA and zero   c) 53 mA and 0.7V   d) 44mA and 1.58V

Ans:  a) 60 mA and zero   

Explanation:  V=0;  I=6/100= 60 mA

19) Consider the following statements:
A clamper circuit 
1. adds or subtracts a DC voltage to or from a waveform
2. does not change the shape of the waveform
3. amplifies the waveform

Which of the following statements given below are correct?
a) 1 and 2               b) 2 and 3             c) 1 and 3            d) 1,2 and 3

Ans: a) 1 and 2

20) A half wave rectifier having a resistance load of 1kOhm rectifiers an AC voltage of 325V peak value and the diode has a forward resistance of 100 Ohm. What is the RMS value of the current?

a) 295.4 mA     b) 94.0 mA      c) 147.7 mA        d) 208.0 mA 


Ans:c) 147.7 mA
Explanation:  Rl=1kOhm, Rf=100 Ohm  Vm =325 V

                       Im=Vm/(Rl+Rf)= 325/(1000+100) = 295.45 mA.

                       Irms=Im/2= 295.45/2=147.7 mA

21) The PIV rating of the diodes used in power supply circuits are chosen by which one of the following criteria?
a) The diodes that are to be used in a fullwave rectifier should be rate 2Vm and in bridge rectifier equal to Vm
b) The diodes that are to be used in a full wave rectifier should be rated Vm and in bridge rectifier equal to 2Vm
c) All diodes should be rated for Vm only
d) All diodes should be rated for 2Vm

Ans: a) The diodes that are to be used in a fullwave rectifier should be rate 2Vm and in bridge rectifier equal to Vm
Explanation: Refer above topic

22) Consider the following rectifier circuits
1. Half wave rectifier without filter
2. Full wave rectifier without filter
3. Full wave rectifier with series inductance filter
4. Full wave rectifier with capacitance filter
The sequence of these rectifier circuits in decreasing order of their ripple factor is
a) 1,2,3,4               b) 3,4,1,2             c)  1,4,3,2            d) 3,2,1,4
Ans: a) 1,2,3,4

23) The use of a rectifier filter in a capacitor circuit gives satisfactory performance only when the load 
a) current is high   b) current is low   c) voltage is high    d) voltage is low 

Ans: b) current is low