1. (b)JFETs operate in depletion mode only i.e.VGS < 0V. This condition is necessary so asto reverse bias the gate junctions and thuscreate depletion region in the channel whichcontrols the drain current flowing through thechannel.
2. (a)Ebers Moll model is used to represent thelarge signal model of BJT, that involves twoideal diodes placed back-to-back with reversesaturation currents –IEO and –ICO and twodependent current controlled current sourceshunting the ideal diodes.
EIE IC
–(IE0) –(IC0)
IIC IIC
B
VE VC
IB
C
3. (b)Fermi-Dirac probability function, f(E), is theprobability that an energy level E beingoccupied by an electron and it is given by
f(E) = FE E /KT1
1 e
Probability that an energy level E is occupiedby a hole is given by
1 f E = FE E /KT11
1 e
=
F
F
E E /KT
E E /KTe
1 e
= FE E /KT1
1 e
4. (a)Graph IC versus VCE for constant IB representthe output characteristics of a BJT in common-emitter configuration.
From this graph, can be directly determinedby finding ratio of the change in collectorcurrent to the change in base current.
5. (d)For the Bipolar Junction Transistor (npn typeor pnp type) to operate in cut-off, both theEmitter junction (JE) and collector junction (JC)must be reverse biased.
Circuit in option (d) uses a p-n-p BJT andbiasing used reverse biases both JE and JC.Thus, this circuit represents the cut-off of p-n-p transistor.
6. (a)Given,Collector current, IC = IE + ICO
This relationship holds good for the BJToperating in active region of operation.
7. (c)In a MOSFET, source region supplies thecharge carriers to the drain region throughthe channel formed between the two regions.Thus, the inversion layer formed between thesource and drain has the polarity same asthat of the majority carriers in the source.
=CE
C
B V constant
II
8. (a)Carrier concentration in a semiconductor is afunction of both time and distance andcontinuity equation is derived to establish therelationship between carrier concentration andtime & distance. It is given by
p p p0 n
n
dn n n dJ1dt q dx
For a p typesemiconductor
2
pp p p0 pn n2
n
d n Edn n n d nD
dt dxdx
...(i)
pn p n n
dnJ n q E q D
dx
When light is made to fall on a p-typesemiconductor continuously with electric field(E) = 0, Equation (i) becomes
(D) and mobility is given by the Einsteinequation as
p nT
p n
D D kT Vq
Where VT is the Volt-equivalent oftemperature.
At equilibrium, the product of free negativeand positive charge concentrations is aconstant independent of the donor andacceptor impurity doping. This relationship iscalled the Mass-action law and is given by
2inp n , where
ni is the intrinsic carrier concentration.
Charge neutrality equation: A semiconductoris neutral, that is, the magnitude of positivecharge density is equal to that of the negativecharge concentration, or
p + ND = n + NA,
where NA (ND) is the acceptor (donor) impurityconcentration.
10. (b)The effective mass of a charge carrier is given
by
2*
2 2
h 2m
d E dk
Thus, the curvature of the band determines theeffective mass.
From the graph shown,
2 2
2 2conduction valenceband band
d E d Edk dk ...(1)
Free electrons in the conduction band and holesin the valence band serves as carriers of current.
2 2* *n p2 2 2 2
Conduction ValenceBand Band
h 2 h 2m m
d E dk d E dk
11. (b)Here ND >> ni, thus free electron density in
equilibrium is given by
n = ND = 1014/cm3
According to mass action law,
np = 2in
p = 2 2i i
D
n nn N
= 210
141010
= 106/cm3
12. (c)Indirect transition via a discrete level :
E
Et
k2
1
In an indirect bandgap semiconductor, electrontransition from the conduction band to the valenceband occurs in two steps : One, change inmomentum & energy and second, change in energy.
In the indirect transition involving a change inmomentum, part of the energy is given as heat tothe lattice rather than as an emitted photon. Thus,indirect bandgap semiconductors are not used inoptoelectronic devices.
13. (c)The resistivity of a semiconductor is given by
= n p
1 1nq pq
D n
1N q
n D n D
for an n type semiconductorn N and p N
= 16 191
2 10 1.6 10 1300 = 0.24 cm
14. (d)Gold is extensively used as a recombination
agent in semiconductor devices. This is because,by introducing gold into silicon under controlledconditions, designers can obtain desired carrierlifetimes.
15. (d)The carrier concentration in the semiconductor
is a function of both time and distance. Thedifferential equation which governs this functionalrelationship is called the continuity equation. Thisequation is based on the fact that charge can neitherbe created nor destroyed. Thus, continuity equation
16. (c)Let the reverse saturation current of a pn junction
diode at a temperature (T1) of 25°C be I0,25. Then,Reverse saturation current at a temperature (T2) of80°C is given by
I0, 80 =2 1T T10
0,25I 2
=80 25
100,25I 2
= 5.50,25I 2
= 45 × I0,25
17. (b)
8V
100
R
IS
IL
IZ
V =4VZ
+
–
From the circuit,
IS =
8 4 40mA100
Load current, IL = IS – IZThus, IL,max = IS – IZK = 40m–10m
= 30mA
Minimum value of R so that the voltage acrossit does not fall below 4V is given by
Rmin =Z
L,max
V 6 200I 30m
18. (a)
Cut-in voltage of various types of diodes is givenbelow:
1. Germanium diode:- 0.2V
2. Silicon diode :- 0.6V
3. Schottky diode :- 0.3 V
4. Tunnel diode :- 0V
19. (b)Transition region in a p-n junction diode consists
of immobile acceptor and donor ions in p-side andn-side respectively.
+ ++ ++ ++ ++ ++ +
p-type n-type
x = –Wp x = Wnx = 0
Since the net charge must be zero,
e NA Wp = e ND Wn
p
n
WW =
D
A
N 1N
Wp < Wn20. (b)
The expression for the depletion layer width isgiven by
jA D
2 1 1W V ,q N N
Where Vj = Vo+VR, for reverse bias voltage of VR
= Vo–VF, for forward bias voltage of VF
21. (d)Forward current, ID = 5mA
VT 26mV (at room temperature)
Dynamic or AC resistance of a forward biasedSilicon diode is given by
rac =T
D
VI
=2 26m
5m 2 for Silicon and
1 for Germanium
= 10.422. (c)
Depletion capacitance:- It is due to the presenceof uncovered immobile charges present in thedepletion region. It comes into effect when the diodeis reverse biased. In forward biased diode, it isnegligible. It is also called as Space-charge /Transition capacitance.
Diffusion capacitance:- It occurs due to thestorage of minority carriers outside the depletionregion. It comes into effect when the diode is forwardbiased. It is also called as storage capacitance.
23. (d)Symbols for various types of diodes is shown
below:
Varactor diode :-
It represents the variable capacitance of thediode.
Ls Series inductance and dependsupon the lead length and geometryof the diode package
CS Junction capacitance, dependsupon the applied bias and ismeasured usually at the valley point.
–Rn Negative resistance of the Tunneldiode.
Equivalent circuit of a Varactor diode:-
CT
Rr
RS
(Reverse resistance)Ohmic/Contactresistance,
(Transition capacitance)Small-signal equivalent of normal p-n junction
diode:-
r
CD Tor C
(Dynamic or ACresistance)
D
T
C Diffusion capacitanceC Transition capacitance
25. (a)Pinch-off voltage (VP) is the diode reverse
voltage that removes all the free charge carriersfrom the channel. It is given by
VP = 2Dq Na ,
2Ewhere, ND is the dopant atom concentration in
the n-channel, a is half of the distance between thetwo gate terminals.
26. (b)
RC
10V
5V +–
200k CB
E
+
–VCE,sat+
–VBE,sat
Applying KVL in Base-emitter loop,
5 – 200K × IB – VBE,sat = 0
IB =BE,sat5 V 5 0.8
200 k 200 k
IB = 21 ACollector saturation current,
IC,sat = B.I 100 21 A
= 2.1 mA
For transistor to remain in saturation,IC IC,sat
CE,sat
C
10 VR
IC,sat
RC CE,sat
C,sat
10 V 10 0.2I 2.1m
RC 4.67 k
RC, min = 4.67 k27. (d)
Doping concentration of emitter region is veryhigh because it supplies the charge carriers.
Doping concentration of Base region is very lowso as to prevent recombination of charge carrierssupplied by the emitter in the Base region.
Doping concentration of collector region ismoderate.
Thus,18 35 10 cm Emitter region
17 310 cm Collector region7 32 10 cm Base region
28. (c)
Biasing of a transistor is done so as to establishoperating point (Quiescent point or Q-point) in theactive region of operation of BJT. This is done soas to improve the stability so that Q-point is stableand BJT can work as an amplifier.
29. (b)In semiconductors, mobility of electrons is
approximately 2.5 times that of holes.
In semiconductors, increase in temperatureincreases more number of electron-hole pairs(EHPs). Thus, its conductiv ity increases andresistivity decreases with increase in temperature.
Metals have positive temperature coefficient ofresistance and are of much smaller magnitude i.e.+ 0.4%/°C
30. (d)Collector leakage current in common-emitter
configuration, ICEO = 500 A
ICEO = CBO1 I ,where ICBO is the collector leakage current in
common-Base configuration
500 = (1 + 99) × ICBO
ICBO = 5 A31. (d)
Given, = 0.98
=0.98 49
1 1 0.98
Collector current, IC is given by
IC = B CBOI 1 I
= 49 100 (1 49) 5
= 4900 250
= 5150 A 5.15mA
32. (d)
There are two types of breakdown possible in atransistor : Avalanche breakdown and Punchthrough. In the Avalanche breakdown mechanism,the Avalanche multiplication factor depends on thevoltage VCB between collector and base and is givenby
nCB CBO
1M ,1 V V
where VCB is the collector-to-base voltage,
VCBO is the maximum reverse biasing voltagethat can be applied before breakdown between thecollector and base terminals
and n is the parameter that controls thesharpness of the onset of breakdown and rangesbetween 2 to 10.
33. (d)When the emitter-base junction (JE) is forward
biased and the collector-base junction (JC) is reversebiased, transistor is said to be working in the activeregion of operation. In the active region, transistoris used as an amplifier.
Transistor is said to be working in cut-off regionof operation, when both JE and JC are reverse-biased.
Transistor is said to be working in saturationregion of operation, when both JE and JC are forwardbiased.
NOTE:- Transistor is used as a switch, whenit is made to operate in the saturation and cut-
off region.Transistor is operating in inverse-active region,
when JE is reverse biased and JC is forward biased.In this region of operation, it works as a low gainamplifier.
34. (c)Input static characteristic:- Plot of input voltage
versus input current for different values of outputvoltage.
Output static characteristic:- Plot of outputcurrent versus output voltage for different values ofinput current.
Transistor Input Output Input static Output staticConfiguration Side Side Characteristics Characteristics
Plot of V Vs. Plot of I Vs.cEBCommon Base Emitter Collector I for different V for differentE CBconfiguration values of V . vCB
alues of I .EPlot of I Vs. VPlot of V Vs.Common C CEBE
emitter Base Collector I for different for differentBconfiguration values of Ivalues of V BCE
Plot of V Vs. PCommon CBCollector Base Emitter I for differentBconfiguration values of VCE
CElot of I Vs. V forEdifferent valuesof I .B
35. (a)If IC,sat and IB can be determined independently
from the circuit under consideration, the transistor
is said to be in saturation if C,sat BI · I
36. (c)The increase in magnitude of collector reverse
voltage increases the space-charge width at theoutput junction diode. Such an action causes theeffective Base width to decrease. This phenomenonis called as Early effect or Base width modulationConsequences of Early effect :-
(i) Less chance of recombination within the
Base region. Hence, the transport factor * and
increases with an increase in magnitude ofjunction voltage.
(ii) The charge gradient is increased within thebase and consequently, the current of minoritycarriers injected across the emitter junction increase.
(iii) At a certain collector voltage, the transitionlayer covers whole of the base region and thuscollector and emitter are effectively shorted. Thisphenomenon is called punch-through or Reach-through and due to this, the transistor action ceases.
37. (b)
There is no channel present between the sourceand drain in an Enhancement type MOSFET (E-MOSFET). Thus, no drain current flows at VGS = 0VTransfer characteristic of E-MOSFET :-
Threshold voltage of N-channel E-MOSFET ispositive and that of p-channel E-MOSFET isnegative.
From the transfer characteristic, drain currentflows only for +ve values of VGS (n-channel E-MOSFET) and only for –ve values of VGS (p-channelE-MOSFET). Thus, E-MOSFET works only inEnhancement mode.
38. (a)In most MOS fabrication process, the threshold
voltage can be adjusted by selective dopant ionimplantation into the channel region of the MOSFET.
For n-channel MOSFETs, the threshold voltagecan be increased (made more positive) by addingextra p-type impurities (acceptors) into the channelregion.
Alternatively, the threshold voltage of n-channelMOSFET can be decreased (made more negative)by implanting n-type impurities (dopant ions) intothe channel region.
39. (d)In the transistor operation, temperature of the
collector junction increases because of self-heating.This leads to increase in collector current and thusincrease in power dissipation. More junction powerdissipation further increases the junctiontemperature.
This is a cumulative process and to preventexcessive temperature rise, a heat sink is generallyused with a transistor.
40. (d)JFET has the main drawback of small gain-
bandwidth product. The gain-bandwidth product ofthe JFET is limited by the presence of Millercapacitance, which comes into effect at highfrequency.
41. (b)Field-effect transistor is a semiconductor device
which depends, for its operation, on the control ofcurrent by an electric field.
In JFET, the change in drain current is due tothe applied electric field between gate and source.
42. (b)For the MOSFET, drain current is given by
ID = 2DS
n ox GS T DSVWC . V V V
L 2
, in linear
region
ID= 2n oxGS T DS
C W. V V 1 V2 L
,
in saturation region
43. (d)In a semiconductor, current flow due to drift and
diffusion phenomena.
Total current, due to holes, is given by
Jp = Jp,drift + Jp,diffusion
= p pdppq E q Ddx
Total current, due to free electrons, is given by
Jn = Jn,drift + Jn,diffusion
= n ndnnq E q Ddx
44. (c)The diode current of p-n junction is given by
ID = TV VoI e 1
where Io is the reverse saturation current,
V is the voltage across the diode,
VT is volt equivalent of temperature
and = 1 (for Germanium) and =2 (forSilicon)
45. (a)RMS output voltage of the bridge full-wave
rectifier is given by
Vrms = mV2
Vm = rms2V Vm = 2 20V
= 28.2 V
Peak Inverse voltage (PIV) across the diode,for a bridge full-wave rectifier, is given as
58. (c) For equivalent resistance between P & S, thegiven network can be redrawn as,
40 40
40
40 20
20 S
P
RPQ = 20 + (80 || 80) + 20
= 20 + 40 + 20
= 80
59. (c)In Thevenin’s equivalent circuit:
(1) Independent voltage sources are replacedwith their internal resistances or, idealvoltage sources are short-circuited.
(2) Independent current sources are replacedwith their internal resistances or, idealcurrent sources are open-circuited.
(3) Dependent voltage or, current sources arekept as it is.
60. (b)61. (b)
+–
ZTh+
–
+
–
VTH ZN
N =Th
Th
VZ & ZTH = ZN
62. (b)For RTh:Short-circuiting the ideal voltage sources,
A
B
10 5
RTH = (10) || (5) = 5015 =
103
For VTh:
+– 10V5V +
–A
B
10 5
I
I =
(10 5) 510 5 15
1 A3
VAB – 10I – 5 = 0
VAB =
110 53 =
25 V3
63. (b)For voltage soruce to be an ideal voltagesource, the equivalent impedance connectedin series with voltage source to be zero, sothat there will be no voltage drop across it.For this condition, the circuit to be inresonance,
The current flowing in the loop will be 1Aclockwise.
so, Vab – 2Vab + 10 × 1 – 5 = 0
–Vab = –10 + 5 = –5
Vab = 5 Volt
67. (d)Given circuit can be redrawn as,
R1
I
V V
I
Shortcircuit
So, the current delivered by source,
= V0 (Infinity)
68. (a)The circuit shown in question, can be redrawnas,
b
d
2
2
2 2
I
a n c
3
10V
Here, Rbn = Rcn = Rdn =
6 6
6 6 6
= 36 218
Req = [(4) || 5] + 2
= 20 3829 9
Current, =eq
VR
= 10
(38 / 9)
= 90 2.34 A38
69. (b)For the circuit shown,
(i) The current through load is 5A as thecurrent source is in series to the load.
(ii) Voltage across the load cannot be find, asvoltage across the current source is not
given.
(iii) Power delivered by the voltage source
P = VI = (20) × (–5)
= – 100W
i.e., Power absorbed by voltage source =100W
so, 2 and 3 are correct.
70. (b)Since, the current ‘i’ leaves the coil L1 throughthe dot end and enter the coil L2 through thedot end, hence the mutual inductance betweenthe coils will be negative.So, net inductance,
115. (d)Electric field exists from high potential to lowerpotential.
VP > VQ and hence VPQ is negative (VPQ= VQ – VP)Therefore VPQ is negative, there is a loss inpotential energy in moving unit charge from Pto Q and hence work is being done by theelectric field.
116. (d)Equipotential surface is a surface which iscut by the electric field perpendicularly.
Sphere
+q
The equipotential surface due to a pointcharge will be concentric spheres as shownabove.
117. (d)
W = F.ds
W = q E.ds
W = 9 ˆ ˆ ˆ ˆ ˆ ˆ10 4i 3j 2k . 10i 2j 7k m
W = 40 6 14 nJ
W = 20 nJ118. (b)
Energy stored in the field is given by
U =1 2
0
Q Q14 r
U =
129
2 2 2
4 1 109 101 2 3 1 1 5
U =336 10 36 mJ
749
U = 5.14 mJ119. (b)
Electric flux through the complete sphere willbe :
=0 0
encQ QE.ds
Therefore through the hemisphere will be
0
Q2 .
120. (d)Z
E1
E2
r1 = 4
r2 = 3
Normal component of 1 inE E 3
.
E = n tE E
1tE = 1 nˆ ˆE E 5i 2j
Tangential component of E is continuous
across the boundary.
E1t = E2t
E2t = ˆ ˆ5i 2j
Also, normal component of electric flux densityis equal across the boundary.
E1t = E2t
E2t = ˆ ˆ5i 2j
Also, normal component of electric flux densityis equal across the boundary.
123. (c)Electric field inside a conductor is always zeroand electric field lines can cut the conductorsurface at right angle only.Electric field exists inside a dielectric material.
124. (b)
Net potential at 0 = 0 0
2QQ4 a 4 b
V =0
Q 1 24 a b
125. (c)
Applying Gauss’ law to a arbitrary Gaussiancylindrical surface of radius a b , wehave
Q = E.ds E 2 L
E =Q a
2 L
V =a a
b b
Q ˆ ˆE.d a .d a2 L
l
V =a
b
Q d Q bn2 L 2 L a l
Capacitance, C = Q 2 L
bV lna
126. (d)127. (d)128. (b)
Ampere’s Law :
I = lC
HdBiot-Savart Law :
H =
l r2
Id a4 r
Coulomb’s Law :
F =1 2 1 2
21 122 20 0
Q Q Q Qa or a4 R 4 R
where a12 and a21 are unit vectors along theline joining Q1 and Q2 .Gauss’s Law :
132. (c)Bolts are insulated, so as to reduce the eddycurrent losses as the bolts are made ofconducting materials, and they are placed influx, there is voltage induced. They are insulatedto recduce the eddy current and eddy currentlosses.
133. (b)
This is a cruciform, here the shape tends tocircle. For a given area, circle occupies leastcircumference, so the amount of copper is less,
134. (b)
Efficiency =output
output losses
In the case of two winding and auto-transformer,the losses remain same. But as the output MVAof an auto-transformer is more than two windingtransformer, the efficiency increases
135. (d)
Resistance, R =VI =
lE d
J ds
=
lE d
E ds
Capacitance, C =QV =
l
E ds
E d
So, RC =
l
l
E d E ds
E ds E d
i.e. RC =
136. (d)If the pole tips are chamfered, there is leastgap between pole shoe and armature at the
polar axis, and the gap increases while movingto the pole tips. Correspondingly, reluctance isminimum at the polar axis and increases whileapproaching to pole tips. So, the flux density ismaximum at polar axis and minimum atinterpolar region and the air-gap flux densitydistribution achieve nearly a sinusoidalwaveform.
137. (a)
q1 q2
q3q4
0.04m
0.04m0.04 2 m
W =4
n nn 1
1 q V2
V1 = V21 + V31 + V41=
0
q 1 1 14 0.04 0.04 0.04 2
=0
q 124 (0.04) 2
For the given distribution of charges,V1 = V2 = V3 = V4
Voltage measured by Pressure coil will beVPC = Vload + ICC RCC
= 200 + (20) (0.02)= 200 + 0.4= 200.4
Measured power = ICC VPC = (20) (200.4) = 4008
True power consumed by load = ICC Vload= (20) (200)= 4000 V
% Error =
4008 4000 100 0.2%4000
163. (b)(1) High torque to weight ratio leads to higher‘accuracy’.(2) In PMMC jewels are used in bearing toreduce wear and tear.(3) Swamping resistance is used to negatethe effect of temperature in an ammeter as ithas low thermal emf, hence it provides thermalstabilisation.(4) Overdamped galvanometer is sluggish innature. Hence it should be either criticallydamped or slightly under-damped.
164. (c)In MI instrument, deflecting torque
Td =
21 dLI2 d
and controlling torque TC = K At equilibrium Td = Tc
21 dLI
2 d = K
…(1)For a linear scale,
I = C (i.e. I ) …(2)
from (1) and (2)
2 21 dLC2 d = K
dLd = 2
2 KC
( = constant)
Hence plot of
dLd constant will be a
rectangular hyperbola.165. (c)
Voltage Sensitivity (SV) =20000 / VV = 100 V
Vfullscale = 1000
Rm = V· Sv = 2 MRm + Rs = (Vfull scale) × (Sv)
= 20 M
Rs = 18 M
166. (d)i. The sacle is uniformly divided
ii. The power consumption is very low (as25 W to 200 W )
iii. The torque to weight ratio is high whichgives high accuracy
iv. Since the operating forces are large onaccount of large flux densities which maybe as high 0.5 Wb/m2 the errors due tostray magnetic fields are small
167. (c)168. (c)
LOAD
CC
Supply
Compensatingcoil
PCIP
I+Ip
The compensating coil is connected in serieswith the pressure coil circuit and is made asnearly as possible identical and coincident withthe current coil. It is so connected that itopposes the field of the current coil. Thecompensating coil carries a current Ip andproduces a field corresponding to this current.This field acts in opposition to the current coilfield. Thus the resultant field is due to currentI only. Hence the error caused by the pressurecoil current flowing in the current coil is
In the above connection the pressure coil ofthe wattmeter is on the supply side andtherefore voltage applied to the pressure coilis voltage across the load plus the voltage dropacross the current coil.
Hence if load impedance is high i.e., loadcurrent is small the voltage drop in the currentcoil is small, so that the above connection givesvery small error.
170. (a)Tunnel diode is made up of degenerate
semiconductor because of which the depletion layerformed is very narrow. Charge carriers can crossthe junction almost with a speed of light and thisprocess is called as tunneling. Because of thetunneling phenomenon, V-I characteristic possessnegative resistance property. Tunnel diodes can beused as microwave oscillator when working innegative resistance region.
171. (a)In Zener breakdown mechanism, an increase
in temperature increases the energies of the valenceelectrons and hence makes it easier for theseelectrons to escape from the covalent bonds. Lessapplied voltage is therefore required to pull theseelectrons from their positions in the crystal latticeand convert them into conduction electrons.
Thus, Zener breakdown voltage decreases within crease in temperature and is said to have NegativeTemperature Coefficient (NTC) of breakdownvoltage.
172. (a)As the collector junction reverse voltage is
increased, transition layer width increases andpenetrates more into the base region (as base regionis lightly doped). At relatively large voltage, transitionregion will spread completely across the base toreach the emitter junction. This phenomenon iscalled punch-through or reach-through. Thus,collector and emitter are effectively shorted andtransistor action cease. Transistors usefulness isthus terminated due to Punch-through.
173. (b)Drain current (IDS) has negative temperature
coefficient. This is because mobility of the chargecarriers decreases with increase in temperature. Dueto this, thermal runaway is not encountered in field-effect transistors.
174. (a)Reverse saturation current (I0) in a silicon diode
is in the range of nA, whereas that in a germaniumdiode is in the range of A .
An increase in temperature f rom roomtemperature (25°C) to 90°C, I0 increases to hundredsof A for a Ge diode, whereas in silicon diode, I0rises only to tenths of A .
Thus, silicon diodes are preferred to germaniumdiodes for high temperature operation.
175. (d)176. (d)177. (b)
In PMMC, Magnetic field produced is quite
strong and about 0.1 to 1 T 2Wb
m
Electrodynometer instrument produces weak
magnetic field of about 0.005 to 0.006 2Wb
mand hence needs to be shielded from straymagnetic fields.But ‘R’ is not correct explanation of ‘A’ henceoption ‘b’.
178. (b) Shunt enhancement allows the ammeter tomeasure currents higher than the metercurrent.
I Im
Rsh
I sh
Rm
A
I is measured, which is larger than the fullscale meter currents Im.The voltage drop across the meter for fullscale deflection always remains ImRm.
179. (c)PMMC instruments can be used only for dcbecause the torque reverses if the currentreverses. If the instrument is connected to a.c.,the pointer cannot follow the rapid reversalsand the deflection corresponds to the mean
torque, which is zero. Hence, the instrumentscannot be used for a.c.
180. (b)Since the current of the moving coil is carriedby the instrument springs, it is limited to valueswhich can be carried safely by springs withoutappreciable range. Therefore, a series resistoris used in the voltage circuit and the current islimited to a small value, usually upto 100 mA.
The fixed coils are used as current coilsbecause they can be made more massive andcan be easily constructed to carry considerablecurrent since they present no problem ofleading the current in or out.
