single phase transformer
DESCRIPTION
By gurpreet Singh from FaridkotTRANSCRIPT
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TransformerTransformer
Muhamad Zahim SujodExt : 2312A1-01-06
BEE2123 ELECTRICAL MACHINES
BEE2123 ELECTRICAL MACHINES
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Learning OutcomesLearning Outcomes
At the end of the lecture, student should to: Understand the principle and the nature of principle and the nature of
static machines of transformer.static machines of transformer.
Perform an analysis on transformers which their principles are basic to the understanding of electrical machineselectrical machines.
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IntroductionIntroduction A transformer is a static machinesstatic machines. The word ‘The word ‘transformertransformer’ comes form the word ‘’ comes form the word ‘transformtransform’.’. Transformer is not an energy conversion devicenot an energy conversion device, but is a
device that device that changeschanges AC electrical power at one voltage AC electrical power at one voltage level into AC electrical power at another voltage level level into AC electrical power at another voltage level through the action of magnetic field, without a change in through the action of magnetic field, without a change in frequency. frequency.
It can be either to It can be either to step-upstep-up or or step downstep down..
Generation Generation StationStation
TX1 TX1
DistributionsDistributions
TX1
TX1
Transmission Transmission SystemSystem
33/13.5kV33/13.5kV 13.5/6.6kV13.5/6.6kV
6.6kV/415V6.6kV/415V
ConsumerConsumer
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Transformer Construction Two types of iron-core construction:
a) Core - type construction
b) Shell - type construction
Core - type construction
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Transformer Construction
Shell - type construction
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Ideal Transformer
An ideal transformer is a transformer which has no loseswhich has no loses, i.e. it’s winding has no ohmic resistance, no magnetic leakage, and therefore no I2 R and core loses.
However, it is impossibleimpossible to realize such a transformer in practice.
Yet, the approximate characteristic of ideal transformer will approximate characteristic of ideal transformer will be used in characterized the practical transformer.be used in characterized the practical transformer.
VV11 VV22
NN1 1 : N: N22
EE11 EE22
II11 II22
VV11 – Primary Voltage – Primary Voltage
VV22 – Secondary Voltage – Secondary Voltage
EE11 – Primary induced Voltage – Primary induced Voltage
EE22 – secondary induced Voltage – secondary induced Voltage
NN11:N:N22 – Transformer ratio – Transformer ratio
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Transformer Equation
Faraday’s Law states that, If the flux passes through a coil of wire, a voltage will be
induced in the turns of wire. This voltage is directly proportional to the rate of change in the flux with respect of time.
If we have NN turns of wire,
dt
tdEmfV indind
)(
dt
tdNEmfV indind
)(
Lenz’s Law
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Transformer Equation
For an ac sources, Let V(t) = Vm sint
i(t) = im sint
Since the flux is a sinusoidal function;
Then:
Therefore:
Thus:
tt m sin)(
tNdt
tdNEmfV
m
mindind
cos
sin
mmindind fNNEmfV 2(max)
mmm
rmsind fNfNN
Emf
44.42
2
2)(
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Transformer Equation For an ideal transformer
In the equilibrium condition, both the input power will be equaled to the output power, and this condition is said to ideal condition of a transformer.
From the ideal transformer circuit, note that,
Hence, substitute in (i)
m
m
fNE
fNE
22
11
44.4
44.4
1
2
2
1
2211 coscos
I
I
V
V
IVIV
poweroutputpowerInput
………………… (i)
2211 VEandVE
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Transformer Equation
aI
I
N
N
E
ETherefore
1
2
2
1
2
1,
Where, ‘aa’ is the Voltage Transformation RatioVoltage Transformation Ratio; which will determine whether the transformer is going to be step-up or step-down
EE11 > E > E22For a >1For a >1
For a <1For a <1 EE11 < E < E22
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Transformer Rating
Transformer rating is normally writtenwritten in terms of Apparent Apparent PowerPower.
Apparent power is actually the product of its rated current its rated current and rated voltageand rated voltage.
2211 IVIVVA Where,
I1 and I2 = rated current on primary and secondary winding.
V1 and V2 = rated voltage on primary and secondary winding.
Rated currents are actually the Rated currents are actually the full load currentsfull load currents in in transformertransformer
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Example
1. 1.5kVA single phase transformer has rated voltage of 144/240 V. Finds its full load current.SolutionSolution
AI
AI
FL
FL
6240
1500
45.10144
1500
2
1
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Example
2. A single phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60m2. If the primary winding is connected to a 50Hz supply at 520V, calculate:a) The induced voltage in the secondary winding
b) The peak value of flux density in the core
SolutionSolution
N1=400 V1=520V A=60m2
N2=1000 V2=?
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Example 2 (Cont)a) Know that,
b) Emf,
2
520
1000
400
V
2
1
2
1
V
V
N
Na
VV 13002
2
21
/976.0
)60)()(400)(50(44.4520
44.4
1300,520,
44.4
44.4
mWbB
B
ABfNE
VEVEknown
ABfN
fNE
m
m
m
m
m
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Example
3. A 25kVA transformer has 500 turns on the primary and 50 turns on the secondary winding. The primary is connected to 3000V, 50Hz supply. Find:
a) Full load primary current
b) The induced voltage in the secondary winding
c) The maximum flux in the core
SolutionSolution
VA = 25kVA
N1=500 V1=3000V
N2=50 V2=?
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Example 3 (Cont)a) Know that,
b) Induced voltage,
c) Max flux
VI
IEE
AI
I
I
N
Na
3003.83
33.83000
3.8350
33.8500
2
112
2
1
2
2
1
AV
VAI
IVVA
FL 33.83000
1025 3
11
mWb
fNE
27
)50)(50(44.4300
44.4
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Practical Transformer (Equivalent Circuit)
V1 = primary supply voltage
V2 = 2nd terminal (load) voltage
E1 = primary winding voltage
E2 = 2nd winding voltage
I1 = primary supply current
I2 = 2nd winding current
I1’ = primary winding current
Io = no load current
Ic = core current
Im = magnetism current
R1= primary winding resistance
R2= 2nd winding resistance
X1= primary winding leakage reactance
X2= 2nd winding leakage reactance
Rc= core resistance
Xm= magnetism reactance
VV11
II11 RR11XX11
RRCC
IIcc
XXmm
IImm
IIoo
EE11 EE22
VV22
II11’’
NN11: N: N22
RR22
XX22
LoadLoad
II22
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Single Phase Transformer (Referred to Primary) Actual MethodActual Method
22
22
2
2
12 '' RaRORR
N
NR
22
22
2
2
12 '' XaXORX
N
NX
a
II
aVVORVN
NVE
22
2222
1'21
'
'
VV11
II11 RR11 XX11
RRCC
IIcc
XXmm
IImm
IIoo
EE11 EE22 VV22
II22’’ NN11: N: N22
RR22’’ XX22
’’
Load
II22
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Single Phase Transformer (Referred to PrimaryPrimary) Approximate MethodApproximate Method
22
22
2
2
12 '' RaRORR
N
NR
VV11
II11 RR11XX11
RRCC
IIcc
XXmm
IImm
IIoo
EE11 EE22 VV22
II22’’ NN11: N: N22RR22
’’ XX22’’
Load
II22
22
22
2
2
12 '' XaXORX
N
NX
a
II
aVVORVN
NVE
22
2222
1'21
'
'
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Single Phase Transformer (Referred to PrimaryPrimary) Approximate MethodApproximate Method
22
22
2
2
12 '' RaRORR
N
NR
VV11
II11RR0101 XX0101
aVaV22
22
22
2
2
12 '' XaXORX
N
NX
'
'
2101
2101
XXX
RRR
2222
1'2 ' aVVORV
N
NV
In some application, the excitation branch has a small current compared to load current, thus it may be neglected without causing serious error.
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Single Phase Transformer (Referred to SecondarySecondary) Actual MethodActual Method
21
11
2
1
21 ''
a
RRORR
N
NR
a
VVORV
N
NV 1
111
21 ''
I1’ R1’X1’
RC’
Ic
Xm’
Im
Io
I2 R2
X2
VV22
21
11
2
1
21 ''
a
XXORX
N
NX
a
V1
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Single Phase Transformer (Referred to SecondarySecondary) Approximate MethodApproximate Method
21
11
2
1
21 ''
a
XXORX
N
NX
2102
2102
'
'
XXX
RRR
II11’’ RR0202 XX0202
VV22
a
V1
21
11
2
1
21 ''
a
RRORR
N
NR
a
VVORV
N
NV 1
111
21 ''
11 ' aII
Neglect the excitation branch
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Example
4. For the parameters obtained from the test of 20kVA 2600/245 V single phase transformer, refer all the parameters to the high voltage side if all the parameters are obtained at lower voltage side.
Rc = 3.3, Xm =j1.5, R2 = 7.5, X2 = j12.4
Solution
Given
Rc = 3.3, Xm =j1.5,
R2 = 7.5, X2 = j12.4
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Example 4 (Cont)
i) Refer to H.V side (primary)
R2’=(10.61)2 (7.5) = 844.65,
X2’=j(10.61)2 (12.4) = 1.396k
Rc’=(10.61)2 (3.3) = 371.6,
Xm’=j(10.61)2 (1.5) = j168.9
61.10245
2600
2
1
2
1 V
V
E
Ea
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Power Factor
Power factor = angle between Current and angle between Current and voltagevoltage, cos
V
I
= -veV
I
= +ve
VI
= 1
Lagging Leading unity
Power factor always lagging for real transformer.
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Example
5. A 10 kVA single phase transformer 2000/440V has primary resistance and reactance of 5.5 and 12 respectively, while the resistance and reactance of secondary winding is 0.2 and 0.45 respectively. Calculate:
i. The parameter referred to high voltage side and draw the equivalent circuit
ii. The approximate value of secondary voltage at full load of 0.8 lagging power factor, when primary supply is 2000V.
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Example 5 (Cont)
SolutionSolution
R1=5.5 , X1=j12
R2=0.2 , X2=j0.45
i) Refer to H.V side (primary)
R2’=(4.55)2 (0.2) = 4.14,
X2’=j(4.55)20.45 = j9.32
Therefore,
R01=R1+R2’=5.5 + 4.13 = 9.64
X01=X1+X2’=j12 + j9. 32 = j21.32
55.4440
2000
2
1
2
1 V
V
E
Ea
V1 aV2
R01 X01
21.329.64
I1
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Example 5 (Cont)
SolutionSolution
ii) Secondary voltage
p.f = 0.8
Cos = 0.8
=36.87o
Full load,
From eqn. cct,
AV
VAIFL 5
2000
1010 3
1
o
oo
oo
V
Vj
aVIjXRV
8.06.422
)55.4()87.365)(32.2164.9(02000
))((0
2
2
2101011
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Transformer Losses Generally, there are two types of losses;
i.i. Iron lossesIron losses :- occur in core parameters
ii.ii. Copper lossesCopper losses :- occur in winding resistance
i.i. Iron LossesIron Losses
ii.ii. Copper LossesCopper Losses
circuitopenccciron PRIPP 2)(
022
2012
1
22
212
1
)()(,
)()(
RIRIPreferredifor
PRIRIPP
cu
circuitshortcucopper
Poc and Psc will be discusses later in transformer test
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Transformer Efficiency
To check the performance of the device, by comparing the output with respect to the input.
The higher the efficiency, the better the system.
%100cos
cos
%100
%100,
22
22
cuc
lossesout
out
PPIV
IV
PP
P
PowerInput
PowerOutputEfficiency
%100cos
cos
%100cos
cos
2)(
)(
cucnload
cucloadfull
PnPnVA
nVA
PPVA
VA
Where, if ½ load, hence n = ½ ,½ load, hence n = ½ , ¼ load, n= ¼ ,¼ load, n= ¼ , 90% of full load, n =0.990% of full load, n =0.9Where Pcu = Psc
Pc = Poc
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Voltage Regulation
The purpose of voltage regulation is basically to determine the percentage of voltage drop between no load and full load.
Voltage Regulation can be determine based on 3 methods:
a) Basic Defination
b) Short – circuit Test
c) Equivalent Circuit
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Voltage Regulation (Basic Defination)
In this method, all parameter are being referred to primary or secondary side.
Can be represented in either Down – voltage Regulation
Up – Voltage Regulation
%100.
NL
FLNL
V
VVRV
%100.
FL
FLNL
V
VVRV
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Voltage Regulation (Short – circuit Test)
In this method, direct formula can be used.
%100
cos.
1
. V
VRV fpscsc
If referred to primary side
%100
cos.
2
. V
VRV fpscsc
If referred to secondary side
Note that:Note that:
‘–’ is for Lagging power factor‘+’ is for Leading power factor Isc must equal to IFL
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Voltage Regulation (Equivalent Circuit )
In this method, the parameters must be referred to primary or secondary
%100
sincos.
1
.01.011
V
XRIRV fpfp If referred to
primary side
If referred to secondary side
Note that:Note that:
‘+’ is for Lagging power factor‘–’ is for Leading power factor j terms ~0
%100
sincos.
2
.02.022
V
XRIRV fpfp
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Example
6. In example 5, determine the Voltage regulation by using down – voltage regulation and equivalent circuit.
SolutionSolution
Down – voltage Regulation
Know that, V2FL=422.6V
V2NL=440V
Therefore,
%95.3
%100440
6.422440
%100.
NL
FLNL
V
VVRV
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Example 6 (Cont)
Equivalent Circuit
I1=5A R01=9.64 X01 = 21.32 V1=2000V, 0.8 lagging p.f
%12.5
%1002000
)6.0(32.21)8.0(64.95
%100sincos
.1
.01.011
V
XRIRV fpfp
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Example
7. A short circuit test was performed at the secondary side of 10kVA, 240/100V transformer. Determine the voltage regulation at 0.8 lagging power factor if
Vsc =18V
Isc =100
Psc=240W
SolutionSolutionCheck: Check:
Hence, we can use short-circuit methodHence, we can use short-circuit method
,
100100
10000
2
2
scFL
FL
II
AV
VAI
%100
cos.
2
. V
VRV fpscsc
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Example 7 (Cont)
o
scsc
scsc
scscscsc
ofp
fpscsc
IV
P
IVP
thatKnow
Hence
fpGiven
V
VRV
34.82)100)(18(
240cos
cos
cos
,
87.368.0cos,
8.0.
%100cos
.
1
1
1.
2
.
%62.12
%100100
87.3634.82cos18.
oo
RV
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Example
8. The following data were obtained in test on 20kVA 2400/240V, 60Hz transformer.
Vsc =72V
Isc =8.33A
Psc=268W
Poc=170W
The measuring instrument are connected in the primary side for short circuit test. Determine the voltage regulation for 0.8 lagging p.f. (use all 3 methods), full load efficiency and half load efficiency.
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Example 8 (Cont)
.72.786.34.6364.8
64.833.8
72
4.63)33.8)(72(
268cos
cos
cos
,
87.368.0cos,
8.0.
%100cos
.
0101
1
1
1.
2
.
sideprimarytoconnectedbecausejXRjZ
I
VZ
IV
P
IVP
thatKnow
Hence
fpGiven
V
VRV
osc
sc
scsc
o
scsc
scsc
scscscsc
ofp
fpscsc
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Example 8 (Cont)
%68.2%100
2400
)6.0(72.7)8.0(86.32400
20000
%100sincos
.,.2
%68.2%1002400
87.364.63cos72.
%100cos
.,.1
1
.01.011
1
.
V
XRIRVcircuitEquivalent
RV
V
VRVmethodCircuitShort
fpfp
oo
fpscsc
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Example 8 (Cont)
%68.2
%100240
58.233240
%100.
79.058.233
240
24004.6364.887.36
2400
2000002400
,.3
2
2
20111
NL
FLNL
o
ooo
V
VVRV
VV
V
aVZIV
DefinationBasic
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Example 8 (Cont)
%12.97%100)268()5.0(170)8.0)(20000)(5.0(
)8.0)(20000)(5.0(
%34.97%100)268()1(170)8.0)(20000)(1(
)8.0)(20000)(1(
2)(
2)(
loadhalf
loadfull
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Measurement on Transformer
There are two test conducted on transformer.i.i. Open Circuit TestOpen Circuit Test
ii.ii. Short Circuit testShort Circuit test
The test is conducted to determine the parameter of to determine the parameter of the transformerthe transformer.
Open circuit testOpen circuit test is conducted to determine magnetism parameter, RRcc and X and Xmm.
Short circuit testShort circuit test is conducted to determine the copper parameter depending where the test is performed. If performed at primary, hence the parameters are RR0101 and XX0101 and vice-versavice-versa.
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Open-Circuit Test
Measurement are at high voltage sidehigh voltage side From a given test parameters,
Rc Xm
Voc
Ic Im
Voc
Ioccosoc
Ioc
Voc
Ic
Im
Iocsinoc
oc
Note:If the question asked parameters referred to question asked parameters referred to Low voltage sideLow voltage side, the parameters (Rc and Xm) obtained
need to be referred to low voltage sideneed to be referred to low voltage side m
ocm
c
occ
mc
ococm
ococc
ococ
ococ
ococococ
I
VX
I
VR
XandRThen
II
II
Hence
IV
P
IVP
,
,,
sin
cos
,
cos
cos
1
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Short-Circuit Test
Measurement are at low voltage sideat low voltage side If the given test parameters are taken on primary taken on primary
side, Rside, R0101 and X and X0101 will be obtained will be obtained. Or else, vice-versa.
X01R01
For a case referred to Primary side
010101
01
1
,
cos
cos
jXRZ
I
VZ
Hence
IV
P
IVP
scsc
sc
scsc
scsc
scscscsc
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Example
9. Given the test on 500kVA 2300/208V are as follows:
Poc = 3800W Psc = 6200W
Voc = 208V Vsc = 95V
Ioc = 52.5A Isc = 217.4A
Determine the transformer parameters and draw equivalent circuit referred to high voltage side. Also calculate appropriate value of V2 at full load, the full load efficiency, half load efficiency and voltage regulation, when power factor is 0.866 lagging.
[1392, 517.2, 0.13, 0.44, 202V, 97.74%, 97.59%, 3.04%]
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Example 9 (Cont)
A
II
A
II
IVP
o
ococm
o
ococc
ooc
ococococ
2.49
6.69sin5.52
sin
26.18
6.69cos5.52
cos
6.69)208)(5.52(
3800cos
cos
1
Ioccosoc
Ioc
Voc
Ic
Im
Iocsinoc
oc
From Open Circuit Test,
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Example 9 (Cont)
23.421.49
208
39.1126.18
208
m
ocm
c
occ
I
VX
I
VR
Since Voc=208Vall reading are taken on the secondary side
Parameters referred to high voltage side,
21.517208
230023.4'
1392208
230039.11'
22
2
1
22
2
1
E
EXX
E
ERR
mm
cc
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Example 9 (Cont)
AV
VAIFL 4.217
2300
10500 3
11
osc
scscscsc IVP
53.72)4.217)(95(
6200cos
cos
1
From Short Circuit Test,
First, check the First, check the IIscsc
Since IFL1 =Isc , all reading are actually taken on the primary side
42.013.0
53.7244.053.724.217
95
01
j
I
VZ
oo
scsc
sc
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Example 9 (Cont)
Equivalent circuit referred to high voltage side,
VV22’=aV’=aV22VV11
RRcc
13921392 XXmm
517.21517.21
RR0101
0.130.13 XX0101
0.420.42
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Example 9 (Cont)
Efficiency,
%59.97
%1003800)5.0)(6200()866.0)(10500)(5.0(
)866.0)(10500)(5.0(
%100cos
cos
%74.97
%10038006200)866.0)(10500(
)866.0)(10500(
%100cos
cos
23
3
22
1
3
3
ocscL
ocscFL
PPnnVA
nVA
PPVA
VA
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Example 9 (Cont)
Voltage Regulation,
%04.3
%1002300
3053.72cos)95(
%100cos
.1
E
VRV pfscsc