direct current machine first part about machine.pdf

it contains about dc machine

2. « Armature mmf or armature flux is termed as armature reaction.

« Field winding produces magnetic Field (action) which causes emf to bi
induced in armature causing current to flow & production of armature flux

(reaction).

Lu]

+ It has 2 undesirable effects.
(i) net reduction in flux/pole (demag. effect)
(ii) distortion of main field fluxe Reduction in flux per pole reduces generated

voltage & torque.
Shape changes

3. & T under leading pole tip for
(emf & torque proportional to flux) motor emf 1
+ Distortion of main field flux causes sparking during commutation.

* GNA is axis perpendicular to field flux (q-axis)

+ MNA is axis perpendicular to resultant flux moe 0 a Ena
+ coil being SC has non zero
* at no-load I, =0, f,, = 0, fres = fr emf

MNA = GNA

4. Brushes SC the coil placed at GNA

Peal Motor rotation SHA Gen" rotation
8

MNA (on-load) 0 <MNA (on-load)
á v
field current

1
D
pa

LA

q always . (b) 9
lies along brushes
(b) Resultant flux phasoı

(a) Resultant magnetic field lines

At loaded condition

5. for gen”

Motor ¡Generator

Resultant field lines pole &p
B = max, e = max

leading pole tip
For motor & trailing pole tip for gen"

Magnetic field lines in a loaded DC Machine

6. During rotation, the pole edge at which conductor enters the influence of a
pole is called as leading pole edge & the edge at which conductor leaves the
pole is called trailing pole edge.

Magnetic field is strengthened at trailing pole tip in a generator.

Magnetic field is strengthened at leading pole tip in a motor.
At no-load, GNA = MNA

Under loaded condition, MNA is shifted by 6 in the direction of generator
rotation or opposite to motor rotation.

The shift 6 «I, higher the armature current more is the shift.

7. (1) Punching holes in field poles Interpoles (improve comm)
(2) High S pole tip (chamfered poles) | + cancel armature reac” at q-axis
& overcome reactance emp.

(3) Brush shift in dir” of gen rotation
+ long & narrow (tapered shape)

20 fla Z
far(demag ) = 5/4 ap + wd connected in series with
80-20 fla Z arm.
Far( cross ) = 228 N u
fint = Hg(net) : 8 + far — feomp
8 = brush shift (elec. deg)
/

In Zz

Nint ‘la -.
A

8. Embedded in pole shoe
long bars
Cancels AR under pole
_(_Polearc Y (1g, 2)
fcomp = ( pole pitch ) @ 2)
wdg in series with armature
Hg : net mag. field in interpolar air gap

+ lap wdg:
brushes = parallel path = mP
m = degree of multiplexity
m = 1, simplex m = 2, duplex etc
due to circulating current
between paths equalizer Ying is

regd.

* wave winding
brushes = parallel path = 2 m
no circulating current.

10. 1 boosh vemoved

P= Vv Lag

11. ta
pis NV fay,
=M

brush removal
Vra Eng : Same

Pau rating : changes

2 brushu removed
Dr > PxaR
fi: ty,

12. Assume = flux per pole = Bay X area per pole (=)

P = total number of poles PZ
5 E= Ira OO) = Know
Z = total number of conductors

N = rotor speed in rpm Km = machine const =

2, for wave winding

A = total number of parallel paths = P forlap winding

2100

«Nob

PZ
2nA

13. = flux per pole (wb)

N = speed of machine (rpm)
P = number of poles

A = number of parallel paths
A = 2 for wave winding

A =P for lap winding

Z = number of conductors

If speed is given in rad/sec,
Where o = speed (rad/s)

_ gat

27

G)= a) ($a) = Know

14. T= Kmóla

EZ
ER

machine constant ,@ = flux per pole, I, = armature current

Developed power: power that actually gets converted from elect. to mech
or vice versa.

Paw = Ebla = Tag: ©

15. Separately Excited Machine <field wdg is excited by a Separate source
R,

+

R, Ve = Ea — lara — Ven (gen)
V V, = Ea + lara + Vep (motor)
t

Vpp= 2x drop of each brush
= 21 (usually)

16. motor: I, = I, — Ish

Ish = Vt/Rsh
E, = Ve — lata — Ven

gen: l, =I, +lsn
Ish = Ve/Rsh
Ea = Vi + lara + Ven

17. motor: I, = I, — Ish

Ish = Vt/Rsh
E, = Ve — lata — Ven

O
A
Ve

Parallel to armature —

gen: l, =I, + Ign
Ish = Ve/Rsh
Ea = Vi + lara + Ven

18. Generator Motor

| VR; | WR

KCL hake h = h-hh

KVL | Es = V¿+ la ra +Vep | Es = V¿— la Ta Veo

19. field w' in series with motor armature

| fa

Dr: E= V¿ lata — Van — lars

R

Ve Re 4 Armature V

Va = Ea fir, — Vpp (gen)
V, = Ea + lara + Vgp (motor)

20. field windy in series with motor armature

a | 2 E=Ve-Iara — Van — lars
R

Ve Es a Armature V

Va = Ea ra — Vpp (gen)
VA = Ea + lara + Vgp (motor)

21. Series with parallel comb" of shunt field & armature

Va = Vr + Ise (gen)
rer
L, =I, + If (gen")

E, = Va + Iara + Vep (gen)

Va = Vt — IL Tse (motor)
I, = I, — If (motor)
Ea = Va — lata — Ven

22. Series with parallel comb" of shunt field & armature

Va = Vr + Ise (gen)
A
I, =I, + If (gen")

E, = Va + Iara + Vep (gen)

Va = Vt — IL Tse (motor)
I, = I, — If (motor)
Ea = Va — lata — Ven

23. Solve like shunt me
but ore (rotte) Parallel with series comb of armature & series field
Vv irabead of Vo
Ron
L =1, +1
E, = Ve + lara + large + Ven

Long shunt

Iq =1, — If
2 Motor
Eq = Vitara — Tate — Veo

24. * Over Compound (Nse
+ Under Compound (Nse
+ Level Compound (Nge

Pr = Ps + Dre

Series flux aids shunt flux

Dr = Don - Dre

1. Cumulative compound DC machine

Series flux opposes shunt flux

high Ve, > Vi)
low Ve. < VnL)

moderate Ur, = Vwi)

2. Differentially compound DC machine

25. A 4 pole dc generator has 48 slots and 8 conductors per slot. The useful flux
per pole is 30 mWb and speed is 800 rpm the generator emf is E, if the machine
is lap would and it is E, if the machine is wave wound. The values of E, and E,
are respectively Eon: 2 Etap

(a) 307.2 V,153.6V (b) 307.2 V, 307.2 V = 304 2V

(c) 153.6 V,153.6V (dŸ153.6V, 307.2 V

S= 48 N= In rpm Etap = (ng)
Z= 48x8 = 324 a

= 4x 384 (2wx0:03)
fre P= 4 6oxy
$= 003 = 152.6V

26. eT
shunt: Ve: same = same

Ya not au

amume Ta=0
230V, DC Shunt Motor draws an armature current of 50A and runs at a
peed of 1250 rpm. Now if a resistance is connected in series with the armature
e motor runs at a speed of 1150rpm drawing same armature current. Assume
erminal voltage remains same, find the value of external resistance.

Cone-ol V= 2304 fa=s0A E¡= 230- 50 Ya = 220V

d- same EXN

N,= (2890%prv Gy = N
i E 7 UN
Cove-02: N2= I1sDrpm €, = USD y 230 = 2%0-50%;
1250

Tex = 0-368

27. | 2 EE.)

DS
A 4 pole, 220V, 5kW shunt motor gives 5kW when running at 1000 r.p.m
which draws armature and field currents of 25 A and 0.5 A and the armature
resistance is 0.1 Q and a drop of 1 volt per brush. The armature torque of the
machine is N-m.

E= Vi - Lara -Veo
= 220 -25x 0-1-2
= DISSV
Nam: Elo. 2SCx 20 - SEUUEN-m
w 277, x ooo

28. tums = 32X6 =192
Ze 2xturm =284 »A=2
A 8 poles, DC generator has a simplex wave-wound armature containing 32
oils of 6 turns each. Its flux per pole is 0.06 Wb. The machine is running at 250
pm. The induced armature voltage is
(b) 192 V
(d) 768 V

EL

8x 250x 0:06 x 384
60 2

334V

29. Z= SIX20 = l020

k=2

2

A 4 pole dynamo with wave wound armature has 51 slots containing 20
conductors in each slot. The induced emf is 357 volts and the speed is 8500
rpm. The flux per pole will be

(d) 21 mWb 3s+4= 4x B80 xP x (020
60 2

b= (-2mwb

30. 200 he ul
rhe Palo 00 eond. Be in lo parallel path
20 cond” per in Sag
A 10-pole lap wound DC armature has a total of 200 conductors each with a
current carrying capacity of 10 A. The average induced emf per conductor is 10
V. The power rating of the DC machine is kw.

Epash = lox 20 = 200 V

(ac Epath = 200V

a= Fono: Az lox lo= (mA

P= Eg f= 20xleo
= 20KW

31. | Question -07 | Level 2 (Classwork

Fone Ves = 230 = LISA
. Ren 20

50 kW, 230 V DC shunt motor has an armature resistance of 0.1 Q and a field)
esistance of 2000. It runs on no-load at a speed of 1400 rpm, drawing a
urrent of 10A from the mains. When delivering a certain load, the motor,
raws a current of 200A from the mains. Assume that the armature reaction]
ause a reduction in flux/pole of 4% of its no-load value. Then the torque
eveloped at this load in N-m is 228-115 « Dux 1460 - ch
no lad Las fi-fsn

Ree lo-(15= 8-250
Eure 230- O1x LES
=225-115V

32. Hooded faz 980-115 A
= 198:25A 15 Sa

E= 230 - faxo:] = 2l0Sx 1988

= 210-I11ISV « Öxm, 214, 512374

2I0-UC _ d y Na = 299-39 Nm

29105 Oy wo

due to AR = 0:96 ui.

N= 1334-394

33. Bau x oneo./pole =
b=rsx Caxirs)xio4t - /Smwb

e flux density in the pole cores of a four pole d.c. generator is 1.5
b/m?. The size of the pole is 8 cm x 12.5 cm and the leakage coefficient is

.2. If there are 1000 conductors in the lap re armature, calculate the

mf generated, when the speed is 600 rpm. Es eu $(=4)
125 V (b) 150 V
(d) 200 V = Ax ver 0: al
s

= 125V

34. ; Question-09 |

b= 0-9x TX 0-2x02S - 23:56mwb
6

A-P=6

A 10 kW, 6 pole dc generator develops an emf of 200 V at 1500 rpm.
The armature has a lap connected winding. The average flux density over a
pole pitch is 0.9 Tesla. The length and rn E ES armature are 0.25 m and
0.2 m respectively. pied (5)
Calculate : a Tol
(1) the flux per pole > = Baux TI = 6344 Nm
(2) the total number of active conductors in the armature.
(3) the torque developed by the machine when the armature supplies a current

of 50 A. Z= 329.56
(9) 260 = re Ÿ @ %) =240

200= 6x 1S00x 0022356, Z
SA

35. Armature of DC machine is connected as shown below. At no-load speed of
the motor is 1000 rpm. Value of R; = 100 Q, R, = 100 Q. Armature resistance
is 0.50. It is given that at T, = 100 Nm speed of motor is 800 rpm find
R,/R2 = (upto two decimal places. Given R; + R, = 250 Q.

mo Load Ins 0 Ca = Relwonr
Va= 40x Im = 20V
400 „un loot lw
400V le ee

Seria E-20V
Bjr 1000: Em= 20V at lmvpme

36. Loaded. cond" N=gwrpm CE«N) Vas E+ ha
E- Box 800 = l6oV = l60+ sory x OSs 186-12

looo Va-k vv o
Q-tw + Va + So =0
ff = com = corut R ic Y
T= loo Nm -213:32 + 18618 5 om
R Ro
160 Lo = Im (Ay x 800 '
(re > Rı+R, = 250
ni à Rin
R
4w
Ri sort
Ve a

R

37. vod cond. E= 2170-10 x2=180V
T=Ef/, = Box -8ss4N-m
Aygox 2050
A 200 V, 2000 rpm, 10A, separately exec dc motor has an armature resistanc
f 2Q. Rated dc voltage is applied to both the armature and field winding of thi
otor. If the armature draws 5A from the source, the torque developed by th

(b) 4.77 Nm
(d) 0.50 Nm

sep excited =covut
Ta to
= VEN
Te = So Tr = 43 Nem

38. Efs=Tuw J90x20 = 4oxw
w=/95 7)
separately excited motor is connected as a load is drawing of a
oad current of 20 Amp. at load torque of 40 N.m. Find the speed of motor
motor Ya not given

2 Voc= Av (gnore Ya

Vv

Re 055 V.= Mum -20x0:5 = 390V
Ve = E =390V

Electric

network Íoc= gmA

39. Fis Asxio*. xa Ej= bw + #50 Co-oIs+o015)
Ew = 622:5V

A separately excited dc generator, operating with fixed excitation, delivers 450
kW to a dc bus at 600 V. Estimate the percentage change in generator speed
required so that 180 kW is delivered to dc bus. Resistance between bus bar
terminals and the armature terminals is 0.0150 and armature circuit resistance
is 0.0150. Pa = [Roxlo*. 380A &- 600 +30 x(0-015 + 0-015)
(a) -1.1687% Cy = 609V

Ex N

RE EE = x tof
Ni E, 6225 x
= -2-16€),

40. soo-Im(o-os+025) N: rpm

= 44ov
A series motor when working at full load of a 500 V supply takes a current of
100 A and runs at 800 rpm . The resistance of field winding is 0.052 and

armature resistance is 0.250 respectively.
To develop one half of full load torque at a speed of 900 rpm , the required
resistor in parallel with armature is a.
(Neglect saturation and armature reaction)
(Give the answer up to two decimal places)

41. 2
N,= Impm
005
+ si ta
IR 025
Soov R
Ta k fa
Pots fa. Iw
2
tp Th. = £000

Eand dxf,
Ex iN

E zZ N.
%, ° Me A

o:0s fp - OS a . Ex de

goo= Bas Te -
4to
5.3373 % +025 À, = So
sr E 40:25 XS0O = fo
ERBFS ff se ff +129 =0

e
pra = 91-106, 2-54

43. fut Load: Es Ve-Íaro= 240-25 Fe 4gmRT

IS) Fsh = fe =24 00
A 240 V dc long shunt cumulative compound motor has the following
magnetization characteristics at 850 rpm.

Total Excitation

1200 00 3600
(AT/pole) |
Induced emf(volts) | 80 180

The resistance drop in armature at full load is 25 V and no-load is 3.125 V.
At full-load, the shunt and series field MMF's are equal. The no-load speed of
the motor is rpm

45. Improved commutation in d.c. machines cannot be achieved by
(a) Use of inter poles
(b) Using brushed of higher resistivity
creasing reactive voltage Cowes delayed comm.
(d) Decreasing reactive voltage

46. lectromagnetic torque in rotating electrical machinery is present when
(a) air gap is uniform
(b) stator winding alone carries current
(c) rotor winding alone carries current
oth stator and rotor windings carry current

47. The function of pole shoes in the case of D.C. machine is.
(a) To reduce the reluctance of the magnetic path
(b) To spread out the flux to achieve uniform flux density
(c) To support the field coil

o discharge all the above functions

48. ir gap at the pole tips of a DC machine is kept more than that at the center o:
he pole mainly to reduce
a) Reactance voltage
Effect of armature reaction
‘c) Losses of armature core

d) Noise of the machine

49. A commutator in de machines can
1. Provide half-wave rectification

2. Provide full-wave rectification 7
3. Convert ac to de Y

4. Convert de to ac 4

5. Provide controlled full-wave rectification
Form these, the correct answer is

50. (d) Anywhere on stator or rotor

51. ich of the following part is used in construction of DC machine but not in
C machine?
a) Armature Winding

52. For C coils and P poles, the commutator pitch for simplex-wave winding is
Gt Cil
P/2 % foe)

C+1

oe

(c) £-1

P/2
C42
ir

53. Air gap at the pole tips of a DC machine is kept more than that at the center of
the pole mainly to reduce
(a) Reactance voltage
ffect of armature reaction fe
(c) Losses of armature core

(d) Noise of the machine

54. 'ommutator action is analogous to in analog electronics.

) half wave rectifier
(©) comparator
d) amplifier

55. pure rea tke more time
thnk : sparking

Under-commutation results in

(a) Sparking at the middle of the brush

parking at the leading edge of the brush
(c) Stop the DC machine. {8 Ÿ high
(d) No sparking

56. e direction of rotation of a DC motor can
e determined using the:
a) Coulomb's Law
b) Biot-Savart Law

57. Level 2(Assignment)

n the magnetic circuit of fig., the coil is excited by a current I. choose the

orrect statement form the following:

6% Most of the magnetic energy will get stored in the air gap

(b) Most of the magnetic energy will get stored in the iron core.

(c) The magnetic energy stored will be stored almost equally in the iron core

and the air gap
(d) All the magnetic energy will be stored in the iron core.

Series CKE
b= same
Wed: 4 Yxs

58. ough the cost of single layer winding is less than that of the double layer
inding, single layer winding is not generally used for dc machines because it
leads to high reactance voltage and sparking commutation.
(b) increases the effect of armature reaction under pole.
(c) reduces the efficiency of operation.

(d) leads to less number of segments.

59. Tr: when au gap offered to mmf vara
with rotos "

Four singly-excited electromagnetic structures are shown in fig. and labelled I, II,
III and IV (R: Rotor; S : Stator).

I u m
The reluctance torque can be developed in
(a) land IV (b)IandIII (c) lll and IV # III alone

60. = =2x (lo) - 20
= op ox tot A on

Given figure shows as electromagnet to lift a load of W gram. Assume infinite
permeability for iron. The ampere turns of the exciting coil to maintain an

equilibrium at an air-gap of x mm would be given by D 02 JA
= 1 = eu
We Ley = NER /4
4ox Iron

Me

= NA - wlio?
Aox>

61. Wmech= 0 MWelec= Wed

A physical system of electromechanical energy conversion, consists of a
stationary part creating a magnetic field with electric energy input, and a
moving part giving mechanical energy output. If the movable part is kept
fixed, the entire electrical energy input will be

(stored in the magnetic field

(b) stored in the electric field

(c) divided equally between the magnetic and electric fields

(d) zero

62. An electromagnetic relay has linear magnetic circuit. If the armature moves

slowly to shorten the air gap, then

1. mechanical word done, Wiech
=W,

elec

= ; (electrical energy input, Wejec )

3. field energy stored, Wad = 2 Welec ”
4. Wald = Wmech Y
From these, the correct answer is

(b) 1,2,3

63. When a transformer winding suffers a short-circuit, the adjoining turns of the same
winding experience + care] Current in Same seme

1. an attractive force”

2. a repulsive force

3. no force

4. an attractive force proportional to currents“

5. a repulsive force proportional to currents

6. an attractive force inversely proportional to distance between adjoining turnsw”
7. an attractive force directly proportional to distance between adjoining turns

8. a repulsive force inversely proportional to distance between adjoining turns From
these, the correct statements are

146 (6) 14,7

64. xt aix gap Y S= reduces We FZs ?

e following statements relating to the singly-excited electro-mechanical
system of Fig.

Here i and Y are current and flux linkages of the coil respectively x is the position of
the armature from a reference.

1. Magnetic stored energy decreases with increased value of x 2

2. Force on the armature acts in a direction to increase x Y.

3. Electromagnetic force in equal to the rate of decrease of stored magnetic field
energy with displacement at constant flux “ -AW¢id/gy

4. Force in equal to the rate of increase of co-energy with displacement at constant
ve Fixed 2
current Y AWAY, Coll
Of these statements
(a) 1,3 and 4 are correct
(b) 1, 2 and 3 are correct
G53 and 4 are correct

65. The outer conductor is showing the polarity of induced voltage and the
inner conductor in the rotor is showing the polarity of current.
The direction of rotation is

66. A DC motor (separately excited) is connected to DC supply. The motor is
running clockwise at 1200 rpm. Now the supply terminals are interchanged
and the motor runs anti-clockwise with speed 1100 rpm. Which of the
following statement is correct regarding the motor.

(a) Brushes are at GNA

(b) Brushes are shifted clockwise from GNA
(c) Brushes are shifted anti-clockwise from GNA
(d) Armature current charges due to reverse connections.

67. If the machine is working as a motor then the direction of rotations of the
motor will be

(a) Clockwise

(b) No rotation
(c) Anti-clockwise

B(a) = stator flux density

(d) Rotation will be first clockwise

& then anti-clockwise

a a increasing

68. A 6-pole, 148 A D.C. shunt generator has 240 conductors and is wave wound.
Its field current is 2A. The demagnetizing ampere turns per pole at full load if
brushes are shifted from GNA by 5° mechanical

(a) 3000 ATs/ pole

(b) 166.67 ATs/ pole

69. In question 3 if the interpoles are fitted then proper polarity of interpoles will be
Interpole-1

Interpole-2

70. A long-shunt compound generator fitted with interpoles is cumulatively
compounded. With the supply terminals uncharged, the machine is now run
as compound motor. If shunt field is stronger than the series field then which
of the following statement is correct.

(a) Motor is differentially compounded & interpoles have improper polarity

for good commutation

(b) Motor is cumulative compounded & interpoles have improper polarity for
good commutation.

(c) Motor is cumulative compounded & interpoles have proper polarity for
good commutation.

(d) Motor is differentially compounded & interpoles have proper polarity
good commutation.

71. A 4 -pole, DC machine has a lap-connected armature winding having 60 slots
and 8 conductors per slot. The flux per pole is 30 mWb . If the emf available
across its armature terminals is 240 V . The frequency of emf in the armature

coils is Hz.

72. A short compound wound de generator supplies a load current of 150 A at 230
V. The armature, series field and shunt field resistance are 0.150, 0.10 and
1000. If the brush drop is 2 V per brush, the emf generated is v.

73. In a long shunt compound generator, the terminal voltage is 230 V when it
delivers 150 A . The shunt field, series field, diverter and armature resistance
are 920, 0.0150, 0.030 and 0.0320. The total power generated by the armature
is kw.

74. A 4 pole, 100 kW, 400 volt wave connected DC machine have 400 conductors.
Find no. of turns required on each interpole, if interpolar air gap is 1 cm and
interpolar flux density is 0.25 T. Neglect the effect of iron ports of the circuit
and of the leakage

75. A 200 kW, 400 V, 14-pole DC machine has a wave wound armature with 1200
conductors. Pole arc to pole pitch ratio is 0.8. The no. of pole face conductors
of the compensating winding in each pole, so as to obtain uniform air-gap
flux density under the pole faces

76. A 250 V, 10 kW D.C shunt generator has 1400 turns on each pole. At rated
speed, a shunt current & 2A produces a no-load voltage of 250V, but at rated
load the same load voltage of 250 V can be produced by a field current of 2.2
Amp. It is required not to change series field winding. Calculated the number
of series field turns per pole required for short shunt connection

77. A separately excited DC generator has terminal characteristics and armature
resistance drop characteristics as shown in the figure. At no-load flux per pole
is 0.02 wb and 2 = 100.

Find the speed of the prime mover in rpm (A.R. neglected) at load current of
22 A

78. A DC generator has 24 armature conductors. Avg. emf induced in one
conductor is 2V and each conductor is designed to handle a current of 5A. A
load of 240 watt is connected at 24 volt at the terminals with no. of parallel
path of 2 in armature circuit. Now if the same load (240 watt, 24V) is
connected to DC generator terminal but now the no. of parallel path is 4. The

current in the armature conductor will be
(a) 5 Amp.

(b) 1.25 A

(c) 2.5 A

(d) None

79. The power drawn from 200 V source is used to supply losses only. Assum
brush volage drop is 2 V for each machines. % Efficiency of motor

80. For the above question % efficiency of generator

81. A 250 v shunt motor drawing an armature current of 40 A at full load has an
armature current of 0.5 Q. If the developed torque is constant, then by how
much the main flux must be reduced to raise the speed by 50% in %

82. A separately excited DC generator is operating at voltage 200 V and armature
current of 20 A. Armature resistance is 2 Q. Generator has constant loss & 200
watt. The maximum efficiency of generator is %.

83. A 50 kw. 20 short-circuit compound generator has armature resistance, shunt
field resistance and series field resistance of 0.5 Q, 40 Q, 0.2 Q respectively. It
supplies a load at its maximum efficiency at the rated voltage. The rotational
losses are 2 kw. The maximum efficiency of the generator is

84. 2-pole, wave wound DC machine has 24 slots with 20 turns per coil. The
ffective length of the machine is 20 cm and radius is 8 cm. The magnetic pole
‘overs 80% of armature periphery. The average flux density per pole is 1.5 T
nd the armature angular velocity is 183.2 rad/sec. The induced emf per coil
early equal to volt.