Transcript 5. Capacitance & Inductor

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A device that can hold or store a reasonable amount of
electric charge
It is made of two parallel plates separated by insulator(
dielectric) or air
It has two leads that can be connected directly to other
components
The ability to store amount of charges for a particular
values of voltage is called capacity or capacitance and
measured in Farad (F) . 1 Farad is defined as 1 Coulomb
(C ) of electricity capacity at 1 V potential difference
picofarads (pF)
nanofarads (nF)
microfarads (F)
millifarads (mF)
Farad
= 10-12 F
= 10-9 F
= 10-6 F
= 10-3 F
(F)
= 100 F
d
d
A
Insulator
(dielectric)
C
39nF
fixed capacitor
Insulator
(dielectric)
C
100pF
+
Polarized
capacitor
Cross section
C
470pF
Variable
capacitor
Plate A
Plate B
Capacitor
R
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V
Positive terminal of a battery repels the positive charges
(positive
ions) towards plate A and attracts negative charges (electrons) towards
it – Plate A then becomes positive.
Negative terminal of a battery repels negative charges (electron)
towards plate B and attracts positive charges (positive ion) towards it
– Plate B then becomes negative charge.
Positive charges accumulate in plate A reduces more positive charges
from the battery terminal to enter it and at the same time negative
charges in also reduces more negative charges from the negative
terminal of the battery.– the current flow from the battery the plate
will be reduced.
Plate A
Plate B
Capacitor
R
V
•Charges develop the potential different between the
plates
and increase with the increase of charges.
•When the potential different is same as the voltage of the
battery, the entering of charges stop.
•Charges are stored in the capacitor plates after the
connection to the battery is disconnected.
•Ratio of Q:V is constant and is called as capacitance, thus
C = Q/V
or
Q = CV
Capacitance of a capacitor depends on its physical construction
and given as:
C = A/d
[F]
where  = ro
 = absolute permittivity [F/m]
r = relative permittivity [no unit]
o = free space permittivity [8.854 x 10-12 F/m]
A = area of a plate [m2]
d = distance between two plates [m]
A capacitor made of two parallel plates of area 10 cm2 each.
The plates are separated by an insulator of 0.5mm thickness and
a relative permittivity equal to 5. Determine its capacitance.
C
= roA/d
= (5 x 8.854 x 10-12 x 10 x 10-4)/0.5 x 10-3
= (5 x 8.854 x 10-12 x 10)/5
= 88.54 pF
When the voltage applied to the capacitor changing with time ,
the charges also will change with time . Thus Q = CV
becomes
dq = C dv
Then differentiate with time , we have
note
dq
i
dt
integrates
therefore

1
v
idt
C
dq
dv
C
dt
dt
dv
iC
dt
Power in capacitor is given by
dv
p  vi  Cv
dt
Energy for time dt is given by
dw  pdt  Cvdv
V
Total Energy
W  C  vdv
Thus Energy stored in capacitor is
0
V
v 
 C 
 2 0
2
W
1
CV 2
2
i(t)
Using Kirchhoff‘s voltage law we have:
1
1
1
Both side have
idt 
idt 
idt



C1
C2
Same integration CT
1
1
1


C T C1 C 2
Generalized for n capacitors
1
1
1
1


 ...... 
C T C1 C 2
Cn
C1
v2(t)
C2
vT(t)
vT(t) = v1(t) + v2(t)
Thus
v1(t)
Note from previous slides

1
v
idt
C
i(t)
Voltage across each capacitor
Q
Q
V2 
V1 
C2
C1
Taking the ratio
V2 C1

V1 C 2
(*)
v1(t)
C1
v2(t)
C2
vT(t)
But summation of voltage
V1  V2  V
or
Substitute in (*)
V2  V  V1
V-V1 C1

V1
C2
Then we have
V1  V 
C2
C1  C 2
Similarly
V2  V 
C1
C1  C 2
iT(t)
Using Kirchoff’s current law
i1(t)
iT(t) = i1(t) + i2(t)
CT
v(t)
i2(t)
C1
dv(t)
dv(t)
dv(t)
 C1
 C2
dt
dt
dt
CT = C1 + C2
To generalized for n capacitor , thus
CT = C1 + C2 + ……. + Cn
Note from previous slides
dv
iC
dt
C2
A
C2
220pF
C1
3.5nF
Total of series capacitors C2 and C3
C2C3
220  470
Cs 

 150 pF
C2  C3 220  470
Cs
C3
470pF
B
The overall total
CT  C1  Cs  3500 pF  150 pF  3650 pF  3.65nF
A
C1
4.67uF
Charges in the capacitors C1 and C2 are Q1 = 20
C and Q2 = 5 C respectively, determine the
energy stored in C1, C2 and C3 and the total
B
energy in all capacitors.
C1 = 4.67 F; Q1 = 20 Q
Therefore
V1
And
W1
= Q1/C1
= (20 x 10-6)/(4.67 x 10-6)
= 4.3 V
= ½C1(V1)2
= ½ x 4.67 x 10-6 x 4.32
= 43.2 J
C2
980nF
C3
392nF
C2 = 980 nF; Q2 = 5 Q
Then V2
W2
W3
= Q2/C2
= (5 x 10-6)/(980 x 10-9)
= 5.1 V
= ½C2(V2)2
= ½ x 980 x 10-9 x 5.12
= 12.7 J
= ½C3(V2)2
= ½ x 392 x 10-9 x 5.12 = 5.1 J
Electric current passing through a conductor will produce
magnetic field or flux around it as shown in Figure.
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If the wire conductor is wound around a core,
magnetic filed /flux will resemble like permanent
magnetic bar.
Magnitude of flux produced depends on magnitude
of current ,I, properties of core , ,and physical
construction (length  , area of cross-section A and
number of turn N)
NI

S
V
I
R
l
S
μA
R
l
σA
where
flux  is equivalent to current I
m.m.f. NI equivalent to e.m.f. V
reluctance S equivalent to resistance R
and  is an absolute pemeability of core material and given by :
 = ro where o is a permeability of air (4 x 10-7 H/m)
Substituting S in equation  = NI/S: we have
 r  o ANI

l
If the current in the inductor is varied with time (t), flux  will
also varied with time. Variation of flux in the windings will
induce voltage.
d
The induced voltage is: v  N ()
dt
 r  o AN 2 di

l
dt
 r  o ANi

l
where:
By introducing
d    ANi 
N  r o

dt 
l
 r  o AN 2
L
l
then
di
vL
dt
1
Or for current i   vdt
L
The inductance L is measured in Henry defined as a coil that
induce 1 V when rate of current variation is 1 A/s.
H
mH
H
Symbol for inductors:
The units are:
(a) – air cored
inductor
(c) –ferritecored inductor
= 10-6 H
= 10-3 H
= 100 H
(b) –iron cored
inductor
(d) – variable
inductor
An inductor is built from a coil of 180 turns and the core is of
iron having a relative permeability of 1500. The length of the
core is 30 mm and area of cross section is 78.5 mm2. Calculate
the value of the inductance.
r = 1500; o = 4 x 10-7 H/m; A = 78.5 mm2;
N = 180; l = 30 mm
 r  o AN 2 1500  4  10 7  78.5  10 6  180 2

L
30  10 3
l
 160 mH
Voltage
Power
di
vL
dt
I
di
p  iv  iL
dt
L
For duration of dt sec
di
dw  pdt  Li dt
dt
 Lidi
For current changes between i = 0 to i = I
I
W  L  idi
0
I
1 2
i2 
 L    LI
2
 2 0
Inductor stores energy in the form of magnetic fields.
di
vL
dt
vT  v1  v2
di
di
di
L T  L1  L 2
dt
dt
dt
L T  L1  L 2
In general
LT  L1  L2  ........Ln
i
v1
L1
vT
v2
L2
1
i   vdt
L
iT i1
i2
iT  i1  i2
1
1
1
vdt

vdt

vdt



LT
L1
L2
1 1 1
 
LT L1 L2
In general
1
1 1
1
   ........ 
LT L1 L2
Ln
L1
L2
Effective inductance between A
and B is:
A
25 mH
L2
15 mH
L3
60 mH
L1
Le = L2 + L3 = 25 + 15 = 40 mH
Lt = (L1 x Le)/(L1 + Le)
= (60 x 40)/(60 + 40) = 24 mH
B
A
I = 600
mA
By supplying a total of constant current
L2
at 600 mA, L1 is found to store an
350 mH
L
energy of 28 mJ in its magnetic field.
L3
Calculate the total energy stored in all
R1
R2
three inductors L1, L2 dan L3?
W1 = ½L1I12
B
I1= ( 2W1/L1) = [(2 x 28 x 10-3)/(350 x 10-3)] = 400 mA
150 mH
1
I2 = I – I1( Kirchoff’s current law) = 600 – 400 = 200 mA
Le= L2 + L3 = 150 + 100 = 250 mH
We= ½LeI22 = ½ x 250 x 10-3 x (200 x 10-3)2 = 5 mJ
Total energy W = W1 + We = 28 + 5 = 33 mJ
100 mH