a device that can hold or store a reasonable amount of electric charge it is made of two parallel...
TRANSCRIPT
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) = 10-12 F
nanofarads (nF) = 10-9 F
microfarads (F) = 10-6 F
millifarads (mF) = 10-3 F
Farad (F) = 100 F
dA
Insulator(dielectric)
d
Insulator(dielectric)
Cross section
C39nF
fixed capacitor
+
C100pF
Polarized capacitor
C470pF
Variable capacitor
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.
R
V
Capacitor
Plate A Plate B
•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
R
V
Capacitor
Plate A Plate B
Capacitance of a capacitor depends on its physical construction and given as:
C = A/d [F]
where = ro
= 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 = roA/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 havedt
dvC
dt
dq
dt
dvCi
idtC
1v
therefore
integrates
note idt
dq
dt
dvCvvip Power in capacitor is given by
Cvdvpdtdw Energy for time dt is given by
VvdvCW
0
Vv
C0
2
2
Total Energy
2
2
1CVW Thus Energy stored in capacitor is
n21T C
1......
C
1
C
1
C
1
idtC
1idt
C
1idt
C
1
21T
21T C
1
C
1
C
1
C1
C2
i(t)
vT(t)v1(t)
v2(t)
Using Kirchhoff‘s voltage law we have:
vT(t) = v1(t) + v2(t)
Generalized for n capacitors idtC
1v
Note from previous slides
Both side haveSame integration
Thus
2
1
1
2
C
C
V
V
VVV 21
21
21 CC
C
VV
21
12 CC
C
VV
2
1
1
1
C
C
V
V-V
12 VVV
C1
C2
i(t)
vT(t)v1(t)
v2(t)
22 C
QV
11 C
QV
Voltage across each capacitor
Taking the ratio (*)
But summation of voltage
or
Substitute in (*)
Then we have
Similarly
v(t)
iT(t)
i1(t) i2(t)
C1 C2
Using Kirchoff’s current law
iT(t) = i1(t) + i2(t)
CT = C1 + C2
To generalized for n capacitor , thus
CT = C1 + C2 + ……. + Cn
dt
dv(t)C
dt
dv(t)C
dt
dv(t)C 21T
dt
dvCi
Note from previous slides
C3470pF
C13.5nF
C2220pF
A
BpFCC
CCCs 150
470220
470220
32
32
nFpFpFpFCCC sT 65.3365015035001
Total of series capacitors C2 and C3
The overall total
Cs
C3392nF
C2980nF
C14.67uF
A
B
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 energy in all capacitors.
C1 = 4.67 F; Q1 = 20 Q
Therefore V1 = Q1/C1
= (20 x 10-6)/(4.67 x 10-6)= 4.3 V
And W1 = ½C1(V1)2
= ½ x 4.67 x 10-6 x 4.32
= 43.2 J
C2 = 980 nF; Q2 = 5 Q
Then V2 = Q2/C2
= (5 x 10-6)/(980 x 10-9)= 5.1 V
W2 = ½C2(V2)2
= ½ x 980 x 10-9 x 5.12
= 12.7 J
W3 = 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.
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)
whereflux is equivalent to current Im.m.f. NI equivalent to e.m.f. Vreluctance S equivalent to resistance Rand is an absolute pemeability of core material and given by : = ro where o is a permeability of air (4 x 10-7 H/m)
Substituting S in equation = NI/S: we have
S
NI
R
VI
μAS
l
σAR
l
l
ANIor
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.
l
ANior
The induced voltage is: )(dt
dNv
l
ANi
dt
dN or
dt
diAN2
or
l
dt
diLv vdt
L
1i
l
2
or ANL
where:
then
By introducing
Or for current
The inductance L is measured in Henry defined as a coil that induce 1 V when rate of current variation is 1 A/s.
The units are: H = 10-6 HmH = 10-3 HH = 100 H
Symbol for inductors:
(a) – air cored inductor
(b) –iron cored inductor
(c) –ferrite-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
l
2
or ANL
3
267
1030
180105.781041500
mH160
dt
diLv
dt
diiLivp
dtdt
diLipdtdw Lidi
I
0
idiLWI
0
2
2
iL
2LI
2
1
Inductor stores energy in the form of magnetic fields.
I
L
For duration of dt sec
For current changes between i = 0 to i = I
Voltage
Power
21 iiiT
vdtL
1vdt
L
1vdt
L
1
21T
vdtL
1i
L2L1
i1 i2iT
21
111
LLLT
nT LLLL
1........
111
21
In general
Effective inductance between A and B is:
Le = L2 + L3 = 25 + 15 = 40 mH
Lt = (L1 x Le)/(L1 + Le)
= (60 x 40)/(60 + 40) = 24 mH
L3
L2
L1
25 mH
15 mH
60 mH
A
B
L1
L2
L3
R1R2
I = 600 mAA
B
350 mH150 mH
100 mH
W1 = ½L1I12
I1= ( 2W1/L1) = [(2 x 28 x 10-3)/(350 x 10-3)] = 400 mA
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
By supplying a total of constant current at 600 mA, L1 is found to store an energy of 28 mJ in its magnetic field. Calculate the total energy stored in all three inductors L1, L2 dan L3?