8/11/2019 Chap 6 DC Current Handout
1/10lied Physics (PU-1113) Faculty of Science, Telkom Institute of Technology
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
ELECTRIC CURRENT
There are free electrons in the conductors that moving in
random motion
Conductor
Number of electron that passing the plane from right side andfrom left side is the same
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Positive pole ofBaterry
+
Negative pole ofBaterry
-
If two side of conductor (length: d) is connected to baterry
Eo
Then there is electric field inside
Va VbV=Va-Vb
== drEVVV
bA
.
lEV=l
VE =
d
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
This force is causing current inside conductor
oEqF .= eelectronq =
oEeF = EwithoppositeF
oo EiEbecauseand =
oeEiF =
The electric field causes a force at the electron inside
conductor
8/11/2019 Chap 6 DC Current Handout
2/10lied Physics (PU-1113) Faculty of Science, Telkom Institute of Technology
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
-qi +qi
Positive pole Negative pole
+ -
Eo
Va VbV=Va-Vb
Ei
F
There are moving electron to the left and moving positronto the right
This Moving positron is defined by current
I
Negative and positive charge will be separated in both sidesof conductor induktion charge
And there is induction field Ei Due to this induction charge
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
If the net of q charge through the plane of surface area
of conductor in t secon the the current is
t
qi=
i current (A)q net of charges (C)t time (s)
If the net of charge depend on time then the current is
dt
dqi=
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
If induction charge qi is increasing then induction field
Ei is increasing too, it means Ei ~ qi.
If Ei = Eo then total field inside conductor becomes
0EiEiEEE ioio ==+=
r
From the relation
= dr.EV0Vthen0Eif ==
(Thre is no potential difference between both sides of
conductor
8/11/2019 Chap 6 DC Current Handout
3/10lied Physics (PU-1113) Faculty of Science, Telkom Institute of Technology
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
It means there is no moving charge inside conductor
(the current is stoping).
In order to continuously moving charge, we must passing
induction charge from both sides of conductor so there is no
induction field inside conductor
In the looping circuit, this emf must be resulted from the
baterry then the charge is continuously moving around the
loop
Needed a force to pass this charge
This force is called electromotive force.
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Suppose there is a conductor and a baterry.
-qi +qi
+ -
+
-
Eo
Ei
Collection ofNegative charge
Collection ofpositive charge
+ -
E
F
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
If there is no emf (potential difference) then the work
to pass the electron is not needed
EqFdlFW == ,.
0. == dlEqW
8/11/2019 Chap 6 DC Current Handout
4/10lied Physics (PU-1113) Faculty of Science, Telkom Institute of Technology
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Electron has a extra energy that proportional to emf that is in
magnitudo of q.
== dlEqqW .
= dlE.
If there is emf (potential difference) then the work
to pass the electron is needed
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Ohm Law
Current in the metal conductor is constant .
It means the density current is constant to.
And because the velocity of carrier charge is proportional with
density current so this velocity is constant too
The velocity of carrier charge is proportional with electric field,
then
vkonstan~ Ej ~ E
Final result is :
j = E (Ohm law)
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
where
is electric conductivity
Suppose there is a silindrical conductor :
A B
L
VA VB = V =L
dlE
0
.
For homogen E :L
VE=
8/11/2019 Chap 6 DC Current Handout
5/10lied Physics (PU-1113) Faculty of Science, Telkom Institute of Technology
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
EJ =L
V=
AJi .= AL
V= V
L
A=
iA
L
V =
iRV .=
A
LRwhere =
thenyresistivit
1i =definedwef
LR =
(Ohm Law)
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Joule Law
i
Potensial at both sides:
VA VB
VA > VB
Current at conductor:R
Vi=
Because current is constant then vA = vB
vA vB
potential
Drift speed
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Because the velocity is constant va = vb , so the kinetic
energy of the carrier charge at both sides is the same
Carrier charge move in the presence of potential difference V
= VA - VB
kBkA EE =22
2
1
2
1BA mvmv =
So carrier charge dq has extra energy dU= dq.V
Because the kinetic energy of carrier charge in both sides is not
changing along carrier charge moves so there is no extra
energy for carrier charge
Then where does the energy dU run out ?
8/11/2019 Chap 6 DC Current Handout
6/10lied Physics (PU-1113) Faculty of Science, Telkom Institute of Technology
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
This energy is changing in the form of heat. The magnitude
Of heat power is
dt
dU
P= dt
dqV
= Vdt
dq
= iV=
Because of V = i R
so :
This power is called disipation power
P = i2R ( Joule Law)
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Kirchoff Law
For electric circuit 2 loop, the magnitude of current can be
found usingKirchoff law.
Kirchoff law I
Number of current at a node point is equal to zerro (Number of
current that come in to the node is the same with number of
current that come out from the node
=0i(Conservation of current)
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
nodei1
i2i3
0= i0321 =+ iii
Some illustration
321 iii =+so
nodei1
i2i3
0= i0321 =++ iii
123 iii =+so
8/11/2019 Chap 6 DC Current Handout
7/10lied Physics (PU-1113) Faculty of Science, Telkom Institute of Technology
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Kirchoff Law II
Number of electric potential at every loop is equal to zerro
= iR
in positive sign if direction of emf is the same with loop
direction and in negative sign if direction of emf is in the
opposite direction with loop direction
i in positive sign if direction of current is the same with loop
direction and i in negative sign if direction of current is in the
opposite direction with loop direction
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Some illustration
Direction ofloop
i
so: In negative signi in positive sign
Direction ofloop
i
In positive signi in positive sign
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Example : 1 2
R1
R2 R3
3
1 = 2 V
2 = 6 V3 = 10 V
R1 = 1 R2 = R3 = 2
Find disipation power at every R
Suppose that the curren at every R are : i1 , i2 , i3
i1
i2
i3
Loop 1 Loop 2
Solution:
8/11/2019 Chap 6 DC Current Handout
8/10lied Physics (PU-1113) Faculty of Science, Telkom Institute of Technology
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
From Kirchoff I
=0ii2 + i1 = i3 .(1)
From Kirchoff IIAt loop I :
= iR1 = i1 R1 i2 R2
2 = i1 2 i2 ..(2)
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
At Loop II :
= iR3 - 2 = i2 R2 + i3 R3
10 6 = i2. 2+ i3 .2
4 = 2 i2+ 2 i3..(3)
Substitute i2 from (1) to (2)
i2 = i3 i1 2 = i1 2 (i3 i1)
2 = 2 i1 2 i3 (4)
Substitute i2
from (1) to (3)
i2 = i3 i1 2 = (i3 i1) + i32 = 2 i3 i1 (5)
2 = i2 + i3
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
We found
2 i1 2 i3 = 2
2 i3 i1 = 2_____________ +
i1 = 4 Ampere
4 2 i2 = 2 i2 = 1 Ampere
i3 = i2 + i1
i1 2 i2 = 2
i3 = 1 + 4 = 5 Ampere
8/11/2019 Chap 6 DC Current Handout
9/10lied Physics (PU-1113) Faculty of Science, Telkom Institute of Technology
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
So the disipation powers :
at R1 : P1 = i12 R1
P1 = 42 . 1 = 16 J/s
At R2
: P2
= i2
2 R2
P2 = 12 . 2 = 2 J/s
At R3 : P3 = i32 R3
P3 = 52 . 2 = 50 J/s
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
RC Circuit
Suppose there is a circuit with resistor, capacitor, and
baterry
R C
At t = 0 capacitor is still empty sopotential difference between bothsides of capacitor is zerro
After current through the circuit, the charge start to colect
in the capacitor so
= VR + Vc
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Dimana VR = R I dan
C
QV c =
Sehingga
C
QRi +=
Diferensialkan persamaan ini terhadap t :
dt
dQ
Cdt
diR
10 +=
C
i
dt
diR +=0
RC
i
dt
di=
8/11/2019 Chap 6 DC Current Handout
10/10
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
RC
dt
i
di=
Use integral to equation above
= RCdt
i
di
21ln KRC
t
Ki +=+
ieK
RC
t
=+
12ln KKRC
ti += Def K2 K1 = K
With the fact that at t = 0 the current at circuit is
maximum so
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
RK
Ritat
=== '0
ieeie KRCt
KRC
t
== +
. ', Kedef K =
RC
t
eKi
= '
soRC
t
eR
ti
=
)(
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Remember dq = i(t) dt
dteRdq RC
t
=
Integrate that equation
=t
RC
t
dteR
tq
0
)(
t
RC
t
eRCR
tq
0
)(
=
=
1)( RC
t
eCtq