Chap 6 DC Current Handout

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


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


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




-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


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=


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




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.


Suppose there is a conductor and a baterry.

-qi +qi

+ -

+

-

Eo

Ei

Collection ofNegative charge

Collection ofpositive charge

+ -

E

F


If there is no emf (potential difference) then the work

to pass the electron is not needed

EqFdlFW == ,.

0. == dlEqW




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


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)


where

is electric conductivity

Suppose there is a silindrical conductor :

A B

L

VA VB = V =L

dlE

0

.

For homogen E :L

VE=




EJ =L

V=

AJi .= AL

V= V

L

A=

iA

L

V =

iRV .=

A

LRwhere =

thenyresistivit

1i =definedwef

LR =

(Ohm Law)


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


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 ?




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)


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)


nodei1

i2i3

0= i0321 =+ iii

Some illustration

321 iii =+so

nodei1

i2i3

0= i0321 =++ iii

123 iii =+so




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


Some illustration

Direction ofloop

i

so: In negative signi in positive sign

Direction ofloop

i

In positive signi in positive sign


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:




From Kirchoff I

=0ii2 + i1 = i3 .(1)

From Kirchoff IIAt loop I :

= iR1 = i1 R1 i2 R2

2 = i1 2 i2 ..(2)


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


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




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


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


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=


10/10


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


RK

Ritat

=== '0

ieeie KRCt

KRC

t

== +

. ', Kedef K =

RC

t

eKi

= '

soRC

t

eR

ti

=

)(


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

Documents

Chap 6 DC Current Handout