electric current note

7/21/2019 Electric Current Note

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Electric Current. Symbol I

Electric current is the flow of electric charge.

An electric current is a flow of electrons from one

point to another. For an electric current to flow there

are two requirements

(i) A source of electrons e.g. electric cell which is a

source of electrons and also electrical energy whichcauses the electrons to flow

(ii) Current will only flow in an electric circuit only if

there is a complete circuit.

S.I. unit of current : Ampere Amp A

If energy is used in when a current flows through a

material then that material has what we call resistance.

esistance is defined as that which opposes the flow of

electric current.

S.I. unit of electric resistance : !hms Ω

Electrons flow from the negati"e side of an electric

cell to the positi"e side.

http://en.wikipedia.org/wiki/Electric_charge

http://en.wikipedia.org/wiki/Electric_charge



#owe"er in electric circuits we show the current

flowing from the $"e to %"e

$ e &

I

esistance

Ammeter measures current

'oltmeter measures "oltage

attery or Electric Cell

A single cell or other power source is represented y along and a short parallel line. A collection of cells or

attery is represented y a collection of long and short

parallel lines. In oth cases* the long line is

representati"e of the positi"e terminal of the energy

source and the short line represents the negati"e

terminal.

http://www.clker.com/clipart-13195.html



esistance shown as either a rectangular o+ or ,ig,ag

line

R

Resistance (Ohms)

Ampere : Is defined as that constant current which is

maintained in two infinitely long conductors of

negligile cross&section one metre apart in a "acuumwould produce etween the conductors a force of - +

/&0 1ewton2s per metre.

Coulomb : Is defined as the quantity of charge which

passes when a steady current of one ampere flows for

one second.It can e shown that

C#A3E (4) 5 C6E17 (I) 8 7I9E (t) It Q =

( ) JoulesVIt Energy

V It V Q EnergyWork

=

=== ..

)(time

EnergyWattsVI

t

VIt Power ===

E+amples.

. Calculate the amount of energy supplied y a ;<

heater in a time of minutes.



-. #ow long will it ta;e a - ;< ;ettle to oil -// g of

water initially of -/o C.

=. If the p.d. across an electrical heater is -/ ' and its

electrical power is ; < calculate the current flowing.

>. If the power rating of a heater is // < and the

current flowing through it is A calculate the p.d.

across it.

OHM’S LAW.

7he current flowing through a conductor is directly

proportional to the potential difference across its ends



pro"ided that the temperature remains constant I

∝ ' at constant temperature

7he constant of proportionality is the resistance of the

conductor IRV =

Consider the case of a current I flowing through a

resistor from A to . 7he "alue of the current

flowing in and out of the resistance is I.#owe"er since wor; is done is mo"ing the current

through the resistor the potential at A and are

different from one another.

A

I R I

Current always flows from a higher to a lower

potential. In this case

"alue positi"eais

sinceand '

resistortheacross...

AB

B A B A AB

R

V

V V V V V

d p R I V

>−==

==

KIRCHOFF’S FIRST (CRRE!T" LAW.

States that the sum total of the current flowing into a

?unction in an electric circuit is equal to the sum total

of the current flowing out of the ?unction.

I I-



I=

In the ao"e circuit I 5 I- $ I=

Electric Cell# in Serie#.

' '- '=

7he total "oltage ( p.d.) is equal to the sum of the

indi"idual "oltage "alues '7 5 ' $ '- $ '=

7his is also true of all potential differences in series

i.e. for electric cells as ao"e and also for potential

differences across electrical components (resistors).

$olt%&e %n' Current or component# in Serie#

$olt%&e# %'' up for components connected in series.

Current# %re t)e #%me through all components

connected in series.

Cell# in *%r%llel ($olt%&e# in p%r%llel"

'

'-

'=



7he total p.d. here is the "alue of only one of them

'7 5 ' 5 '- 5 '=

Again true for oth electric cells and also for potential

difference "alues across electrical components

*OTE!TIAL +IFFERE!CES I! *ARALLEL

ARE E,AL

$olt%&e %n' Current or component# in *%r%llel

$olt%&e# %re t)e #%me across all components

connected in parallel.

Current# %'' up for components connected in

parallel

RESISTORS I! SERIES

= esistors in series across a "oltage source '7



1ote that since there is only one path for thecurrent to tra"el* the current through each of the

resistors is the same. =-. I I I I ===

Also* the "oltage drops across the resistors must

add up to the total "oltage supplied y the

attery:=- V V V V T ++=

Since IRV = for each resistor then

( )=-.

=-.

==--..

R R R I V

IR IR IRV

R I R I R I V

T

T

T

++=

=++=

++=

!hm@s aw must also e satisfied for the

complete circuit

( )=-. R R R I IRV equivalent T ++==



7herefore the e+pression for equi"alent resistance

for = resistors connected in series is

=- R R R Requivalent ++=

In general* the equi"alent resistance of resistors

connected in series is the sum of their resistances.

7hat is for n resistors in series the equi"alent

resistance is

∑=

=n

i

iequivalent R R.

R ESISTORS I! *ARALLEL

esistors can e connected such that they ranch

out from a single point (;nown as a no'e)* and

?oin up again somewhere else in the circuit. If we

lael the point at the $ side of the attery A and

the point at the % side of the attery and



connect three resistors in parallel across the

attery the circuit is

A

B

1ote that the node does not ha"e to physically e

a single pointB as long as the current has se"eral

alternate paths to follow* then that part of the

circuit is considered to e parallel.

At A the potential must e the same for each

resistor. Similarly* at - the potential must also e

the same for each resistor. So* etween points A

and -* the potential difference is the same.

7hat is* each of the three resistors in the parallel

circuit must ha"e the same "oltage.

=- V V V V V T ====

Also* the current splits as it tra"els from A to -.

So* the sum of the currents through the three



ranches is the same as the current at A and at -

(where the currents from the ranch reunite).

=-. I I I I ++=

y !hm@s aw and recalling that all the "oltages

are equal the equation is equi"alent to:

=-.=

=

-

-

.

.

R

V

R

V

R

V

R

V

R

V

R

V

R

V

equivalent

++=++=

As the "oltages are equal. So the 'oltages cancelout* and we are left with

∑=

=

++=

n

i iequivalent

equivalent

R R

or

R R R R

=-

7he general equation for any numer of resistors

connected in parallel.

!ne prolem with this equation is that the

solution gi"e the in"erse "alue of equi"alent and so

you must rememer to re&in"ert the "alue to

determine equi"alent

#owe"er If the numer of resistors in parallel is

- the equation ecomes



aluesresistor "twotheof sum

aluesresistor "twoof tionmultiplica

parallelinresistors7<!for7herefore

.

gi"esin"erting

-

-

-

-

-

=

+=

+=

+=

equivalent

equivalent

equivalent

equivalent

R

R R

R R R

R R

R R

R

R R R

7his is the equation for equi"alent and so we don2t

ha"e to rememer to rein"ert ut the equation can

only e used for TWO resistors in parallel.

*roce'ure or #olin& Electric Circuit#.

. Find the equi"alent resistance for any parallelcominations in the circuit* and replace the

comination with its equi"alent. 7he circuit is now a

series circuit.

-. edraw the SERIES circuit.

=. Find the total resistance of the series circuit.

>. edraw the circuit.

. 6se !hm2s aw to calculate the total or circuit

currentT

T T

R

V I =



. 7he total current flows through each resistor that

is in series with the attery using !hm2s aw

IRV =

the potential difference across each resistor in series

with the attery can e calculated.

0. Also in the solution we ;now that the total current

flows through the equi"alent resistance of any of

the parallel cominations. So we can use !hms

aw in the same way as section to calculate the

p.d. across the equi"alent resistance.

eqT eqeqeq R I R I V ==

Since p.d in parallel are equal this "oltage "alue is

equal to the p.d. across each component

(resistance) in a parallel comination.

D. 7hen use ;nowing the "oltage use !hm2s aw

n

eq

n R

V I =

to calculate the current flowing through each

resistor n in the parallel comination.



. 7o calculate the power dissipated through any

component (resistance) use the equation.

VI Power =



*roblem S)eet O)m# L%/

For each of the circuits shown elow calculate

(i) total circuit current

(ii) current through each resistor(iii) p.d. across each resistor.

Circuit .V1

12V

R1

100ohm

R3

200ohm

R2

200ohm

R4

400ohm

Circuit -. 3i"en 5 / Ω* - 5 // Ω*

= 5 / Ω* E 5 / '

E

-

=



Circuit =

4uestion -. Find the reading of the ammeter A when

the switch (s) in the circuit is (a) open and () closed

-/ '

-/ ohms S

=/ ohms A -/ ohms



4uestion =. Sol"e the following circuit i.e. find the

current through* the potential difference across and the

power of each of the components (resistors) in the

circuit.

'7

-

> =

3i"en that '7 5 ' 5 >// Ω - 5 -/ Ω

= 5 // ohms and > 5 / Ω.

4uestion >.In an electric circuit two parallel cominations of

(a) a >// Ω resistor in parallel with a // Ω resistor

and () 7wo // Ω resistors in parallel are in series

with a // Ω resistor across a - ' supply.

raw the circuit diagram and sol"e the circuit.



Intern%l Re#i#t%nce o %n electric cell %n' e.m..

eal atteries are constructed from materials which

possess non&,ero resisti"ities. It follows that real atteries are not ?ust pure "oltage sources. 7hey also

possess internal resistances. A pure "oltage source is

usually referred to as an e! (which stands forelectrootive !orce).

Electric cells contain an internal resistance and the cellloose some of its e.m.f. when it dri"es current through

this internal resistance. It is more accurate to

represent an electric cell as follows

E 5 e.m.f. of the electric cell

r 5 internal resistance of the cell



If this is connected up to an e+ternal resistance then

a current I flows

5 E+ternal circuit resistance

( )r R I E

r R

E e! I

+=

+==

resistancetotal

Ir E V

Ir V E

Ir IR E

−=

+=

+=

<here ' 5 the terminal "oltage across the e+ternal

"oltage the potential difference

I r is the potential ("oltage) lost inside the electric cell

E 5 emf of the attery

E+ample . An electric cell supplies a current of /. A

when a -// ohm resistance is connected in series with

it. 7he same cell supplies /.- A thro2 a 0// ohm

resistance. Calculate the internal resistance and the

e.m.f. of the cell.



-. A cell can supply(a) a current of .- A thro2 two -/

ohm resistors when they are connected in parallel and

() a current of /.> A when they are in series.

Calculate the e.m.f. internal resistance of the cell andthe p.d. of the attery for (a) and ()

4uestion =.

Sol"e the following circuit. i.e. find the current

through* the potential difference across and the powerof each of the components ( resistors ) in the circuit.

/ ' /.- Ω

-/Ω 40Ω

40Ω

20 Ω 60 Ω

4uestion >. Sol"e the following circuit i.e. find the

current through* the potential difference across and the power of each of the components ( resistors ) in the

circuit.

E7

-

>



=

3i"en E 5' * internal resistance r 5 Ω

5 > ; Ω - 5 ; Ω

= 5 ; Ω and > 5 ; Ω.

electric current note

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