===Electric current, ohm's law ......===Electric current, ohm's law=====Unit ( 1 ) Chapter ( 1 )===...

===Electric current, ohm's law==================================Unit ( 1 ) Chapter ( 1 )===

Mr. Yousri Abdel-Gaber The 1st Physicist 1

Lesson one Electric Current When a conductor is placed in an external electric field (battery) the electrons flow from

its negative terminal to its positive terminal. This direction of electrons is known as the

electronic current direction.

At the same moment t flow of positive changes from the positive terminal to the negative

terminal of conductor. This is known as the conventional current direction.

electronic direction

The electric Current:- 1- It is the flow of electric charges in a conductor.

The conditions for the flow of electric current: Presence of a conductor , Source and A closed circuit

Metals are electrically good conductors but nonmetal is not.

Metals are rich in free electrons but nonmetal is poor in free electrons . It is preferable to use copper wires in the electrical connection.

Because copper is rich of electrons

The electric current intensity (I)

It is the quantity of electric charges in coulombs passed in a given cross section in

one second.

I = t

Q

Unit of I :- Ampere, Coulomb/sec Q = N.ē where N = no of electrons ē = change of 1 electron

I = N. ē

+ -

Conventional direction

Q

I t

t



The graphical relation between ( Q – t )

Slope of line = = I

Is the electric current intensity through a circuit when a charge of 1 coulomb passed in a given

section in one second.

Is the quantity of charge passed a given cross-section in when electric current intensity is 1 A

in one sec.

Note 1 ) The charge of electron = 1.6 x 10-19 C

2 ) Number of electrons in one coulomb. =

N = 19-10 x 6.1

1 = 6.25 x 1018 electrons

3

Electric current intensity is measured by Ammeter

which connected in the circuit in series.

What is meant by? The intensity of an electric current = 2 A.

o It means that the quantity of electricity which passes through cross sectional of a

conductor in 1 sec = 2 C

Example 1

Find the current intensity flowing through a certain conductor when 1018 electrons pass

through its cross section per minute if you know that the charge of an electron = 1.6 x

10-19 Coulomb.

Solution:

N=1018 t=1 min=60sec

I = = = = 0.26 A

Q

t

Q

t

1 C

Charge of 1 electron

t

Q t

N. ē 60

1018X1.6x10-19



Example 2

What is the number of electrons passing through cross sectional area of a conductor in 1

sec,knowing that the current intensity in circuit is 20A and the charge of an electron =

1.6 x 10-19 Coulomb.

Solution:

N=1018 t=1 min=60sec

I =

20 A= N=20x60/1.6x10-19=1.25x1020 electros

The electric current flows, when the electric potential at a point must be greater than it at

another point which gives ability to do work to overcome the resistance of the path.

So all substance will permit the passage of some electric current, provided the potential

difference is high.

It is the work done in Joules to transfer a unit charge ( 1 C ) between them.

V = Q

W

Unit of V :- Joule / Coulomb, Volt

The graphical re;ation betn ( W- Q )

Slope of line = = V

It is the potential difference between two points when work done of 1 J to transfer a

charge of 1 C between them.

The electric potential

difference is measured by

using voltammeter which is

connected in parallel.

W

V Q

w

Q

w

Q

t

N. ē

60

N x 1.6x10-19



What is meant by? The potential difference between two points = 10 V.

o It means that the work done to transfer a charge of 1 C between two points is 10 J.

This work is called the electric energy consumed.

V = Q

W W = V.Q where Q = I.t. W = V. I. t

E.E =V.I.t Unit of E. E :- Joule = Kw.hr= volt.ampere.sec

The electrical power of a device:

It is the electric energy consumed in 1 sec

Pw=t

EE.=

t

tIV ..= IV . = I2R =

Unit of Pw: joule/sec , watt

When the more of the electrical instruments uses in houses the electric current

intensity flowing through the fuse increases

• More electrical devices mean more power consumed where the electric current

intensity I is directly proportional to the electric power.so the current passes

through the fuse that connected in series with electric instruments .

Example1 If the potential difference between terminals of a conductors in a circuit is 10V

Calculate the work done to transfer 6.25x1020 electrons if the charge of an electron = 1.6

x 10-19 Coulomb.

Solution .

V=10v N=6.25x1020 electrons e = 1.6 x 10-19 Coulomb.

W = V.Q Q=N.e

W=V.N.e=10x6.25x1020x1.6 x 10-19=1000j

Example 2 An electric lamp written on it (220 volts – 100 watt )

a)what is the meaning of the written numbers?.

b) Calculate;

-the intensity of the current of the lamp.

-the quantity of electricity passing through the lamp in this period of time .

- the energy consumed in the lamp in 0.5 hour

Solution . a)220 volt means potential difference needed,100 watt means power or the rate of

consuming the electric energy

- P = IV I = 220

100=

V

pw = 0.454 Amp .

-Q = It = 0.45x1= 0.454 C

- w= V.I. t =220 x 0.454 x 0.5 x60 x 60 = 1.79 x 105 Joules



It is the total work done to transfer a charge of 1 C through the whole electric circuit

i.e. through the source as well as the external circuit.

Unit of e. m. f. :- Volt , joule/coloumb

Is measured also by Voltammeter

What is meant by? The electromotive force of an electric cell = 3 V.

o It means that the total work done to transfer a charge of 1 C through whole circuit

is 3 J.



Lesson two

Ohm related the potential difference

applied at the terminals of a

conductor and the current flowing

through it.

The current intensity in a conductor is directly proportional to the potential difference a

cross its terminals at a constant temperature.

I α V V = I R

Where R is a constant of proportionality known as conductor resistance.

The graphical relation between ( V – I )

the slope of line = R

Resistance a condutor

= potential difference / current

intensity

I

The resistance of a material (R) It is its opposition to the flow of electric current.

OR: the ratio between the potential difference across its terminals and electric current intensity

that flow.

Unit of resistance :- Ohm(Ω),volt/amp Ohm:

It is the resistance of a conductor carries a current of one amp through potential

difference of one volt.

V

I



Different substances offer different resistances to the flow of electric current.

Generally, metals offer little resistance and are good conductor because they are rich

in free electrons.

Example 1 Fuse is designed for maximum current 1 Ampere and 6 volts what is the smallest

resistance can be connected in series with it in this circuit without the fuse melts.

Solution .

R = = = 6 Ω

The factors affecting the electric resistance of a conductor: 1. Temperature where resistance directly to temperature, by increasing temperature the

kinetic energy , the amplitude of vibration of the metal atoms the speed of electrons and

molecules of conductors increases which leads to increase the rate of collision that increases

opposition to the flow of electrons.

At constant temperature R depends on:

2. The length of the conductor (R is directly proportional to L).

3. The cross sectional area of the conductor(R is inversely proportional to A).

4. The material of the conductor .

R α A

L R = ρ

A

L

Where ρ is the constant of proportionality Is called the resistivity of the

material.

The resistivity of a matter is a characteristic physical property of such material that

depends on the type of the material of the resistor and its temperature,

It is the resistance of a conductor of length 1 m and cross sectional area 1 m2 at

constant temperature.

Unit of ρ :- Ohm.m = volt.m/amp

I

V 1

6



It is the reciprocal of the resistivity. or it is the allow of electrons to flow in the conductor.

It is the reciprocal of the resistance of a

conductor of length 1 m and cross sectional

area 1 m2 at constant temperature.

σ = 1

it is a characteristic property of the material

which depends on the material of the resistor

and its temperature,

when the temperature of the conductor increases:

1.its resistivity increases. 2.its conductivity decreases.

Unit of σ :- Ohm-1.m-1 = simon.m-1 = amp/ volt .sec

The product of resistivity and conductivity=1

The electrical conductivity of a material is:very large to some materials (metals) so they

good conductor. very small to some meterials ( insulators) as ZnS2 and CdS2 so they are bad

conductors.

Some materials as germanium and silicon between conductors and isolators, which

is called semiconductors.

1- It is preferable to use copper wires of high conductivity and low resisitivity

in the electrical connection .Give reason

What is meant by ?

1. The resistivity of a conductor = 5 . m.

It means that the resistance of 1m and cross sectional area 1 m2 of this material at

constant temperature = 5

2- The conductivity of a conductor = 5 semon. m-1.

It means that the reciprocal of the resistivity=5

in the electric circuit connected in parallel thick wires are used at the terminals of the

battery, thin wires are used at each resistance…give reason?

Because the current intensity become maximum at the terminals of the battery, thus

thick wires are used due to their small resistance which does not affect the intensity of

the electric, but in the other parts of the circuit the current divides into smaller current

where thinner wires are used.



Example 1 -A power line 5 kilometers long, and of diameter 0.64 mm. Given that the resistivity

of its materials is 1.79 x 10-8 Ω.m, calculate its resistance.

Solution .

L = 5 km

2 r = 0.64 mm r = 0.32 mm

R = ρe x = 1.79 X 10-8 X = 2782 Ω

Example 2 -A copper wire 30 m long and 2 x 10-6 m2 cross sectional area has a voltage

difference of 3 V across. Calculate the current if the copper resistivity is 1.79 x 10-8

.m Solution .

L = 30 m A = 2 x 10-6 m2 V = 3V ρ = 1.79 x 10-8 .m

R = ρ A

L = 1.79 x 10-8

6-10 x 2

30 = 0.268

I = R

V =

0.268

3 = 11.1 A

Example 3 -A wire 2 meter long and cross sectional area 0.1 cm2 , electric current its intensity

1.5 a,pere pass through it when potential difference between the ends of wire is 7.5

volts , find the electric conductivity of the wire material.

Solution .

R = = = 5 Ω

σ = = = = 4 X 104 Ω-1.m-1

A

L

Π(0.32)2 X10-6

5 x103

I

V

1.5

7.5

ρe

1

RA

L 5X0.1X10-4

2



Example 4 -A wire the resistance of 1 meter of it is 35Ω, it is required to be used in making a heater to obtain heat energy of 25200 joule per minute. Calculate the length of the

wire needed if the potential difference is 210 volts.

Solution .

E.E = V.I.t = tR

V2

25200 = 60210 2

xR

==> R = 105 Ω where R of 1 m = 35 Ω

2

11

L

L

R

R=

135

105 1L= L = 3 m

Example 5 -Find the mass of A copper wire of resistance 1.5 Ω its length 50cm,density of its

material=8600kg/m3khnowing that its resistivity =1.79x10-8 Ω.m

Solution .

R = ρ A

L A = ρ

R

L=1.79x10-8x

5.1

1050 2−x

=5.9x10-9m2

m= ρxVol= ρxA.L=8600X5.9x10-9 X50X10-2=2.5x10-5kg

The graphical relation between the resistance of the conductor and its length

R α L R is directly proportional to L at constant c.s.a and temp

2

1

R

R =

2

1

L

L

if length of conductor is doubled the resistance is doubled and when length halved

resistance halved and so on.

R

The slope of line=A

L



The graphical relation between the resistance of the conductor and its cross

sectional area

R α A

1 R is inversely proportional to A at constant L and temp

2

1

R

R =

1

2

A

A =

2

1

2

2

r

r

If c.s.a is doubled the resistance halved and when c.s.a halved resistance doubled

but if radius doubled resistance quartered and radius halved resistance increases four

times where R α 1/r2

The slope of line= ρ L R

A

1

Thick wires (low resistance) are used at the terminals of the battery to allow electrons

to flow and thin wires(high resistance) are used at the terminals of the resistance to

control the flow of electrons in it Give reason

The ratio between the resistances of 2 wires of lengths L1, L2. their radii are r1, r2 and

of resistivities ρ1, ρ2

2

1

R

R =

122

211

A L

A L

= 2

122

2

211

r L

r L

Example 1 A wire of length 30cm,its cross sectional area 0.5cm2its resistance 20 Ω.what is the

resistance of the another wire made of the same materialist length 10cm and cross

sectional area 0.3cm2

Solution .

L1=30cm A1=0.5cm2 R1=20 Ω

L2=10cm A2=30cm2 R2=?

2

1

R

R =

122

211

A L

A L

2R

20 =

0.510x

0.330x R2=11.11 Ω



Example 2 -A student wound a wire of a finite length as a resistor. Then, he made another of the same

material but half the diameter of the first wire and double the length. Find the ratio of the

two resistances. Solution :

r1 = 2 r2 1

2

r

r =

2

1 , L1 =

2

1 L2

2

1

L

L =

2

1

2

1

R

R =

22

11

L

L

2

1

2

2

r

r =

2

1 x

4

1 =

8

1

Example(3) -You have a metallic wire uniform cross-sectional area, if the wire is stretched so that its

diameter reduced to one half of its initial diameter. Calculate the ratio between the two

lengths and the ratio between the two resistances of the wire before and after stretching it.

Volume of wire material in constant.

Solution :

L1 A1= L2 A 2 = = ( )2 = ( )2 =

= ( )2 = ( )2 =

Example(4) -In an experiment for verification of ohm's law, connected a wire of 2 meter length and cross

sectional area 0.1 cm2. The following were obtained.

Potential difference

( Volt ) 3.5 3.00 2.5 2 1.5 1.0 0.5

Current intensity

( Ampere ) 1.75 1.50 1.25 1 0.75 0.5 0.25

Draw a graph for the results and find:

Resistance of the wire – The resistivity of it – The conductivity of the wire.

Slope = = = 2

R = slope = 2 Ω

ρe= =

= 10-5 Ω . m

σ = = 105 Ω -1 . m-1

L2

L1

A1

A2

r1

r2

2

1

4

1

R2

R1

L2

L1

r1

r2

4

1

16

1

2

1

V

I 0.5- 0.25

1- 0.5

L

R.A

2

2 X 0.1 X 10-4

ρe

1

10-5

1

0.25 0.50 0.75 1 1.25 1.5 1.75

3.5

3

2.5

2

1.5

1

0.5



Lesson three

1-Connecting resistors in series: The aim of connecting resistors in series is to obtain a large resistance from small

ones.

==The electric current intensity is the same in all resistors .

==the total potential difference a cross the group is equal to the sum of the potential

difference a cross the resistors (kerschofs law)

V = v1 + v2 + v3

But V = IR

V1 = IR1

V2 = IR2

V3 = IR3

Rt = R1 + R2 + R3

Thus, the equivalent resistance Rt of a group of resistors connected in series equals

the sum of these resistances.

If N resistance are equal each R,

We conclude that if we want a large resistance out of small resistances, we simply

connect them in series.

IRt = IR1 + IR2 + IR3

Rt = NR



2-Connecting resistors in parallel : The aim of connecting resistors in parallel is to obtain a small resistance from a

group of large ones.

The potential difference is the same a cross in all resistors and the electric current

is divided such that:

I = I1 + I2 + I3

tR

V =

1R

V +

2R

V +

3R

V

tR

1 =

1R

1 +

2R

1 +

3R

1

Rt is the equivalent resistance.Rt is

small than the smallest resistance of

any the individual resistors.

The reciprocal of the equivalent resistance Rt is the sum of the reciprocal of

resistances in the case of a parallel connection.

In the case of two resistors in parallel, the equivalent resistance Rt is given by:

When N resistances are connected in parallel each equal toR,

tR

1 =

R

N

Therefore, if we wish to obtain a small resistance out of resistors, we simply connect

them in parallel.

Give reason

1. Electric devices in our home circuits is connected in parallel .

a) If they are in series , the equivalent resistance will increase and current intensity

through the circuit decreases.

b) In parallel connection we can control the switching of every device on or off , but if they

are in series they will work together at the same time if we switch one device on or not

work together at the same time if witch one device off.

Rt = 21

21

R R

R R

+

Rt = N

R



2. when the filament of a lamp at home melts the other lamps still lighting .

As they are connected in parallel.

Example(1) 1-Find the equivalent resistance for the resistors in

the circuit shown in figure.

Solution R1, R2, R3 are series.

R± = 1+3+2 = 6 Ω

R± and R 4 Parallel Req= = 2 Ω

Req, R5, R6 series

R = 4+2+2 = 8 Ω

Example (2) -Find the equivalent resistance for the resistors

shown in the figure.

Solution 1 Ω, 2 Ω series

R± = 1+2 = 3 Ω

3 Ω, 6 Ω parallel

R± = = 2 Ω

2 Ω, 3 Ω series

R± = 3+2 = 5 Ω

4, 5 parallel

R± = = 2.22 Ω

Example (3) -Find the equivalent resistance between points a

and b for the combination shown in the figure.

Solution 3, 2 series R±= 3+2 = 5 Ω

5, 6 parallel R± = =

R±, 7 series R± = + 7 =

2

4

3 6

2

4

2

4 2

3

4 5



, 12, 5 parallel

= ==> R± = 2.6 Ω

2.6 Ω, 9 Ω series ==> R± = 9 + 2.6 = 11.6 Ω

Example (4) -Three resistances 10 , 20 , 30 , Ω are connected in parallel. A current of (1) ampere passes through the first resistance . calculate the total current through the resistances

solution

V1 = I1R1 = 1+10 = 10 v

Each one has no ltage 10v bec they are parallel

I2 =

I2 =

It = I1 + I2 + I3 =

Example (5) -Resistances 10, 20, 30 Ω are connected with electric source, so electric currents of 0.15, 0.2 and 0.05 Amperes are passing in the three resistances respectively. Find the

equivalent resistance and draw the connection manner.

solution V1 = I1R1 = 0.15 x 10 = 1.5v

V2 = I2R2 = 0.2 x 20 = 4v

V3 = I3R3 = 0.05 x 30 = 1.5v

R1, R3 has same voltage so they are parallel

Rt =

Rt = 27.5 Ω

Example (6) -Their resistors 25, 70, 85 are connected in series to a 45 volt Calculate:

The current flowing in each resistor.

The potential difference across each resistor.

Solution :

10

30

25 70 85

20



The total resistance of the circuit R is:

R = R1 + R2 + R3 = 25 + 70 + 85 = 180

The current flowing through each resistor is:

I = r R

VB

+ =

0180

45

+ = 0.25 A

V1 = IR1 = 0.25 x 25 = 6.25 V

V2 = IR2 = 0.25 x 70 = 17.5 V

V3 = IR3 = 0.25 x 85 = 21.25 V

Example (7) -If the resistors in the previous example are connected in parallel to the same battery,

calculate:

a ) The current flowing in each

resistor.

b ) The total resistance.

c ) the current through the circuit.

Solution : a) The voltage difference across each

resistor = 45 V, since they are

connected in parallel and the battery is

of negligible internal resistance. The

current flowing through each resistor is calculated separately as follows:

I1 = R

V

1

= 25

45 = 1.8 A

I2 = R

V

2

= 70

45 = 0.643 A

I3 = R

V

3

= 85

45 = 0.529 A

b b) The total ( equivalent or combined ) resistance R is calculated as follows:

R

1

=

1R

1 +

2R

1 +

3R

1 =

25

1 +

70

1 +

85

1

R/ = 15.14

cc ) The current flowing through the circuit I is:

I = rR

V

+ =

015.14

45

+ = 2.972 A

It can be calculated also as the sum of the currents I1, I2, I3 flowing through all

resistors:

I = 1.8 + 0.643 + 0.529 = 2.972 A

=========================



Example(9) -A wire of uniform cross-section, a current of intensity 0.1 ampere passes in it when

the potential difference between ends 1.2 volt. The wire was changed in the form of a

closed square ABCD. Calculate the equivalent resistance of the wire in the two

following cases.

a- the source is connected to the two point A and C.

b- the source is connected to the two points A & B.

Solution .

R= == 121.0

2.1

I

V

Rside= 12/4 =3 Ω

a-R=6/2=3 Ω

b-R=9x3/(9+3)= 2.25Ω



Lesson four

The electromotive force ( VB ) is the total work done to transfer a charge of one

coulomb through the whole circuit.

VB = Potential difference across the external resistance ( Vex ) + Potential

difference across the internal resistance of the source.

VB = Vex + Vin

VB = IR + Ir

VB = I ( R + r )

I = r R

VB

+ ohm’s law for a closed circuit

Where R is the external resistance and r is the

internal resistance.

Note 1- The reading of voltmeter across terminals of battery= Potential difference

across the external resistance ( Vex )

2- E.M.F of a cell is greater than potentional difference between its poles .

According to the relation VB= Vex + Ir, Vex is less than VB due to the internal work

done to overcome the internal resistance (Ir)

3- It of source depends on Rt of circuit. If ( R ) increases by using rheostat:

a- ( I ) decreases (reading of ammeter decreases)

b- the potential difference across the external resistance decreases(reading of

voltmeter decreases)

c- the potential difference across source increases.

4- the potential difference across the source becomes equal to e. m. f when

a- the current vanishes by increasing resistance of rheostat to very large value.

b- r = zero.

Vex = VB - Ir

OOhhmm’’ss llaaww ffoorr aa cclloosseedd cciirrccuuiitt



Electromotive force of a source :

Is the potential difference a cross it when no current is flowing through its circuit.

as the internal resistance decreases which leads to increase Potential difference

across the external resistance or source ( Vex ) The efficiency of the battery

increases

The efficiency of the battery = vex / vB

Example(1) -A cell of e. m. f. 2V and internal resistance 0.1 is connected to a circuit of

external resistance 3.9 . Calculate the current flowing through the circuit. Solution :

I = r R

VB

+ =

0.1 9.3

2

+ = 0.5 A

Example(2) -A 4.7 resistor is connected across the terminals of a battery of 12 v emf and 0.3

internal resistances. Calculate:

a ) The current in the circuit.

b ) The voltage difference across the resistor.

Solution : VB =12 R = 4.7 r = 0.3

I = r R

VB

+ =

0.3 4.7

12

+ =

5

12 = 2.4 A

V = IR = 2.4 x 4.7 = 11.3 v

Example(3) -In the electric circuit in the figure:



a) The external resistance = ………. ohm. b) The total resistance = ………… ohm. c) The total current intensity in the circuit =

………… ampere. d) The intensity of the current in resistance

3 ohm = ………. ampere. e) The reading of the voltmeter V1 = ……., V2 = ………, V3 = ……….

Solution .

4, 4 Ω are parallel Rt =

2, 3 are series R = 5 Ω

I =

I 3 Ω = 2 A

V1 = I1, R1 = 2 x 3 = 6v

V2 = I2, R2 = 2 x 2 = 4v

V3 = It . Rex = 2 x 5 = 10v

Or V3 = VB – Ir = 12 – 2 x 1 = 10v

Example(5) -Two resistors A of 3 and B of 6 are connected in parallel. The combination is

connected in series with a resistor C of 7 and a 18 volt battery of negligible internal

resistance. calculate:

a ) The total resistance.

b ) The current flowing through the circuit.

c ) The current through each of A and B.

Solution : The equivalent resistance for the combination ( A, B ) is:

R/ = 21

21

R R

R R

+ =

63

6 x 3

+ = 2

The equivalent resistance for the combination ( A, B ) and ( C) is:

R = R/ + R3 =2 + 7 = 9

The current I flowing through the circuit is:

I = rR

V

+ =

09

18

+ = 2 A

To calculate the current flowing through each of A and B, we should calculate the

potential difference V/ across the combination ( A, B ):

V/ = IR/ = 2 x 2 = 4 V

Assuming the current flowing through A and B to be I1 and 12, respectively:

I1 = 1R

V =

3

4 = 1.333 A



3x2

5

E

R+r

24

1.2 +0.4

V

R

18

2 12

3

12

6

I2 = 2R

V =

6

4 = 0.667 A

Example(6)

-Calculate the total resistance in the circuit

shown an also the total current intensity if the

internal resistance of each cell is 2

Solution : R ( 20, 40 ) in series = 20 + 40 = 60

R ( 20 , 30 , 60 ) in parallel = 20

1 +

30

1 +

60

1

= 60

6 = 0.1 = 10

R ( 10 , 0.1 ) in series = 10 + 10 = 20

VB (total) = 6 + 12 = 18 V.

r (total) = 2 x 2 = 4

I = r R

VB

+ =

4 20

18

+ = 0.75 A

Example(8) -From the circuit given in the fig .Calculate : 1Ω a- Total resistance b- Total current c- Current passes in each resistor .

Solution .

R6,3 = =2Ω R6,3.1 = 2+1=3 Ω

a)Rt = = 1.2Ω b) I = = = 15A

Vt = IRr = 15x1.2 = 18 Volts

I2 Ω = = =9A I1Ω = 15 –9 = 6A

V3,6 = 6x2= 12 volts I3 = = 4A I6 = = 2A

In case of charging the battery of the car :

The battery is connected by electrical charger is larger than the electromotive force of

the battery.

1. .

Icharging = (in)(ex) R R

)battary ( e.m.f - )charger ( e.m.f.

+



In case of circuit connected to no of batteries :

Example

In the given figure determine the current intensity

and the potential difference across each battery.

(VB) t = 30 + 20 – 10 = 40v

It =

V1 = VB1 – Ir1 = 20 – 4 x 1 = 16v

V2 = VB2 + Ir2 = 10 + 4 x 0.5 = 12v

V3 = VB3 – Ir3 = 30 – 4 x 1.5 = 24v



Lesson five Kirchhoffs Laws

1st Law

( Law of conservation of the charge)

(Kirchhoffs current Law) Kirchhoff s first law;

At any node [junction] in a closed electrical circuit, the sun of currents flowing into the

node is equal to the sum of currents flowing out of that node

Or

The algebraic sum of all current entering and leaving a node must equal zero .

Ienteriny = I leaving

Note: . Kirchhoff s first law is used in the circuits that include resistors connected in parallel

due to the presence of junction for current distribution

. Kirchhoff s first law is an application of the law of conservation of charge[ the quantity

of charge flowing into a node is equal to the quantity of charge flowing out of that

node] because the electric current intensity in [Amperes] equals the quantity of electric

charges in [coulomb] crossing a cross-section area in one second.

I1 + I2 = I3 + I4

Example

From the opposite figure Find I

Solution

4 + 5 + 2 = 8 + I

I = 3 A

Example

in the opposite figure If = I1 = 2 mA

2A 8A

5A 4A

I

I2

I I1 I3

I4



I2 = 5 m A

I3 = 0.5 m A

Find I , I4

Solution

I = 2mA + 5mA = 7mA

I1 = I3 + I4

2mA = 0.5m A + I4 ⎯→⎯ I4 = 1.5 m A

Example

in the opposite figure

Solution

At point b 5=4+I1 I1=1A

At point c I3=4+4=8A

At point a I3=5+I2 8=5+I2 I2=3A

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2nd Law

( Law of conservation of energy)

(Kirchhoffs voltage Law) The algebraic sum of voltage around a closed loop equal zero .

V = I R

The algebraic sum of the electromotive forces in any closed path is equivalent to

The algebraic sum of voltage within that loop.

Note

1) As current passes through a resistor then a voltage is produced .

2) The electron current enters the negative side of a component and exist

the positive side .



3) The source current from positive into negative .

⎯→⎯ Current flow

Note; When solving problems of Kirchhoff s second law, the direction of every closed path

must be determined clockwise or counter clockwise.

The emf of the cell [v]; It is the potential difference across the terminals of the cell in case of no current in the

circuit[switch is opened].

Or

It is the total work done inside and outside the cell to transfer an electric charge of 1 C

[electric charges unit] in the electric circuit.

Example : find the current of battery in the following circuit

Solution

Applying Kcl on left Loop Applying Kcl on right Loop

32 = 2 I1 + 8 I2 20 = -4 I3 + 8 I2

32 = 2 I1 +8 (I1 – I3) 20 = -4 I3 + 8 (I1 – I3)

32 =2 I1 + 8 I1 - 8 I3 20 = -4 I3 + 8I1 - 8 I3

32 = 10 I1 - 8 I3 20 = - 12I3 + 8 I1

I2 32V

2

8 20

I3 4

I1



5 I1 - 4 I3 = 16 2 I1 – 3 I3 = - 20

I1 = 4 A I2 = 1A I3 = 3A

Example : find the current of battery in the following circuit

Applying Kcl on left Loop on right

V = 0

10 + 2 = 5 I1 + 4 I2 + 2 I1 2 + 5= 4 I2 – 3 (I1 – I2)

7= 4 I2 – 3 I1 + 3 I2

12 = 7 I1 + 4 I2 7 = 7 I2 – 3 I1

I2

I1

10 V

I1

2

5

5 I1

I1

4I2 4

3( I1 – I2

2 V

5 V

3 I3= I1 – I2

10V

12

4

5 V 2V

2

3



Example : find the current of battery

in the following circuit

Solution

Applying Kcl on left Loop on right

24 = 2 I1 + 4 I1 + 3 I2 12= - 3 I2+I3+5I3

24 = 6 I1 + 3 I2 eq 1 12 = - 3 I2 +6(I1- I2)

12 = - 3 I2 – 6 I1 -6 I2

12= -6 I1 -9 I2 eq 2

I1

24V

12V

5

1

2

3 I2

4 4 I1

I1 - I2

I3 = I1 – I2

3 I2

2 I1

24V

12V

5

1

2

4

3



Example In the following figure calculate :

1- Current intensity in each branch

2- Pd between a, b

Solution

- in abdea

6 = 2 I1 + 5 I3 = 2 I1 + 5 (I1 + I2)

6 = 7 I1 + 5 I2 ⎯→⎯ (1)

- in c de fc

2 = 3 I2 + 5 (I3) = 3 I2 + 5 (I1 + I2)

2 = 5 I1 + 8 I2 ⎯→⎯ (2)

multiplying eq (1) x 5 and eq (2) x 7

30 = 35 I1 + 25 I2

14 = 35 I1 + 56 I2

16 = - 31 I2

I2 = - 0.516 A

negative sign indicate that I2 in opposite direction sabsitude in eq (2)

2 = 5 I1 8 x (- 0.516)

I1 = 1.226 A

where I1 + I2 = I3

1.226 + (-0.516) = I3

I3 = 0.71 A

V = VB – IR

= 6 – 1.226 x 2

= 3.55 V.

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Example In the opposite ciruit calculate I1 , I2 , I3

Solution

In abcfa

15 + 10 = 1 I1 + 9.5 I1 + 0.5 I2

25 = 10.5 I1 + 0.5 I2 x 2

50 = 21 I1 + I2 ⎯→⎯ (1)

in fcdef

3 + 10 = 0.5 I2 + 0.1 I3 + 1.4 I3

13 = 0.5 I2 + 1.5 I3 x2

26 = I2 + 3 I3 ⎯→⎯ (2)

subsitude I1 = I2 – I3 in eq (1)

50 = 21 (I2 – I3) + 22 I3 – 21 I3 ⎯→⎯ (3)

multiply eq (2) 7 and add with eq (3)

182 = 7 I2 + 21 I3

50 = 22 I2 – 21 I3

232 = 29 I2

I2 = 8 A

substitute in eq (1)

50 = 21 I1 + 8

I1 = 2 A

I3 = 6 A

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Example In the opposite figure find :

1- current intensity in each battery .

2- p. d across terminal of each battery .

3- p.d across 5

Solution

In a e c b a

20 – 30 = I1 x 1 – I2 x 2

- 10 = I1 – 2 I2 ⎯→⎯ (1)

in a e f d b a

20 – 15 = 1xI1 + I3 x 5

5 = I1 + 5 (I1 + I2)

5 = 6 I1 + 5 I2 ⎯→⎯ (2)

Multiply eq (1) x 5 and eq (2) x 2

I1 in opposite (the battery recharging) subsitude in eq (1)

I2 = 3.82 A (The battery discharing).

I3 = 1.46 A

- P.d across 20V

V1 = VB – IR = 20 – (2.35) x 1 = 22.35

- P.d cross 30V

V2 = VB – IR = 3.82 x 2 = 22.35 V.

- VR = IR = 1.46 x 5 = 7.3 V.

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===Electric current, ohm's law ......===Electric current, ohm's law=====Unit ( 1 ) Chapter ( 1 )===...

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