MR.AHMED HEKAL

CHAPTER (1) DYNAMIC ELECTRICITY

Mr. Ahmed Hekal

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Chapter 1

A. Definitions 1- Electrical current:

- Electrons movement through a conductor from negative pole to positive pole in the presence of electric

source

2- Traditional Electrical current:

- Movement of positive charges through a conductor from positive pole to negative pole

3- Current intensity:

- Quantity of electrical charges through a section of a conductor in 1 second

4- Ampere:

- Current intensity results in passing quantity of charges of 1 Coulomb through a conductor in 1 second

5- Coulomb

- The quantity of electrical charges that when passes through a conductor in time of 1 second it generates

current of 1 Ampere

6- Potential difference:

- The work done in joule used to transfer quantity of charges of 1 C between 2 points

7- Volt:

- The potential difference bet. 2 points when work done of 1 J used to transfer quantity of charges of 1 C

8- Electromotive Force:

- The whole work done outside and inside the battery to transfer 1 coulomb of charges in the electrical circuit

- Potential difference between the poles of the battery in case of open circuit

9- Electrical Resistance:

- The opposition الممانعة to the flow of the current

- The ratio between potential difference in volt across the terminals of conductor and current intensity that

flows through this conductor in ampere

10- Ohm’s Law:

- At constant temperature, current intensity is directly proportional with potential difference

11- Ohm:

- The resistance of conductor that allows passage current of 1 Ampere when potential difference across

terminals of this conductor is 1 volt

12- Resistivity:

- It’s the resistance of conductor of length 1 m and cross sectional area of 1 m2 at constant temperature

13- Conductivity:

- The reciprocal of resistivity

- Reciprocal of resistance of conductor of length 1 m and cross sectional area of 1 m2 at constant temperature

14- Kirchhoff’s first low:

- In a closed circuit, summation of currents entering a point is equal to summation of currents exiting this

point

- The algebraic summation of currents in a specific point in closed circuit equals zero

15- Kirchhoff’s second law:

- Algebraic summation of e.m.f. in a closed circuit equals the algebraic summation of potential differences in

this circuit

- Algebraic summation of potential differences in a closed branch equal zero

Mr. Ahmed Hekal

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B. What’s meant by? 1- Current intensity passes through a conductor is 5 A

- The quantity of electrical charges that passes through a section of this conductor in 1second equals 5

coulombs

2- Potential difference between terminals of a conductor is 20 V

- The work done to transfer 1 C of charges bet. terminals of this conductor is 20 Joule

3- Electromotive force of a battery = 1.5 V

- The whole work done outside and inside the battery to transfer 1 coulomb of charges in the circuit = 1.5

joule

4- Electrical Resistance of a conductor 100 Ohm

- The ratio between potential difference across the terminals of conductor and current intensity that flows

through this conductor is 100 V/A

5- Resistivity of a conductor = 6 *10-6 ohm.m

- Resistance of a conductor of this material of length 1 m and cross sectional area of 1 m2 at a constant temp.

is 6 *10-6 ohm

6- Conductivity of material is 5.6 *107 ohm-1.m-1

- Reciprocal of resistance of conductor of length 1 m and cross sectional area of 1 m2 at constant temperature

Is 5.6 *107 ohm-1

C. Deductions 1- Resultant resistance of group of resistances connected in series

- Current intensity is constant through all resistances

- Potential difference is divided between them

V = V1 + V2 + V3

V = I R

I R` = I R1 + I R2 + I R3

Then

2- Resultant resistance of group of resistances connected in parallel

- Potential difference is constant across all resistances

- Current intensity is divided between resistances

I = I1 + I2 + I3

I = V / R

V / R` = V / R1 + V / R2 + V / R3

Then

R` = R1 + R2 + R3

1 / R` = 1 / R1 + 1/ R2 + 1 / R3

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D. Factors that depends on Physical quantity Factors that this quantity depends on

Resistance of a conductor

R = Ꝭe L/A

1- Length of conductor (directly proportional) 2- Its cross sectional area A (inversely

Proportional) 3- Material type of this conductor 4- Temperature of this conductor

Resistivity of conductor 1- Material type of this conductor 2- Temperature of this conductor

Conductivity of a conductor

1- Material type of this conductor 2- Temperature of this conductor

F. Comparisons

resistances connected in series resistances connected in parallel

Shape of connection

Target

To obtain a large resultant resistance from a group of small resistances

To obtain a small resultant resistance from a group of large resistances

Current intensity Equal or constant in all resistances The whole current equals the summation of all currents in all resistances

Potential difference The whole P.D. equals the summation of all P.D.s in all resistances

V = V1 + V2 + V3

Equal or constant across all resistances

Law of resultant

If all resistances are equal

R` = N R Where N is number of them

R` = R / N

R` = R1 + R2 + R3 1 / R` = 1 / R1 + 1/ R2 + 1 / R3

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E. What Happens in these cases 1- Increasing potential difference to double value for current intensity and power consumed?

- Current intensity will increase to double as I = V / R

- Power will increase 4 times as Pw = V2 / R

2- Current intensity increases to double for the resistance value

- Resistance remains constant as it doesn’t depends on current it depends on

- Length of conductor (directly proportional)

- Cross sectional area A (inversely Proportional)

- Material type

- Temperature

3- Increasing cross sectional area of a conductor to double and decreasing its length to half value for the

resistance

- L1 = 2 L , L2 = L , A1= A , A2 = 2 A

- R1 / R2 = L1 A2 / L2 A1 = 2 L * 2 A / L * A

- R2 = ¼ R1

4- Connecting two resistance in parallel on of them has a value of 1 ohm to the resultant resistance

- Resultant resistance will be less than 1 ohm

5- Current doesn’t flow from an electric source to potential difference between the terminals of this electric

source

- Potential difference between the terminals of this electric source will be equal to electromotive force of the

electric source according to this relation

(V = VB – I r) and I = 0 then V = VB

F. Give reason 1- Work should be done to transfer charges from point to point

- To get rid of resistance bet. the two points and current can flow

2- Some materials can conduct electricity but others cannot

- As some materials have a plenty of free electrons so it allows flow of current while other materials don’t

have free electrons or their electrons are strongly correlated to their atoms

3- Increasing radius of a wire of copper leads to decreasing its resistance to quarter value

- According to this relation

R = Ꝭe l/r2

Resistance is inversely proportional with square of radius

4- When a conductor is shaped to be parallelogram its ribs resistances are different while if shaped as a

cube its ribs resistances are equal

- As ribs of parallelogram are different in length so their resistance differs according to relation

R = Ꝭe l/A but cube ribs are equal in length and equal in resistance 5- Resistance increase when increasing temperature

- When raising temperature this increase the speed of vibrating its molecules and increase the rate of

collisions between electrons of current and conductor molecules so the opposition الممانعة of current

increases

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6- Resistivity of a conductor is a physical property

- As it depends on the type of the conductor material at constant temperature

7- Conductivity of a conductor is a physical property

- Because conductivity is the reciprocal of resistivity which depends on the conductor material at constant

temperature 8- Conductivity factor of copper is large

- As resistivity of copper is very small in cause of plenty of free electrons in copper

9- It’s preferred to use wires of copper in the electrical connections

- As resistivity of copper is very small so its resistance is low and this prevent wire from consuming electrical

energy

10- Home devices are connected in parallel

- All devices will work on the same potential difference of the source so each device can work alone and if

one of them damaged it doesn’t affect the others and also to decrease their total resistance which doesn’t

affect the main current

11- Home devices aren’t connected in series

- Because potential difference will be divided across them which leads to a probability of insufficient voltage

on a device that cannot operate and one device cannot work alone also their total resistance will be huge

which prevents the current from passing through the circuit

12- The electrical power increases in case of connecting two resistance on parallel

- When connecting resistances in parallel their total resistance decreases and current will increase and Power

also will increase according to this law Pw= IV

13- In a circuit of parallel connection resistances we use thick wires at the terminals of battery and thin

wires at the terminals of each resistance

- As current intensity should be maximum at the input and output of the battery so we use thick wires of low

resistance while current is divided through each resistance so we can use thin wires in terminals of each

resistance

14- Potential difference between battery poles increases when increasing resistance of its circuit

- According to this relation (V = VB – I r) when increasing resistance current will decrease and the internal

potential difference (I r) will decrease and because VB is constant then potential difference across the

battery will increase

15- E.M.F of a battery is larger than potential difference between its outer terminals when closing the

switch

- Because the internal resistance of the battery consumes power to allow current to flow inside it so

(VB = V + I r) so VB > V

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Mr. Ahmed Hekal

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Laws

To compare between Power consumed in two

resistances

When V is constant

(Pw)1 / (Pw)2 = R2 / R1

When current is constant

(Pw) 1 / (Pw)2 = R1 / R2

Resistance

R = V / I

R = Pw / I2

R= V2/Pw

R = Ꝭe L/A = Ꝭe L/r2

Ꝭ (density) = m (mass) / v (volume)

Ꝭ = m / v

And v = L * A and L = v / A

Then Ꝭ = m / L A then

A = m / L Ꝭ and

L = m / A Ꝭ

R = Ꝭe L2 Ꝭ / m

R = Ꝭe v (volume) / A2

R = Ꝭe L2/ v (volume) Compare between resistances

R = Ꝭe m / Ꝭ A2

R1/R2= Ꝭe1 L1 A2 / Ꝭe2 L2 A1

R1/R2 = Ꝭe1 Ꝭ1 L12 m2 / Ꝭe2 Ꝭ2 L2

2 m1

R1/R2 = Ꝭe1 L1 r22 / Ꝭe2 L2 r1

2

Conductivity

σ = 1 / Ꝭe= L / R A

Laws

Quantity of charges

Q = I t

Q = n qe

Q = W / V

Q: total quantity of charges (electrons)

n: number of electrons

qe: charge of one electron

W: work done

V: potential difference

I: current intensity

t: time

Potential difference V

V = W / Q

V = W / nqe

V = I R

V = Pw / I

Current Intensity

I = Q / t

I = n qe / t

I = V / R

I = Pw / V

Electrical Power

Pw = W / t

Pw = V I

Pw = I2 R

Pw = V2 / R

Mr. Ahmed Hekal

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Mr. Ahmed Hekal

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To Solve Kirchhoff’s Problems follow the following:

1- Divide the circuit to no. of loops (2 or 3 loops)

2- Find a point (junction) in which all currents are entering or leaving it

3- Write the first equation using Kirchhoff’s first law (sum of currents entering a point equals to sum o

currents leaving this point

4- Specify my direction as follow

A. If there is one battery in the loop, specify the direction to be from +ve pole to –ve pole of this

battery

B. If there are two batteries in the loop, specify the direction to be from +ve pole to –ve pole of

the largest battery

5- Write the 2nd and 3rd equations using Kirchhoff’s second law (sum of potential difference inside a

loop equals zero), we have two choices:

A. If there is one battery in the loop, then write its value direct V= IR + IR

B. If there are two batteries in the loop, then we have two choices

If they are connected parallel ( -ve connected to –ve and +ve connected to +ve) then

subtract their values V1-V2 = IR +IR

If they are connected in series ( -ve connected to +ve and +ve connected to negative)

Then add their values to each other V1+V2= IR +IR

C. The sign of IR depends on the direction of the current

If the current passes in the same direction of my direction then put it in +ve (+IR)

If the current passes in the opposite direction of my direction then put it in –ve (-IR)

6- Current that leaves the battery is the same current enters the battery

7- Potential difference between two points is equal to the potential (or voltage) at the point of higher

potential the potential (or voltage) at the point of lower potential

8- Solve the 3 equations using calculator