Current Electricity Notes

Current Electricity

Definitions:

Electronic Current: The flow of electrons from the negative pole to the positive pole inside a solution.
Conventional (Electric) Current: The flow of positive charges from the positive pole to the negative pole outside the solution.
Semiconductors: Substances with conductivity between conductors and insulators, widely used in electronic components.
Conductors: Substances that allow electric current to flow through them.
Insulators: Substances that do not allow the movement of electrical charges through them.
Electric Current Intensity (I): The quantity of charge (Q) passing through a cross-section of a conductor per second. $I = Q/t$
Ampere: The unit of electric current, defined as the flow of one coulomb of charge per second. 1 Ampere = 1 Coulomb/second.
Electrical Resistance (R): The opposition to the flow of electrons in a conductor.
Resistivity ($\rho$): The resistance of a conductor of length one meter and has a cross-sectional area of one square meter. It is an intrinsic property of the material.
Ohm's Law: The current intensity (I) passing through a conductor is directly proportional to the potential difference (V) between the ends of the conductor, at constant temperature. $V = IR$
Electromotive Force (EMF) (E): The potential difference between the two ends of a battery when no current passes through the circuit (open circuit voltage).
Ohm’s Law for a Closed Circuit: Current flows through a conductor when it is connected to a source, causing a potential difference between the ends of the conductor.
Complete Ohm's Circuit: A circuit that contains an electric source (e.g., battery) and resistances.

Rules and Units:

Physical Quantity	Rule	Unit
Electric Current (I)	$Q = It$	Ampere (A) = Coulomb/second
Resistivity ($\rho$)		Ohm.Meter ($\Omega$.m)
Ohm's Law	$V = IR$
Resistance (R)	$R = \rho \frac{L}{A}$	Ohm ($\Omega$)
Ohm's Law for Closed Circuit	$V_O = E - Ir$	Volt (V)
Electromotive Force (E)		Volt (V)
Current in Complete Circuit	$I = \frac{E}{R + r}$	Ampere (A)

Where:

Q is the electric charge in Coulombs (C).
I is the electric current in Amperes (A).
t is the time in seconds (s).
$\rho$ is the resistivity in Ohm-meters ($\Omega$.m).
L is the length of the conductor in meters (m).
A is the cross-sectional area of the conductor in square meters (m²).
V is the potential difference (voltage) in Volts (V).
R is the resistance in Ohms ($\Omega$).
E is the electromotive force in Volts (V).
r is the internal resistance of the source in Ohms ($\Omega$).

Uses:

Semiconductors are used in the manufacture of electronic circuit components.
Ohm's law is used to measure electric current passing through a resistance.

Factors Affecting Resistance:

Length of the wire (L): Resistance is directly proportional to the length of the wire. $R \propto L$
Cross-sectional area (A): Resistance is inversely proportional to the cross-sectional area of the wire. $R \propto \frac{1}{A}$
Kind of material (Resistivity $\rho$): Different materials have different resistivities. Resistance is directly proportional to the resistivity. $R \propto \rho$
Temperature: For most conductors, resistance increases with increasing temperature.

Resistances Connection

A. Series Connection:

Physical Facts:
1. The current is the same through all resistors: $I{tot} = I1 = I2 = I3 = …$
2. The potential difference is divided across the resistors: $V{tot} = V1 + V2 + V3 + …$
3. Total resistance is the sum of the individual resistances: $R{tot} = R1 + R2 + R3 + …$
4. The equivalent resistance is greater than any individual resistance in the circuit.
5. If there is a short circuit in any part of the series circuit, the current stops flowing through the entire circuit.
6. For n equal resistors in series: $R_{tot} = nR$ where $n$ is the number of resistors and $R$ is the resistance of one resistor.

B. Parallel Connection:

Physical Facts:
1. The current is divided among the branches: $I{tot} = I1 + I2 + … + In$
2. The potential difference is the same across all resistors: $V{tot} = V1 = V2 = … = Vn$
3. The reciprocal of the total resistance equals the sum of the reciprocals of the individual resistances: $\frac{1}{R{tot}} = \frac{1}{R1} + \frac{1}{R2} + … + \frac{1}{Rn}$
4. The equivalent resistance is less than the smallest individual resistance in the circuit.
5. If there is a short circuit in one branch of a parallel circuit, the other branches are not affected.
6. Household electrical equipment is connected in parallel to ensure each device receives the same voltage and a short circuit in one device does not affect the others.
7. For n equal resistors in parallel: $R_{tot} = \frac{R}{n}$

Definitions (Scientific Terms):

Materials that do not allow electric current to pass through them: Insulators

Ohm's Law (Words):

The current intensity passing through a conductor is directly proportional to the potential difference between the ends of the conductor, at constant temperature.

Ohm's Law (Equation):

$V = IR$ where V is voltage, I is current, and R is resistance.

Ohm's Law (Symbols):

The electric current (I) passing through a conductor is directly proportional to the potential difference (V) across the conductor. Symbolically: $I \propto V$

Units:

Electric Current: Ampere (A)
Resistivity: Ohm.Meter ($\Omega$.m)

Physical Quantities Measured by Units:

Ohm. Meter: Resistivity
Ampere. Second: Electric Charge (Coulomb) (Since $Q = I \times t$ )
Coulomb /second: Electric Current (Ampere)

Complete the Following:

The electronic current in electric cells moves from the negative electrode to the positive electrode.
The electrons move freely between the atoms when they gain energy or heat.
Substances which prevent current to pass through are called insulators in those substances electrons cannot liberate from their atoms.
From the best electric conductors: 1- Silver 2- Copper
Semiconductors are used in electronic circuits components
The quality of electric conductors depends on resistivity
Electric cells connect in parallel to increase current intensity

True or False:

We can increase electric current intensity passing through a device by:
- Increasing of electric cells in series. (/)
- Increasing of electric cells in parallel. (/)
- Connecting another device in series. (x)
- Connecting another device in parallel. (/)

Multiple Choice:

The ampere unit equals:
- c- (a and b) are corrected (Volt/ohm, Coulomb s-1)
Ohm unit equals:-
- b- volt. Sec/Coulomb
The Ohm unit equal all of the following except:
- d-Joule /Coulomb .Ampere
The voltmeter is connected in parallel with electric devices because:
- a- It has a high resistance.

Short Questions:

Two factors that determine the resistance of conductors:
- (a) Length: Resistance is directly proportional to length.
- (b) Cross-sectional Area: Resistance is inversely proportional to cross-sectional area.
Resistance of a wire with length (L), radius of cross-section (R), and resistivity ($\rho$):
$R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$
Uses:
- Semi-conductors’ substance: Manufacture of electronic circuits components
- Ohm's Law in electricity: To find the electric current, the voltage or the resistance.
Explanations:
- The copper metal is widely used in electric connections in spite of the fact that silver has lower specific resistance than copper.
  Copper is more economical than silver, despite silver's slightly lower resistivity. The cost-effectiveness of copper makes it a preferred choice for widespread electrical applications.
- Home electric appliances are connected in parallel.
  Parallel connections ensure that each appliance receives the same voltage (the mains voltage). Also, if one appliance fails or causes a short circuit, the other appliances will continue to function without interruption.
Other units equal to:
- The volt:
  - 1- Joule/Coulomb
  - 2- Ampere x Ohm
- The electric resistance:
  - 1- Volt / Ampere
  - 2- $\rho$ (L/A)
Scientific term:
- the quantity of electric charge passing through a cross-section of conductor in one second: (Electric Current Intensity)
- the resistant of a conductor if the current passing is one ampere and the potential difference is one volt: (Ohm)
Use the suit able scientific law to find out the unit for measuring the following:
- The electric current Ampere (Coulombs/Second).

Problems:

An electric current of 4 amperes passed through a wire for a period of 4 seconds. Find the number of electrons which have passed through the wire if the charge of the electron is $1.6 \times 10^{-19}$ coulomb.
- $Q = It = 4A \times 4s = 16 C$
- $n = \frac{Q}{e} = \frac{16 C}{1.6 \times 10^{-19} C} = 10^{20} electrons$
Charge of 4C passing through a point in 1/2 second find the current intensity.
- $I = \frac{Q}{t} = \frac{4 C}{0.5 s} = 8 A$
Calculate the time required to pass a number of electrons = $5 \times 10^{20}$ electron through a conductor, take $e = 1.6 \times 10^{-19} C$ , the current passing = one Ampere.
- $Q = ne = 5 \times 10^{20} \times 1.6 \times 10^{-19} C = 80 C$
- $t = \frac{Q}{I} = \frac{80 C}{1 A} = 80 s$
A current of intensity =4A passing through a wire, find the number of electrons, passing through its section within two seconds ( $e= 1.6 \times 10^{-19} C$ ).
- $Q = It = 4A \times 2s = 8 C$
- $n = \frac{Q}{e} = \frac{8 C}{1.6 \times 10^{-19} C} = 0.5 \times 10^{20} electrons$
A wire of metal, its length = 20 cm, the cross-section area=14 millimeter2, its resistivity = $2.8 \times 10^{-7} \Omega.m$ , find the resistance of the wire.
- $L = 20 cm = 0.2 m$
- $A = 14 mm^2 = 14 \times 10^{-6} m^2$
- $R = \rho \frac{L}{A} = 2.8 \times 10^{-7} \Omega.m \times \frac{0.2 m}{14 \times 10^{-6} m^2} = 4 \times 10^{-3} \Omega$
Calculate the resistivity of a wire of length 100 meters and resistance 20 ohm, the cross-sectional area $10^{-8} m^2$ .
- $R=\rho \frac{L}{A}$
- $\rho = \frac{RA}{L} = \frac{20 \Omega \times 10^{-8} m^2}{100 m} = 2 \times 10^{-9} \Omega.m$
calculate the specific resistance of copper wire has a resistance 1.72 $\Omega$ ,length is 10m , cross-section area is $10^{-7} m^2$ .
- $R=\rho \frac{L}{A}$
- $\rho = \frac{RA}{L} = \frac{1.72 \Omega \times 10^{-7} m^2}{10 m} = 1.72 \times 10^{-8} \Omega.m$
A copper wire of length =2m and cross section area = $4 \times 10^{-8}m^2$ ,the current passing =3A , $\rho=2.8 \times 10^{-8}\Omega m$ find the potential difference between its terminal.

R=\rho \frac{L}{A}=(2.8 \times 10^{-8} \Omega m) \times \frac{2m}{4 \times 10^{-8}m^2}

R=1.4 \Omega

V=IR=3A \times 1.4 \Omega = 4.2V

Two wires from copper the length of the 1st is 20m and the 2nd 60 meters and the cross-section area of the 1st is twice the cross area of the 2nd if the resistance of the 1st is 10 $\Omega$ calculate the resistance of the 2nd.

Let $L1 = 20m$ , $L2 = 60m$
Let $A1 = 2A2$
Let $R_1 = 10 \Omega$

R1 = \rho \frac{L1}{A_1}

R2 = \rho \frac{L2}{A_2}

\frac{R2}{R1}=\frac{L2 A1}{L1 A2}

R2= R1 \times \frac{L2 A1}{L1 A2}

R2= 10 \Omega \times \frac{60 \times 2A2}{20 A_2}

R_2 = 10 \times 6 = 60 \Omega

Two copper of wires of the same length, the first one have a cross-sectional area of 3mm2 and resistance of 4 $\Omega$ , find the resistance of the second wire with a cross-sectional area of 5mm2.

R_1=4 \Omega

A_1=3 mm^2

A_2=5 mm^2

\frac{R1}{R2}=\frac{A2}{A1}

R2=\frac{R1 A1}{A2}

R_2= \frac{(4\Omega)(3mm^2)}{(5mm^2)}

R_2=2.4 \Omega

If you have three resistances R1, R2, R3 if R1, R2 connected in parallel and R3 in series.
- (a) Draw the diagram
- (b) If R1= 4 $\Omega$ , R2 = 12 $\Omega$ R3 = 10 $\Omega$ calculate the total resistance.

\frac{1}{R{12}}=\frac{1}{R1}+\frac{1}{R_2}

\frac{1}{R_{12}}=\frac{1}{4}+\frac{1}{12}

R_{12}=3 \Omega

Rt=R{12}+R_3= 3 \Omega +10 \Omega =13 \Omega

Two resistances 3, 6 $\Omega$ find the total resistance in case of.
- a) Series
  
  Rt= R1+R_2 = 3 \Omega + 6 \Omega= 9 \Omega
- b) Parallel

\frac{1}{R{12}}=\frac{1}{R1}+\frac{1}{R_2}

\frac{1}{R_{12}}=\frac{1}{3}+\frac{1}{6}

R_{12}=2 \Omega

Two resistances of two and one ohms are connected in series across a potential difference of 3 volts.
- (a) Draw the electric circuit for this.
- (b)Find the current passing in the resistances.
- $Rt = R1 + R_2 = 2 \Omega + 1 \Omega = 3 \Omega$
- $I= \frac{V}{R} = \frac{3}{3} = 1A$
Two resistances of 6ohm and 12ohm are connected in parallel a cross a potential difference of 12 volts.
- (a) Find the equivalent resistance.
- $\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{12} = \frac{3}{12}$
- $R_{eq} = 4 \Omega$
- (b) Find the total current passing through the resistances.
- $I = \frac{V}{R_{eq}} = \frac{12}{4} = 3 A$
In the circuit below reading of voltmeter =4 volt. Find
- (i) Current intensity.
- (ii) The potential difference across6 $\Omega$ .
- $I= \frac{V}{R} = \frac{4}{2} = 2 A$
- (ii) $V = IR = 3 \times 6 = 18Volts$
An electric charge of 4 Coulombs passed through a conductor within 4 second, find:
- (i) The value of current intensity.
- $I= \frac{Q}{t} = \frac{4}{4} = 1 A$
- (ii) The resistance of the conductor if the P.d. = 6 volts
- $R= \frac{V}{I} = \frac{6}{1} = 6 \Omega$
figure above shows an electric conductor has potential difference at resistant R equivalent to 12 volts ,and at resistant 1 $\Omega$ equivalent to 4 volts find:-
- a-electric current intensity
- $I = \frac{V}{R} = \frac{4}{1} = 4 A$

b-the total resistant of conductor
$R=\frac{V}{I}=\frac{12}{4} = 3 \Omega$
$R_t=R+1=3+1= 4 \Omega$

Two resistors 6 $\Omega$ and 12 $\Omega$ were connected in parallel and then a resistance of 2 $\Omega$ was connected in series with them. If the p.d of the two parallel resistors is 24V, draw the figure and hence calculate:
- 1) The total resistance.
  
  R{parallel} = \frac{R1 \times R2}{R1 + R_2} = \frac{6 \times 12}{6+12}=\frac{72}{18} = 4 \Omega

R{total} = R{parallel}+R_{series}= 4 \Omega + 2 \Omega = 6 \Omega

$R_{total} = 6 \Omega$
2) The current passing in each resistor.

I_6 = \frac{V}{R} =\frac{24V}{6 \Omega} = 4A

I_{12} = \frac{V}{R} =\frac{24V}{12 \Omega} = 2A

3) The total current

I{total}= I6+I_{12}= 4A+2A = 6A

4) The p.d between the ends of the 2 $\Omega$ resistor.

V= I \times R =6A \times 2 \Omega = 12V
5) The total p.d

VT = VP+V_S =24V + 12V = 36V