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AP POLYCET 2018 STUDY MATERIAL FOR
PHYSICS(EM)
SBTET, AP, AMARAVATI
CHAPTER 11
Electric Current:
Summary:
• Lightning is an electric discharge between two clouds or between cloud
and earth.
• This electric discharge through air as an electric spark or lightning.
• Lightning is the motion of charge in the atmosphere.
• The nature of the substance plays an important role in the transfer
of energy from battery to bulb.
• Positive ions in a metal are fixed and negative electrons are free charhe
carriers.
• The fixed arrangement of positive ions is called lattice.
• Electric current = electric charge/time interval
I = Q/t
• The SI unit of electric current is ampere denoted by A.
• 1 Ampere = 1 Coloumb/1 Second
1 A = 1 C/s
• The free electrons in a conductor are accelerated by the electric
field.
• Electrons move in a direction opposite to the direction of the field.
• The electrons in the conductor move with a constant average speed called
drift speed or drift velocity.
• An ammeter is used to measure electric current.
• An ammeter is always connected in series to the circuit.
• This potential difference is also called voltage. The SI unit of potential
difference is “Volt” and it is denoted by V.
• 1 Volt = 1 Joule/1 Coulomb
1V = 1J/C
• Ohm’s law:
Potential difference between the ends of the conductor is directly
proportional to the current passing through it.
V ∝ I
V/I = Constant.
The constant is called resistance of the conductor.
It is denoted by ‘R’. Then we get V/I = R.
V = IR
SI unit of resistance: ohm.
The symbol of ohm is Ί.
1 Ohm = 1 Volt/1 Ampere
1 Ί = 1V/A
Ohm’s law materials are classified into two categories.
1. which obey Ohm’s law are called ohmic materials.
Ex: metals are ohmic materials.
2. which do not obey Ohm’s law are called non ohmic
materials.
Example: LEDs are non ohmic materials.
• Ohm’s law is valid if the temperature remains constant.
• The resistance of the material changes with temperature.
• V-I graph is non-linear when temperature changes.
• Ohm’s law is not applicable to gaseous conductors.
• Ohm’s cannot be applied to semiconductors.
Example: Germanium and silicon.
• The resistance of a conductor is defined as the obstruction to
the motion of the electrons in a conductor.
• The material which offers resistance to the motion of
electrons is called resistor.
Factors affecting the resistance of a material:
• Temperature, length, area and material of the conductor.
• The resistance (R) of a conductor is directly
proportional to its length (l)
• R ∝ l (at constant temperature)
• The resistance of a conductor inversely proportional to area
of its cross section.
i.e.R ∝ 1/A (at constant temperature)
R = ρl/A
Where, ρ is a proportionality constant and is called specific
resistance or resistivity.
• The SI unit of resistivity is ohm-metre
Symbolically Ί - m.
• The reciprocal of resistivity is called conductivity (σ).
• The values of resistivity of material determine their
conductivity.
Equivalent resistance of a series connection:
V = I Req
IReq = IR1 + IR2 + IR3
Req = R1 + R2 + R3
i.e. The equivalent resistance is equal to sum of individual
resistances when the resistors are connected in series.
• One of the resistors in series breaks down, the circuit
becomes open.
• Current flow cannot take place in the circuit.
• Hence, household electrical appliances cannot be
connected in series.
The equivalent resistance of a parallel combination is less
than the resistance of any one of the resistors.
Let three resistors are connected in series,
1/Req = 1/R1 + 1/R2 + 1/R3
Let two resistors R1 and R2 are connected in parallel,
1/Req = 1/R1 + 1/R2
Req = R1R2/(R1+R2)
Kirchhoff’s laws:
Current law:
At any junction point in a circuit, the sum of the currents entering into the
junction must equal to the sum of the currents leaving the junction.
we have I1 + I2 + I3 = I4 + I5
Figure: Current law
Kirchoff’s voltage law or Loop law.
The algebraic sum of changes in potential
differences are equal to zero.
The work done by electric field in time t is given by
W = Q V where
Q = charge in the conductor, V = voltage across the conductor. The work is
equal to the energy lost by the charge when passing through the conductor.
Energy lost by the charge per second = W/t
W/t = QV/t
Q/t = I,
the current flowing through the conductor.
And W/t = the work done per second.
Power: Power is the rate of doing work.
Hence, W/t = electric power (P).
Electric power P = VI
This equation can be used to calculate power consumption by any
electric device that is connected in a circuit.
According to the Ohm’s law,
V = IR
P = I2
R = V2
/R
The equation P = VI can also be used to calculate the power which be extracted
from a battery or any source. In this case modified equation
P = VI
where P = Îľ I.
here Îľ is the emf of the battery.
Example:
A bulb is marked 60W and 120V. This means that if this bulb is
connected to 120V source, it will able to convert 60J of electrical energy
into heat or light in one second.
From the marking of bulb, we can measure the resistance of the bulb. From the
relation P = V2
/R D R = V2
/P
Substituting the values V and P in above equation,
we get R = 120 x 120/ 60 = 240 Ί
Kilowatt is generally used to express power consumption.
1 KW = 1000 W = 1000 J/S
The unit of electric power consumption is equal to 1 KWH (one Kilo Watt
Hour).
1 KWH = (1000 J/S) (60 x 60 S)
= 3600 x 1000 J = 3.6 x 105
J
Multiple choice questions:
1. The kilowatt hour is the unit of ..................
a. Power b. work c. energy d. None of these
2. A thick wire has a--------- resistance than a thin wire.
a. high b. low c. does not depend on thickness d. higher
3. The SI unit of current is_____________
a. ampere b. volt c. ohm d. coulomb
4. A uniform wire of resistance 50 Ί is cut into five equal parts.
These parts are now connected in parallel. Then the equivalent
resistance of the combination is
a. 650 Ί b. 12 Ί
c. 250 Ί d. 2 Ί
5. Check the following statements.
A. In series connection, the same current flows through each element.
B. In parallel connection, the same potential difference gets applied
across each element.
a) both A and B are correct b) A is correct but B is wrong
c) A is wrong but B is correct d) both A and B are wrong
KEY
1. a 2. b 3. a 4. d 5.a
Chapter 11 physics

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Chapter 11 physics

  • 1. AP POLYCET 2018 STUDY MATERIAL FOR PHYSICS(EM) SBTET, AP, AMARAVATI
  • 2. CHAPTER 11 Electric Current: Summary: • Lightning is an electric discharge between two clouds or between cloud and earth. • This electric discharge through air as an electric spark or lightning. • Lightning is the motion of charge in the atmosphere. • The nature of the substance plays an important role in the transfer of energy from battery to bulb. • Positive ions in a metal are fixed and negative electrons are free charhe carriers. • The fixed arrangement of positive ions is called lattice. • Electric current = electric charge/time interval I = Q/t • The SI unit of electric current is ampere denoted by A. • 1 Ampere = 1 Coloumb/1 Second 1 A = 1 C/s • The free electrons in a conductor are accelerated by the electric field. • Electrons move in a direction opposite to the direction of the field. • The electrons in the conductor move with a constant average speed called drift speed or drift velocity. • An ammeter is used to measure electric current. • An ammeter is always connected in series to the circuit. • This potential difference is also called voltage. The SI unit of potential difference is “Volt” and it is denoted by V. • 1 Volt = 1 Joule/1 Coulomb 1V = 1J/C
  • 3. • Ohm’s law: Potential difference between the ends of the conductor is directly proportional to the current passing through it. V ∝ I V/I = Constant. The constant is called resistance of the conductor. It is denoted by ‘R’. Then we get V/I = R. V = IR SI unit of resistance: ohm. The symbol of ohm is Ί. 1 Ohm = 1 Volt/1 Ampere 1 Ί = 1V/A Ohm’s law materials are classified into two categories. 1. which obey Ohm’s law are called ohmic materials. Ex: metals are ohmic materials. 2. which do not obey Ohm’s law are called non ohmic materials. Example: LEDs are non ohmic materials. • Ohm’s law is valid if the temperature remains constant. • The resistance of the material changes with temperature. • V-I graph is non-linear when temperature changes. • Ohm’s law is not applicable to gaseous conductors. • Ohm’s cannot be applied to semiconductors. Example: Germanium and silicon. • The resistance of a conductor is defined as the obstruction to the motion of the electrons in a conductor. • The material which offers resistance to the motion of electrons is called resistor. Factors affecting the resistance of a material:
  • 4. • Temperature, length, area and material of the conductor. • The resistance (R) of a conductor is directly proportional to its length (l) • R ∝ l (at constant temperature) • The resistance of a conductor inversely proportional to area of its cross section. i.e.R ∝ 1/A (at constant temperature) R = ρl/A Where, ρ is a proportionality constant and is called specific resistance or resistivity. • The SI unit of resistivity is ohm-metre Symbolically Ί - m. • The reciprocal of resistivity is called conductivity (σ). • The values of resistivity of material determine their conductivity. Equivalent resistance of a series connection: V = I Req IReq = IR1 + IR2 + IR3 Req = R1 + R2 + R3 i.e. The equivalent resistance is equal to sum of individual resistances when the resistors are connected in series. • One of the resistors in series breaks down, the circuit becomes open. • Current flow cannot take place in the circuit. • Hence, household electrical appliances cannot be connected in series. The equivalent resistance of a parallel combination is less
  • 5. than the resistance of any one of the resistors. Let three resistors are connected in series, 1/Req = 1/R1 + 1/R2 + 1/R3 Let two resistors R1 and R2 are connected in parallel, 1/Req = 1/R1 + 1/R2 Req = R1R2/(R1+R2) Kirchhoff’s laws: Current law: At any junction point in a circuit, the sum of the currents entering into the junction must equal to the sum of the currents leaving the junction. we have I1 + I2 + I3 = I4 + I5 Figure: Current law
  • 6. Kirchoff’s voltage law or Loop law. The algebraic sum of changes in potential differences are equal to zero. The work done by electric field in time t is given by W = Q V where Q = charge in the conductor, V = voltage across the conductor. The work is equal to the energy lost by the charge when passing through the conductor. Energy lost by the charge per second = W/t W/t = QV/t Q/t = I, the current flowing through the conductor. And W/t = the work done per second. Power: Power is the rate of doing work. Hence, W/t = electric power (P).
  • 7. Electric power P = VI This equation can be used to calculate power consumption by any electric device that is connected in a circuit. According to the Ohm’s law, V = IR P = I2 R = V2 /R The equation P = VI can also be used to calculate the power which be extracted from a battery or any source. In this case modified equation P = VI where P = Îľ I. here Îľ is the emf of the battery. Example: A bulb is marked 60W and 120V. This means that if this bulb is connected to 120V source, it will able to convert 60J of electrical energy into heat or light in one second. From the marking of bulb, we can measure the resistance of the bulb. From the relation P = V2 /R D R = V2 /P Substituting the values V and P in above equation, we get R = 120 x 120/ 60 = 240 Ί Kilowatt is generally used to express power consumption. 1 KW = 1000 W = 1000 J/S The unit of electric power consumption is equal to 1 KWH (one Kilo Watt Hour). 1 KWH = (1000 J/S) (60 x 60 S) = 3600 x 1000 J = 3.6 x 105 J Multiple choice questions: 1. The kilowatt hour is the unit of ..................
  • 8. a. Power b. work c. energy d. None of these 2. A thick wire has a--------- resistance than a thin wire. a. high b. low c. does not depend on thickness d. higher 3. The SI unit of current is_____________ a. ampere b. volt c. ohm d. coulomb 4. A uniform wire of resistance 50 Ί is cut into five equal parts. These parts are now connected in parallel. Then the equivalent resistance of the combination is a. 650 Ί b. 12 Ί c. 250 Ί d. 2 Ί 5. Check the following statements. A. In series connection, the same current flows through each element. B. In parallel connection, the same potential difference gets applied across each element. a) both A and B are correct b) A is correct but B is wrong c) A is wrong but B is correct d) both A and B are wrong KEY 1. a 2. b 3. a 4. d 5.a