Saturday, October 10, 2015

Direct Current Circuit




Direct Current Circuit :
Drift Velocity and Electric Current :
Electrical Resistance :






1. Introuduction


Just as the flow of water in a river constitutes a water current, the flow of charge constitutes an electric current. A conductor contains a large number of free electrons. these are moving through the surface of conductor with different speed in different direcions. Motion does not constitute an electric current because billions of electrons pass through any cross sectionof the conductor per second any direction. If a battery is connected across the ends of the conductor, the flow of charges is biased is one direction and as a result there is now a net transport of chrges. This directed flow of electric charges in time t, then electric current.
The electric current is defined as the rate of flow of charge. If a charge Q flows through any cross section in time t, then electric current, I = Qt
If small charge dQ flows in time dt, then I=dQtd
It is a scalar quantity. The unit of electric current is ampere in SI-units.
                1 ampere = 1 coulomb1second
A current flowing through the conductor is said to be one ampere if 1 coulomb of charge flows through the conductor in 1 second. Smaller units of the current are milliampere (10− 3A) and microampere (10− 6A). Electric current is measured using an instrument called ammeter.

Current Carriers

Electric current carries are the charged particles which flow in a definite direction. In mattalic conductors, the current is due to the flow of free electrons in it. In electrolytes such as solution of common salt, both positive and negative charged ions move to produce an electric current.
In gases, the positively charged ions and electrons are current carriers while in semiconductors, the current carriers are electrons and holes.

Direction of Current

The current in a conductor flows due to the flow of electrons in it. It was believed that the current flow is caused by the motion of positive charges. Most of the earlier theories and scientific literature were based on this idea. That is why this usage is followed to these days and the current is known as conventional current. So the direction of the flow of conventional current is opposite to the direction of electron flow.

Types of Current

Electric current in a conductor may pass in different ways. In this regard, current can be divided into two types
  1. Direct current : A current whose magnitude as well as direction remains constant at all the times is called a direct current. The current flowing through a resistor connected to a battery is an example of direct current.
  2. Alternating current : A current whose magnitude and direction changes continuosly periodically is called an alternating current. An ac generator produces an ac current.

Current Density

The current density at any point in a conductor is defined as the current flowing per unit area perpendicular to the direction of the flow. It is denoted by J.
J=IA
or I= JA
It is vector quantity. Its unit is Am − 2 in SI-units. This expressions shows that the electric current I is the dot product of two vectors and, hence it is a scalar quantity.

Mechanism of Metallic Conduction and Drift Velocity :

When no external electric field is applied to a metallic conductor, the free electrons are in thermal equilibrium with the rest of the the conductor and are in random motion. So, the average velocity of the electrons in a direction is zero, and consequently this motion does not make up a net transport of charge across any section of the conductor. Hence, there is no current in the conductor.
When an electric field is applied to the conductor by connecting a battery across it, each electron is acted by an electrostatic force, and the electrons get accelerated in direction opposite to that of the field. Hence, the electrons gain velocity and kinetic energy.
Thus the average velocity acquired by the free electron in a conductor subjected to an electric field is called drift velocity.

Relation between Drift Velocity and Electric Current

Consider a section of a conductor, such as a copper wire, of length l and cross-sectional area A. Let n be the number of free electrons per unit volume in it and e be the charge of each electron. Then                 Volume of the conductor = A l
                total number of free electrons in this volume = n A l
                total charge in this volume, Q = n A l e
When a battery is connected across the conductor, the charges flow through it. Let Vd be the drift velocity of electrons. When a charge, Q drifts through a length l of the conductor in time t, then the current I through the conductor is given by
               I= n e A × l t= n e A vd
                                where, vd=lt
So, the drift velocity, vd=In A e and current density, J=IA= nv de

Ohms Law

It states that the current flowing through a conductor is directly propotional to the potential difference across its ends, provided that the physical conditions (temperature, mechanical strain etc.) of the conductor remian the same. If I is a current flowing through a conductor, and V is potenital difference in it, then
                V α I
                or, V = R I
Where R is a constant of proportioanlity called the resistance of the conductor.
The above equation is the mathematical form of Ohms law.

Electrical Resistance

The resistance of a conductor is defined as its ability to oppose the flow of charge through it. It is represented in circuit as or .
It is measured by the ratio of the potential difference V across its ends to the current I flowing through it.
R =CI

Unit of R is ohm, in SI-units. So,
One ohm=1 volt1 ampere
The resistance of a conductor is said to be one ohm if one ampere current flows through it under a potential difference of one volt. Larger units of resistance are k Ω (103 Ω) and M Ω (106 Ω).

Cause of Electric Resistance


While flowing through a conductor, the electrons frequently collide with the atoms on lattice sites of the conductor. These collisions slow down the motion of electrons and hence increase the resistance. The resistance of the conductor depends upon the number of collisions that the electrons make with the atoms in hte wire. Therefore, the resistance of a conductor depends on the arrangement of the atoms in the materials, i.e. on the material of the conductor. A long wire offers more resistance than a shorter one because the more collisions occur in longer wire. Also a thick wire offers less resistance than a thin one because in the thick wire there is more area of cross-section for the electrons to flow. Thus, we see that the resistance of a wire depends on (i) the material of the wire and (ii) its dimension i.e. length and thickness of the wire.

Resistivity of Specific Resistance

It is observed that at constant temperature, the resistance R of a conductor is directly proportional to its length 'l' and inversely proportional to the cross sectional area A.
Let l be the length of a conductor of resistance R and A be its cross-sectional area. Then
                Rα l ...............(i)
                and Rα lA ..............(ii)
Combining equation (i) and (ii), we get
                          Rα lA
                or R= ρ lA
where ρ is a proportionality constant, called resistivity of specific resistance of the material of conductor. Different materials have different values of resistivity.
If l = 1m, A = 1 m2, then R = ρ
Thus, the resistivity of the material of a conductor is defined as the resistance of the conductor of unit cross-sectional area per unit length. Its unit is ohm metre, Ω m in SI-units.
Note that the value of ρ is independent of the dimension of the wire. It depends only on the materials of the conductor. Thus the resistivity is the characteristics of the material.

Conductance

The reciprocal of resistance of a conductor is called its conductance (C). If a conductor has resistance R then its conductance C is given by,
                C=1R
The unit of conductance is mho or (ohm)− 1 in SI-unit. This is also called Siemen, S.

Conductivity

The reciprocal of resistivity of a conductor is called its conductivity. It is denoted by the symbol σ. If conductor has resistivity ρ, then its conductivity is given by;                 σ=1ρ
The unit of conductivity is ohm − 1 meter− 1 or Ω − 1m− 1 or Sm SI-unit.

Relation between J and E

When a potential difference V is applied to a conductor of resistance R, the current following through the conductor, I=VR
If A be the uniform area of cross-section of the conductor and l be its length, then
                R=Ω lA=lσ A
where σ =1ρ, is the conductivity. Hence                 I=VR=σ A V l
                lA=σ Vl
But current density J=IAand electric intensity, E = Vl
Hence, J = σ E.
This equation gives the relation between J and E.

Effect of Temperature on Resistance

The resistance of a conductor is observed to increase with the increase in the temperature of the conductor.

Wednesday, October 7, 2015

Short question/answer on Wave Motion
Numerical Problems on Electricity and Magnetism :
Solved Numerical Problems on Wave Motion :
Wave Motion :
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Electricity and Magnetism (Numerical Problem)
Q. An electric iron of 1000w is connected in 220v in main line, what should be the capacity of the fuse?
Soln
Power (P) = 1000w
volt (V) = 220v
current (I) = ? By the formula P = I V
or, I = PV
or, I = 1000220v
or, I = 4.54 A
Therefor, suitable fuse is 5 A for this electric heater.
Q. What will be the capacity of a fuse to operate a heater of 0.8 kw with 220v main supplies?
Q. If an iron of 1000w, two bulbs of each 100w, a television of 60 w and a radio of 40w area used in a house, wht should be the capacity of fuse used? the voltage of the supplyied line is 220v.?
Q. A domestic wiring has a fuse of 5A. What is the maximum number of 100watt bulb that can be used in the supply of 220v?
Soln
Maximum current (I) = 5A
volt (V) = 220v
power of bulb (p) = 100w
Suppose maximum number of bulb can be connected is n
By the formula, P = I V
or n.p = 5 × 220
or, n × 100 = 1100
or, n = 1100100or, n = 11
Therefor, 11 bulbs can be connected.

Q. What is the the amount of current needed for a fluorescent lamp of 40w connected to a source of 220v?
Q. Ten electric bulbs of 100 watt each and two electric heaters of 1000 watt each are used for 6 hours continuously. Calculate the unit of electricity consumed?
Soln
Number of bulbs n = 10
Power of bulb (P) = 100w = 1001000kw
                       = 0.1kw
time (t) = 6 hr.
Electricity consumed = P n t
                  = 0.1 × 10 × 6 = 6 unit
Number of heater n = 2
Power of heater (P) = 1000w = 10001000kw
                           = 1kw
time (t) = 6 hr.
Electricity consumed = P n t
                      = 1 × 10 × 6 = 60 unit
Hence total electricity consumed = 6 + 60 = 66 unit.
Q. An electric kettle rated 220v and 2.2kw works for 3 hours. Find the energy consumed and the current drawn by it?
Q. Four electric bulbs of 60w power each and a heater of 1000w power are used in a home for 4 hours daily respectively. Calculate the monthly cost of electricity at the rate of Rs. 7.50 per unit. Claculate the main current when all these appliances run together. (Take the voltage supply is 220v).
Soln
Number of bulbs n = 4
Power of bulb (P) = 60w = 601000kw
                       = 0.06kw
time (t) = 4 hr.
Electricity consumed = P n t
                  = 0.06 × 4 × 4 = 0.96 unit
Number of heater n = 1
Power of heater (P) = 1000w = 10001000kw
                           = 1kw
time (t) = 4 hr.
Electricity consumed = P n t
                      = 1 × 1 × 4 = 4 unit
Hence total electricity consumed = 0.96 + 4 = 4.96 unit.
Rate per unit is Rs. 7.50
Montly cost = (4.96 × 7.50) × 30 = Rs 1116.
Q. Ten bulbs of 100 watt each are used for 6 hrs and 3 electric heaters of 2kw each are used for 2hrs. Calculate the total consumption of electricity for a day.
Q. The number of turns in the primary winding of trnsformer is 200 times more than that in the secondary winding. Calculate the input emf in the primary winding if emf generated in the secondary winding in 220 volt AC.
Soln
Number of turns in primary winding n1 = 200 k
Number of turns in secondary winding n2 = k
Seconadary volt v2 = 220v
Primary volt v1 ?
By the formula, v1n1 = v2n2
                or, v1 200 k = 220k
                or, v1 = 220k × 200 k
                or, v1 = 44000 v
Q. The number of turns in the primary winding of a certian transformer is 150 times more than that in the secondary winding. Calculate the input emf in the primary winding if the emf generated in the secondary winding in 220v AC.
Q. Primary voltage is 220v, what will be secondary voltage if ratio of number turns in secondary coil to primary coil is 1/4?
Soln
Primary voltage v1 = 200v
Secondary voltage v2 = ?
secondary coil n2primary coil n1 = 14
By the formula, v2v1 = n2n1
                or, v2 200 = 14
                or, v2 = 14 × 200
                or, v2 = 50v

Q.To listen 12v radio in a main line 240v with transformer of primary turns 2000. What will be the secondary turns.
Q. ?
Soln

Q. ?
Soln



Monday, October 5, 2015

Numerical Problem on Heat Equation


Numerical Problems on Heat Equation :    



Numerical Problems on Gravitational Force:
Numerical Problems on Electricity and Magnetism :
Numerical Problems on Pressure :
                                                                                                                                ..........ads...........

Saturday, October 3, 2015

Mechanical Wave/Velocity of Sound in different medium/ Factors affecting the velocity of sound


Mechanical Wave
Introduction/Velocity of Waves :    
Velocity of sound Waves in Liquid / Solid / Gas :        
Newton's Formula for velocity of Sound in gas and Laplace Correction :        
Factor's Affecting the velocity of sound of Gas :      


Thursday, October 1, 2015

Numerical Problems on Gravitational Force


Numerical Problems on Gravitational Force :    



Numerical Problems on Pressure:
Numerical Problems on Electricity and Magnetism :
Numerical Problems on Heat Equation :
                                                                                                                                ..........ads...........

Wednesday, September 30, 2015

Pressure/upthrusth numerical


Numerical Problems on Pressure :    



Numerical Problems on Gravitational Force:
Numerical Problems on Electricity and Magnetism :
Numerical Problems on Heat Equation :
                                                                                                                                ..........ads...........