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Saturday, December 30, 2006

ELECTRICITY - III : LONG ANSWERS

GIVE ANSWER IN DETAIL



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  1. EXPLAIN ELECTRIC CHARGE
    ANS:

    • A comb used for combing dry hair picks up small pieces of paper brought near it because it acquires electric charge.
    • Similarly, a glass rod rubbed with silk cloth acquires positive charge.
    • An electric charge is a fundamental property associated with protons and electrons.In addition to mass, protons and electrons possess electrical charge.
    • Electric charge is measured in coulomb(C).
    • The magnitude of charge on both protons and electrons is same, i.e. 1.6 x 10(-19) coulomb.We use negative(-) sign for electron (as it has negative charge) and positive(+) sign for proton (as it has positive charge).
    • Coulomb's law describes the interaction between two charges and gives the magnitude of the force between them.


  2. DEFINE ELECTRIC CURRENT AND EXPLAIN DIFFERENCE BETWEEN ELECTRON CURRENT AND CONVENTIONAL CURRENT
    ANS:

    • Electric current is the net amount of charge that passes through the cross-sectional area of the conductor per unit time.
    • The charge will flow if the electrons move. In fact, electric current is the flow of electrons.
    • Electrons in an atom revolve around nucleus(protons) in different orbits.
    • In metals the electrons in the outermost orbit are relatively free to move and are called FREE ELECTRONS. These 'free' electrons contribute to the flow of charge(electric current).
    • Metals like silver, copper, aluminium, etc. possess maximum free electrons and are good conductors of electricity.
    • If we provide energy to these free electrons, we get flow of electrons in one direction which we call current.
    • The flow of electrons is from cathode(negative terminal) to anode(positive terminal).
    • The conventional current ,however, is shown as flow from anode to cathode as initially it was believed that the positive charge is responsible for electric current.


  3. EXPLAIN : ELECTRIC POTENTIAL AND ELECTRICAL POTENTIAL DIFFERENCE
    ANS:

    • The work done on a charge is stored in it as potential energy.This work done (potential energy) is represented by electric potential.
    • The electrical potential energy per unit charge at a point in an electric field is called the ELECTRICAL POTENTIAL at that point.
    • Electrical Potential = potential energy/charge = work done/charge

      V = W/Q

      The unit of electric potential is joule/coulomb and it is called volt in the memory of Volta.
    • To establish electric current in a conductor we need flow of charge(free electrons).
    • The electrical potential difference across the conductor causes the flow of charge(current).It is the driving force that causes current.
    • The charges flow from the end at a higher potential to the end at lower potential.
    • To create potential difference a cell(source of energy) is used.
    • Electrical potential difference between two points A and B on a conductor is the work done to move a unit charge from A to B.
    • Electrical potential difference is generally called voltage and its unit is volt(V).


  4. EXPLAIN THE CONSTRUCTION AND WORKING OF A SIMPLE CELL (VOLTAIC CELL)
    ANS:


    • Volta's cell(Voltaic cell) is a simple cell which converts chemical energy into electrical energy.
    • As shown in figure, a simple voltaic cell can be made by taking some dilute sulphuric acid in a glass container.Two metallic plates, one of zinc and the other of copper are used as electrodes.
    • When we join two electrodes by a conducting wire and place a lamp in between, we find that the lamp glows as electric current flows.
    • When two plates are placed in dilute sulphuric acid, a chemical reaction takes place between the plates and the acid. Zinc plate gets negatively charged and the copper plate gets positively charged.Thus zinc behaves as cathode and copper as anode.
    • A potential difference is created between the two plates due to this and current flows.
    • Since a chemical reaction causes current, in Volta's cell, chemical energy is converted to electrical energy.


  5. DESCRIBE OHM'S EXPERIMENT AND ITS CONCLUSION
    ANS:

    AIM:To study the relation between voltage(V) and current(I).

    APPARATUS:Conducting wire, ammeter, voltmeter, resistors, battery, key.



    PROCEDURE:Connect various apparatus in a circuit as shown in figure.The ammeter measures current and the voltmeter measures voltage(V) across the resistor(R).

    Press the key(K) and adjust the battery such that the voltmeter shows voltage of 1 volt.Record the value of current(I) as shown by the ammeter.

    Record different values of current obtained for voltage of 2V, 3V, 4V, 5V,etc.

    Record all the observations in the table as shown.

    Calculate the ratio of V and I for each observation and record it in the last column of the table. This gives the value of resistance(R).



    Plot the graph of V v/s I as shown.



    CONCLUSION:

    1. The current(I) increases linearly with voltage(V),i.e, I α V.
    2. The value of V/I (i.e. R) is constant for all observations. R is called resistance of the conductor.
    3. The graph of V v/s I is a straight line passing through the origin.


    Based on above experiment,Ohm formulated the following law:The current passing through a conductor is directly proportional to the potential difference(voltage) across the conductor.

  6. EXPLAIN SERIES CONNECTION OF RESISTORS AND DERIVE THE FORMULA FOR EQUIVALENT RESISTANCE
    ANS:


    • When the resistors are connected end-to-end, we get series connection.
    • For series connection of resistors the current flowing through each resistor is the same as the overall current of the circuit.
    • The overall voltage of the circuit, however, gets divided among the resistors according to their magnitudes.

    • Consider that three resistors R1, R2 and R3 are connected in series in a circuit having overall potential difference(voltage) V and current I.
    • If V1, V2 and V3 are voltage drops across R1, R2 and R3, respectively, then

      V1 = IR1, V2 = IR2, and V3 = IR3

    • Similarly, if R is the equivalent resistance, then

      V = IR.

    • Now,

      V = V1 + V2 + V3

      ∴ IR = IR1 + IR2 + IR3

      R = R1 + R2 + R3

      Thus, equivalent resistance(R) of the resistors(R1, R2, R3) connected in series is the algebraic sum of all the resistances.


  7. EXPLAIN PARALLEL CONNECTION OF RESISTORS AND DERIVE THE FORMULA FOR EQUIVALENT RESISTANCE
    ANS:


    • When resistors are connected between two common points, all having same connecting point on higher potential side as well as on lower potential side, we get parallel connection.
    • For resistors connected in parallel, the voltage drop(V) across all the resistors remains the same as the potential difference(voltage) of the circuit.
    • The overall current, however, gets distributed among the resistors according to their magnitudes.

    • Consider that three resistors R1, R2 and R3 are connected in parallel in a circuit having overall potential difference(voltage) V and current I.

    • If I1, I2 and I3 are the values of current passing through R1, R2 and R3, respectively, then

      I1 = V/R1

      I2 = V/R2 and

      I3 = V/R3

    • Similarly, if R is the equivalent resistance, then

      I = V/R.

    • Now,

      I = I1 + I2 + I3

      ∴ V/R = V/R1 + V/R2 + V/R3

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

      Thus, the reciprocal of equivalent resistance is equal to the sum of the reciprocals of individual resistances.


  8. DISTINGUISH BETWEEN SERIES CONNECTION AND PARALLEL CONNECTION.

    ANS:
  9. EXPLAIN : ELECTRICAL ENERGY AND POWER
    ANS:

    • To maintain the flow of charge(current), a battery has to do work continuously on the charge.
    • This work done on charge is converted into electrical energy.
    • If

      W = work done by the battery

      V = voltage of battery

      Q = electric charge

      I = current

      R = resistance

      t = time

      then we have,

      W = V x Q

      ∴ W = V x I x t        [as Q = I x t]

      ∴ W = I2 x R x t        [as V = I x R]

      Thus, W is the energy consumed if the current(I) flows through resistance(R) for time (t).The unit of electrical energy is joule(J).
    • The rate at which the electrical energy is consumed is known as electrical power.In other words electrical energy consumed per unit time is called power.

    • If P is electrical power, then

      P = W/t

      ∴ P = (I2 x R x t)/t        [as W = I2 x R x t]

      ∴ P = I2 x R

      or P = V x I        [as I x R = V]

      The unit of power is watt.
    • When 1 ampere current passes through a circuit having voltage of 1 V, the power is called 1 watt.
    • Using the definition of power,

      W = P x t

      ∴ joule = watt x second

      Thus, watt-second is a unit of energy.
    • For practical purpose watt-second is a small unit of energy.Therefore, practical unit of energy is kilowatthour(kWh).

      1 kWh = 3.6 x 106 joule
    • The practical unit of energy(kWh) is called 'unit'.


  10. EXPLAIN : CHEMICAL EFFECT OF ELECTRIC CURRENT GIVING EXAMPLE.
    ANS:


    • The phenomenon of a chemical reaction taking place due to electric current is called chemical effect of electric current.
    • The chemical effect of electric current can be utilised to deposit a thin layer of metal on another metal.
    • The process of depositing a desired metal on another metallic object by using electricity is called ELECTROPLATING.
    • As shown in figure, take aqueous solution of copper sulphate in a container.
    • Dip a copper plate and a metal spoon (not of copper) in the solution and connect the circuit such that copper plate works as anode and the spoon as cathode.
    • When we allow electric current to pass through the circuit for some time, the electrolyte (CuSO4) gets ionised and Cu+2 ions formed during ionization travel to cathode (spoon) and accumulate there. Similarly, SO4-2 ions travel to anode and react with copper to form copper sulphate.
    • Thus, copper from anode is removed and it gets deposited on cathode through the electrolyte.
    • Ultimately, the spoon gets electroplated, i.e. the spoon gets coated with a thin layer of copper.


  11. COMPARE THE WORK OF A BATTERY WITH A MECHANICAL PUMP.

    ANS:

    As shown in the figure, a pump supplies the work to maintain a difference of height(h) and thus provides energy to the flow of water. This flow of water can utilise the energy to do work to move some object. For example, it can turn a paddle as shown.

    In electric circuit, we can compare electric current(flow of electrons) with the flow of water, the lamp with the paddle, the wire with the pipe and the battery(source of energy) with the pump.

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