Q.No.13.9: - What is Wheatstone bridge? How it be used to determine an unknown resistance?

Q.No.13.9: - What is Wheatstone bridge? How it be used to determine an unknown resistance?

Ans: - Wheatstone bridge is used to determine unknown resistance of a wire. The circuit is consists of four resistances R₁, R₂, R₃, R₄ connected in such a way as to form a loop or mesh ABCDA. A battery of emf E is connected between points A and C. A galvanometer of resistance R_g is connected between the points B and D.

          When the bridge is balanced, it satisfied the following relation

                                        R₁/R₂      =       R₃/R₄

Or,                                   R₄           =        R₂.R₃/R₁

If the value of R₁, R₂, and R₃ are known then R₄ can be calculated. The galvanometer should not show any deflection by adjusting the potential between terminals B and D. (Sorry for Diagram. Pleas concert with Text Book for diagram.)

Q.No.13.8: - Explain why the terminal potential of a battery decreases when the current drawn from it is increased?

Q.No.13.8: - Explain why the terminal potential of a battery decreases when the current drawn from it is increased?

Ans: - The terminal potential difference of a batter is

                                        IR      =        E – I.r

                                        V_t       =        E – I.r

Where,  E is the emf of the battery, r is the internal resistance of battery, and I.r is the potential difference across internal resistance.

When I increased then I.r becomes larger and terminal potential becomes small. Thus, we can say that when we draw more current from battery then its terminal potential difference will decrease.

Here is a rough example for this concept. The example of current flow is just like a water flow. Let suppose the pressure of water is just like the potential in the electrical system. The dam have a large capacity of water and suppose you are drawing more and more water from the dam and as a result water in the dam becomes low level and now it has low pressure on water to flow. Similarly when we draw more current from the battery then the terminal potential of the battery decreases just like dam pressure on water flow. I just tried to make this answer clear this is not a fit example for it but in my view you can understand something from it.

Q.No.13.7: - Describe a circuit which will give a continuously varying potential?

Q.No.13.7: - Describe a circuit which will give a continuously varying potential?

Ans: - A potential divider or potentiometer is a circuit which can give a continuously varying potential. Consider a resistance R in the form of a wire on which a terminal C can slide. The resistance between A and C can be varied from 0 to R, as C slides from A to B.

          If we connect a battery of emf E across a resistance R. The current flowing thought it is

                                                  I        =        E/R

If we represent the resistance between A and C by r, the potential drop between these points will be

                                                  V        =        rI

Now,

                                                  V        =        r x E/R

Thus as C slides from A to B, r varies from 0 to R, and the potential drop between A and C changes from zero to E. This arrangement by which potential can be varied continuously from 0 to E is known as a potential divider.   (Sorry for Diagram. Pleas concert with Text Book for diagram.)

Q.No.13.6: - Is the filament resistance lower or higher in a 500W, 220V light bulb than in a 100W, 220V bulb?

Q.No.13.6: - Is the filament resistance lower or higher in a 500W, 220V light bulb than in a 100W, 220V bulb?

Ans: -

Power of light bulb                    =        P₁       =        500W

Potential difference        =        V₁       =        220V

Now apply,

                                        P₁       =        V₁²/R₁

                                        R₁       =        V₁²/P₁

                                        R₁       =        (220)²/500

                                        R₁       =        48400/500

                                        R₁       =        96.8 Ω

For second Bulb

Power           P₂       =        100W

Voltage        V₂       =        220V

Now, 

                    R₂       =        V₁²/P₂

                    R₂       =        (220)₂/100

                    R₂       =        48400/100

                    R₂       =        484 Ω

It means that the filament resistance of 100W, 220V light bulb is higher than that of other.

Q.No.13.5: - What are the difficulties in testing whether the filament of a lighted bulb obeys Ohm’s law?

Q.No.13.5: - What are the difficulties in testing whether the filament of a lighted bulb obeys Ohm’s law?

Ans: - According to Ohm when we increase voltage then current will also increases in direct proportion but resistance of the conductor must remain same and hence temperature also must remain same or constant. But when we test the filament of a lighted bulb that either it obeys Ohm’s law or not then with the passage of time its temperature increases and resistance also increases. These are the difficulties in testing whether filament of a lighted bulb obeys Ohm’s law or not.

Q.No.13.4: - Why does the resistance of a conductor rise with temperature?

Q.No.13.4: - Why does the resistance of a conductor rise with temperature?

Ans: - Electrical resistance is due to the collision of free electrons with atoms of the conductors. Atoms in the conductors are in vibratory motion when electrons collide with atoms they transfer some of their energy to the atoms and atoms starts vibrating with larger amplitude as the amplitude of vibration increases the probability of collision of electrons also increases. As a result increases resistance. When temperature increases then also amplitude of vibration increases and then probability of electrons collision also increases and then increases resistance.

Q.No.13.3: - What are the resistances of the resistors given in the figure A and B? What is the tolerance of each? Explain what is meant by tolerance?

Q.No.13.3: - What are the resistances of the resistors given in the figure A and B? What is the tolerance of each? Explain what is meant by tolerance?

(Sorry for Diagram but I’m trying to describe this figure here)

1^st band is Brown 2^nd is Green 3^rd is Red 4^th is Gold. In B fig. 1^st band is Yellow 2^nd is White 3^rd is Orange and 4^th is Silver.

Ans: - For Figure A

First band brown   =        1

            Second band Green =     5

             Third band Red (have number two so)= 00

             Now,       R   =   1500Ω

             Fourth band Gold which shows tolerance = ±5%

So,                       R   =   1500Ω±5%

For Figure B

                    First Band Yellow  =        4

                    Second band White=        9

                    Third band orange =        000 (no. of zeros)

                              R     =     49000 Ω

                    Fourth band silver =        ±10%

Now resistance is

                              R     =     49000±10% Ω

Tolerance: -

       Tolerance means the possible variation from the marked value.

Q.No.13.2: - Do bends in a wire affect its electrical resistance?

Q.No.13.2: - Do bends in a wire affect its electrical resistance?

Ans: - The resistance of wire is

                                       R   =  ρ.L/A

Where “ρ” is the resistivity of the conductor, L is the length and A is the area of cross section of the wire. So according to this equation we can say that bends in a wire will never affect the electrical resistance because we bending the wire its ρ, L and A remain same so its resistance remain same. 

Q.No.13.1: - A potential difference is applied across the ends of a copper wire. What is the effect on the drift velocity of free electrons by (i) Increasing the potential difference. (ii) Decreasing the length and the temperature of the wire.

Q.No.13.1: - A potential difference is applied across the ends of a copper wire. What is the effect on the drift velocity of free electrons by

(i)    Increasing the potential difference.

(ii)   Decreasing the length and the temperature of the wire.

Ans: - (i) It is obvious that drift velocity of free electrons in the copper wire will increase do to increasing potential difference.

(ii) If the drift velocity of free electrons is increasing then it means that resistance in the conductor is low at that time and when we decrease the length of the conductor then basically we are trying to decrease the number of atoms and decreasing resistance. As a result the drift velocity of free electrons increases. We know that resistance depends upon temperature change and if we decrease temperature of wire then it mean we are decreasing resistance and as a result increasing drift velocity. So, we can say that by increasing potential difference, decreasing length and temperature of conductor we are increasing drift velocity of electrons.

Q.No.12.9: - Do electrons tend to go to region of high potential or of low potential.

Q.No.12.9: - Do electrons tend to go to region of high potential or of low potential.

Ans: - By convention we use positive potential as a high and negative potential as low potential. So according to this convention we can say that electrons which have negative charge tend to go to the region of high potential (positive) from low potential (negative).

Q.No.12.8: - Is it true that Gauss’s Law states that the total number of lines of forces crossing any closed surface in the outward direction is proportional to the net positive charge enclosed within surface.

Q.No.12.8: - Is it true that Gauss’s Law states that the total number of lines of forces crossing any closed surface in the outward direction is proportional to the net positive charge enclosed within surface.

Ans: - Yes this is true. According to Gauss’s law the total no of electric field lines coming out of a closed surface is proportional to the total no of charges present in that closed surface.

Mathematically,

The above statement is true.

Q.No.12.7: - Is E necessarily zero inside a charged rubber balloon if balloon is spherical? Assume that charge is distributed uniformly over the surface.

Q.No.12.7: - Is E necessarily zero inside a charged rubber balloon if balloon is spherical? Assume that charge is distributed uniformly over the surface.

Ans: - According to Gauss’s Law if the charge enclosed in the closed surface is zero then flux will be zero out of that surface. In this case the charge is distributed uniformly over the surface and no charge is present

Hence electric field intensity will be zero inside a spherical balloon which has uniformly distributed charge on its surface.

Q.No.12.6: - If a point charge q of mass m is released in a non-uniform electric field, will it make a rectilinear motion?

Q.No.12.6: - If a point charge q of mass m is released in a non-uniform electric field, will it make a rectilinear motion?

Ans: - A rectilinear motion means a motion along a straight line. If a point charge q of mass m is placed at any point in the non-uniform electric field. Suppose this field is caused by a positive point charge then the charge will experience a repulsive force (as the force will exert on the charge along the line joining the two charges) and the charge will follow a straight path which makes a rectilinear motion.

Q.No.12.5: - Electric lines of forces never cross. Why?

Q.No.12.5: - Electric lines of forces never cross. Why?

Ans: - Electric lines of force never cross each other. This is because E has only one direction at any given point. If the lines cross, E could have more than one direction which is physically not correct.

Q.No.12.4: - Describe the force or forces on a positive point charge when placed between parallel plates. (a) with similar and equal charges. (b) with opposite and equal charges.

Q.No.12.4: - Describe the force or forces on a positive point charge when placed between parallel plates.

(a)    with similar and equal charges.

(b)    with opposite and equal charges.

Ans: - (a). If the plates have similar and equal charge in magnitude and we place the positive point charge between such plates then both the plates want to repel this positive point charge with same Coulomb’s force. The net force on the charge will be zero.

          (b). If the plates have similar and opposite charge in magnitude and we place the positive point charge between such plates then positive plate will repel the positive charge and exert a force F₁ while negative plate will attract it and exert an attractive force F₂.

          Net force        =           F        =      F₁   +     F₂