Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance?


Internal resistance of the circuit is negligible.

 So, total resistance is R.

 Therefore, current across the circuit is given bystraight I space equals space E over R

2357 Views

(a) State the working principle of a potentiometer. With the help of the circuit diagram; explain how a potentiometer is used to compare the emf’s of two primary cells. Obtain the required expression used for comparing the emf’s.

(b) Write two possible causes for one sided deflection in a potentiometer experiment.

OR

(a) State Kirchhoff’s rules for an electric network. Using Kirchhoff’s rules, obtain the balance condition in terms of the resistances of four arms of Wheatstone bridge.

(b) In the meter bridge experimental set up, shown in the figure, the null point ’D’ is obtained at a distance of 40 cm from end A of the meter bridge wire. If a resistance of 10W is connected in series with R1, null point is obtained at AD=60 cm. Calculated the values of R1 and R2


(a) Working Principle of Potentiometer

Principle:

Consider a long resistance wire AB of uniform cross-section. It’s one end A is connected to the positive terminal of battery B1 whose negative terminal is connected to the other end B of the wire through key K and a rheostat (Rh).

The battery B1 connected in circuit is called the driver battery and this circuit is called the primary circuit. By the help of this circuit a definite potential difference is applied across the wire AB; the potential falls continuously along the wire from A to B. The fall of potential per unit length of wire is called the potential gradient. It is denoted by ‘k’.

A cell e is connected such that it’s positive terminal is connected to end A and the negative terminal to a jockey J through the galvanometer G. This circuit is called the secondary circuit. In primary circuit the rheostat (Rh) is so adjusted that the deflection in galvanometer is on one side when jockey is touched on wire at point A and on the other side when jockey is touched on wire at point B.

The jockey is moved and touched to the potentiometer wire and the position is found where galvanometer gives no deflection. Such a point P is called null deflection point. VAB is the potential difference between points A and B and L meter be the length of wire, then the potential gradient is given by, 

                                                 straight k space equals space straight V subscript AB over straight L

If the length of wire AP in the null deflection position be l, then the potential difference

between points A and P,

                                                      VAP = kl

Therefore, Emf of the cell  = VAP= kl

In this way the emf of a cell may be determined by a potentiometer.

Comparison of two emfs’ of a cell:

First of all the ends of potentiometer are connected to a battery B1, key K and rheostat Rh such that the positive terminal of battery B1 is connected to end A of the wire. This completes the primary circuit.

Now the positive terminals of the cells C1 and C2 whose emfs’ are to be compared are connected to A and the negative terminals to the jockey J through a two-way key and a galvanometer (fig). This is the secondary circuit.




Distance if P1 from A is l1.

So, AP1 = l1

Emf of cell C1 is given by, straight epsilon1 = kl1                                                    … (1)

Now, plug is taken from terminals 1 and 3 and inserted between the terminals 2 and 3 to bring cell C2 in the circuit. Null deflection point is obtained at P2.

Distance from P2 to A is l2.

Emf of cell C2, straight epsilon2 = kl2                                                                        … (2)

Now, dividing equation (1) by (2), we get

                                                  straight epsilon subscript 1 over straight epsilon subscript 2 equals straight l subscript 1 over straight l subscript 2

Out of these cells if one is standard cell. 


OR

 

Kirchhoff’s rule states that,

(i) At any junction, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction.

(ii) The algebraic sum of the charges in potential around any closed loop involving resistors and cells in the loop is zero.

Conditions of balance of a Wheatstone bridge:

P, Q, R and S are four resistance forming a closed bridge, called Wheatstone bridge.

                                          

A battery is connected across A and C, while a galvanometer is connected B and D. Current is absent in the galvanometer’s balance point.

Derivation of Formula: Let the current given by battery in the balanced position be I. This current on reaching point A is divided into two parts I1 and I2. At the balanced point, current is zero.

Applying Kirchhoff’s I law at point A,

I - I1 - I 2 = 0 or

 I = I1 + I 2                                                              ...(i)

 

Applying Kirchhoff’s II law to closed mesh ABDA,

- I1P + I 2R = 0 or

  I1P = I 2 R                                                            ...(ii)

Applying Kirchhoff’s II law to mesh BCDB,

- I1Q + I 2S = 0 or

 I1Q = I 2S                                                              ...(iii)


Dviding equation (ii) by (iii), we get, 


                                          fraction numerator straight I subscript 1 straight P over denominator straight I subscript 1 straight Q end fraction equals space fraction numerator straight I subscript 2 straight R over denominator straight I subscript 2 straight S end fraction

rightwards double arrow space space straight P divided by straight Q equals space space straight R divided by straight S space ; which is the required condition of balance for Wheatstone bridge.
b) 

             


For space null space point space at space straight D comma space balance space length space straight l subscript 1 space equals space 40 space cm space

So comma space straight R subscript 1 over straight R subscript 2 equals AD over DC equals fraction numerator 40 over denominator left parenthesis 100 minus 40 right parenthesis end fraction equals 2 over 3 space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis

If space resistance space 10 space straight capital omega space is space connected space in space series space of space straight R 1 comma
space balance space point space shifts space towards space ‘ straight y ’ space straight i. straight e. comma space AD space equals space 60 space cm

fraction numerator straight R subscript 1 plus 10 over denominator straight R subscript 2 end fraction equals fraction numerator AD straight apostrophe over denominator straight D straight apostrophe straight C end fraction equals fraction numerator 60 over denominator 100 minus 60 end fraction equals 3 over 2 space space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis

space space space space space space space fraction numerator space straight R subscript 1 over denominator space straight R subscript 2 end fraction plus 10 over straight R subscript 2 equals 3 over 2

From space equations space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space have

2 over 3 plus 10 over straight R subscript 2 equals 3 over 2

rightwards double arrow 10 over straight R subscript 2 equals straight space 3 over 2 minus 2 over 3 equals straight space fraction numerator 9 minus 4 over denominator 6 end fraction equals 5 over 6

rightwards double arrow space straight R subscript 2 straight space equals straight space fraction numerator 10 cross times 6 over denominator 5 end fraction equals straight space 12 straight space ohm
From space equation space left parenthesis 1 right parenthesis comma space we space have

straight R subscript 1 over 12 equals 2 over 3

rightwards double arrow straight space straight R subscript 1 straight space equals straight space fraction numerator 12 straight space cross times 2 over denominator 3 end fraction equals straight space 8 straight space ohm

3380 Views

(i) State the principle of working of a potentiometer. 
(ii) In the following potentiometer circuit, AB is a uniform wire of length 1 m and resistance 10 Ω. Calculate the potential gradient along the wire and balance length AO (= l)


i)

Principale of potentiometer: The potential difference across any two points of current carrying wire, having uniform cross-sectional area and material, of the potentiometer is directly proportional to the length between the two points. 

That is,                                            V Error converting from MathML to accessible text.

Proof:

V = IR = I Error converting from MathML to accessible text. 

i.e., V = Error converting from MathML to accessible text.

For unifrom current and cross- sectional area, we have

Error converting from MathML to accessible text.


ii) 

Given, 

E = 2 V; R = 15 straight capital omega ; RAB = 10 straight capital omega

Potential difference across the wire, = 2 over 25 space x space 10 space equals space 0.8 space V divided by m e t e r
Therefore, potential gradient = 0.8/1 = 0.8 V/ m

Potential difference across AO = fraction numerator 1.5 over denominator 1.5 end fraction space x space 0.3 space equals space 0.3 space V

Therefore, 

Length, AO = fraction numerator 0.3 over denominator 0.8 end fraction cross times 100 
               = 3 over 2 space x space 25 space equals space 37.5 space c m; which is the required balance length of the wire. 

3583 Views

(i) Define the term drift velocity.
(ii) On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend?
(iii) Why alloys like constantan and manganin are used for making standard resistors?

i) 

Drift velocity is the average velocity of the free electrons in the conductor with which they get drifted towards the positive end of the conductor under the influence of an external electric field.

ii) 

Free electrons are in continuous random motion. They undergo changethey undergo change in direction at each collision and the thermal velocities are randomly distributef in all directions. 

straight u space equals space fraction numerator straight u subscript 1 space space plus space straight u subscript 2 space plus space.... plus space straight u subscript straight n over denominator straight n space end fraction space equals space 0 space space space space space space space space space space... left parenthesis 1 right parenthesis



Electric field, E = -eE 

Acceleration of each electron, straight a with rightwards harpoon with barb upwards on top space equals space fraction numerator negative e E over denominator m end fraction         ... (2) 
Here, 

m = mass of an electron
e = charge on an electron

Drift velocity is given by, straight v subscript straight d space equals space fraction numerator straight v subscript 1 space plus space straight v subscript 2 space plus space.... plus space straight v subscript straight n over denominator straight n end fraction

straight v subscript straight d space equals space fraction numerator left parenthesis straight u subscript 1 space space plus aτ subscript 1 right parenthesis plus space left parenthesis straight u subscript 2 plus aτ subscript 2 right parenthesis plus....... plus thin space left parenthesis straight u subscript straight n space plus space aτ subscript straight n right parenthesis over denominator straight n end fraction space

straight v subscript straight d space equals space fraction numerator left parenthesis straight u subscript 1 space plus space straight u subscript 2 plus.... plus space straight u subscript straight n right parenthesis over denominator straight n end fraction plus fraction numerator straight a left parenthesis straight tau subscript 1 space plus space straight tau subscript 2 plus.... plus straight tau subscript straight n right parenthesis over denominator straight n end fraction

Since comma space straight v subscript straight d space equals space aτ space space space space space space space space space space space space space space space space space space... space left parenthesis 3 right parenthesis space

Here comma space

straight tau space equals space fraction numerator straight tau subscript 1 space plus space straight tau subscript 2 plus..... plus space straight tau subscript straight n over denominator straight n end fraction comma space is space the space average space time space elapsed
Substituting space for space straight a space from space equation space left parenthesis 2 right parenthesis comma space

straight v subscript straight d space equals space fraction numerator negative eE over denominator straight m end fraction space straight tau space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 4 right parenthesis 

Electrons are accelerated because of the external electric field. 

They move from one place to another and current is produced.

For small interval dt, we have

I dt = -q ; where q is the total charge flowing

Let, n be the free electrons per unit area. Then, total charge crossing area A in time dt is given by,

Idt = neAvd dt

Substituting the value of vd, we get 

I.dt = neA open parentheses eE over straight m close parentheses d t 
Current density, J  = straight n space straight e squared over straight m open vertical bar straight E close vertical bar straight tau
From ohm's law, we have

J = straight sigma space straight E
Here, straight sigma is the conductivity of the material through whic the current is flowing,

Thus, 

straight sigma space equals space straight n space straight e squared over straight m straight tau
straight sigma space equals space 1 over straight rho space or space straight rho space equals space 1 over straight sigma

Substituting space the space value space of space conductivity comma space we space have

straight rho space equals space fraction numerator straight m over denominator ne squared straight tau end fraction space semicolon space straight tau space is space the space relaxation space time

iii) 

Alloys like constantan and manganin are used for making standard resistors because:

a) they have high value of resistivity

b) temperature coefficient of resistance is less. 

2532 Views

During a thunderstorm the 'live' wire of the transmission line fell down on the ground from the poles in the street. A group of boys, who passed through, noticed it and some of them wanted to place the wire by the side. As they were approaching the wire and trying to lift the cable, Anuj noticed it and immediately pushed them away, thus preventing them from touching the live wire. During pushing some of them got hurt. Anuj took them to a doctor to get them medical aid.

Based on the above paragraph, answer the following questions:

(a) Write the two values which Anuj displayed during the incident.

(b) Why is it that a bird can sit on a suspended 'live' wire without any harm whereas touching it on the ground can give a fatal shock?

(c) The electric power from a power plant is set up to a very high voltage before transmitting it to distant consumers. Explain, why. 

a) Anuj displayed concern for others lives, presence of mind and a selfless attitude.

b) When the bird perches on a live wire, body becomes charged for a moment and has a same voltage as the live wire. However, no current flows in its body. Body is a poor conductor of electricity as compared to the copper wire. So the electrons do not travel through the bird’s body.

On the other hand, if the bird touches the ground while being in contact with the high voltage live wire, then the electric circuit gets complete and high current flows through the body of the bird to the ground, giving it a fatal shock.

c) Inorder to reduce the loss of power transmission, the power plant is set up at high voltages. Power loss during the transmission is I2R. Therefore, by reducing the value of current, power loss can be minimized. So, the value of the voltage should be kept high.

1581 Views

a) State Kirchhoff's rules and explain on what basis they are justified.

(b) Two cells of emfs E1 and E2 and internal resistances r1 and r2 are connected in parallel.

Derive the expression for the

(i) Emf and

(ii) internal resistance of a single equivalent cell which can replace this combination.

OR

 

(a) "The outward electric flux due to charge +Q is independent of the shape and size of the surface which encloses is." Give two reasons to justify this statement.

(b) Two identical circular loops '1' and '2' of radius R each have linear charge densities -straight lambdaand +straight lambda C/m respectively. The loops are placed coaxially with their centre distance apart. Find the:

magnitude and direction of the net electric field at the centre of loop '1'.

a)

i) Junction Rule: The algebraic sum of currents meeting at a point in an electrical circuit is always zero.

                                      sum I space equals space 0

This law is in accordance with law of conservation of charge.

ii) Loop Rule: In a closed loop, the algebraic sum of emfs is equal to the algebraic sum of the products of the resistances and the current flowing through them.

                                Error converting from MathML to accessible text.

This law is based on the conservation of energy.

b) Consider the circuit, 

 

Here, E1 and E2 are the emf of two cells,

r1 and r2 are the internal resistance of cell,

I1 and I2 current due to two cells.

Terminal potential difference across the first cell is given by, 

space space space straight V space equals space straight E subscript 1 space minus space straight I subscript 1 straight r subscript 1 space

rightwards double arrow space space straight I subscript 1 space equals space fraction numerator straight E subscript 1 space minus space straight V over denominator straight r subscript 1 end fraction

For the second cell, terminal potential difference will be equal to that across the forst cell.

So, 

V space equals space E subscript 2 space minus space I subscript 2 r subscript 2
rightwards double arrow I subscript italic 2 space equals space fraction numerator E subscript italic 2 space italic minus space V over denominator r subscript italic 2 end fraction 

Let E be the effective emf and r the resultant internal resistance.

Consider, I as the current flowing through the cell.

Therefore, 

I thin space equals space I subscript 1 space plus space I subscript 2 italic space

rightwards double arrow space I thin space equals space fraction numerator E subscript italic 1 italic minus V over denominator r subscript italic 1 end fraction plus space fraction numerator E subscript italic 2 italic minus V over denominator r subscript italic 2 end fraction italic space

rightwards double arrow space I thin space equals space fraction numerator r subscript italic 2 space italic left parenthesis E subscript italic 1 italic minus V italic right parenthesis italic plus r subscript italic 1 italic left parenthesis E subscript italic 2 italic minus V italic right parenthesis over denominator r subscript italic 1 r subscript italic 2 end fraction italic space

rightwards double arrow space I r subscript italic 1 r subscript italic 2 space equals space E subscript italic 1 r subscript italic 2 space plus space E subscript italic 2 r subscript italic 1 minus space left parenthesis r subscript italic 1 plus r subscript italic 2 right parenthesis V

rightwards double arrow space V space equals space fraction numerator E subscript italic 1 r subscript italic 2 italic plus space E subscript italic 2 r subscript italic 1 over denominator r subscript italic 1 space italic plus space r subscript italic 2 end fraction minus space fraction numerator I r subscript italic 1 r subscript italic 2 over denominator r subscript italic 1 space italic plus space r subscript italic 2 end fraction

Now, comparing the equation with V = E –Ir, we have

straight E space equals space fraction numerator E subscript 1 r subscript 2 plus space E subscript 2 r subscript 1 over denominator r subscript 1 space plus space r subscript 2 end fraction space and
Internal space resistance comma space straight r space equals space fraction numerator r subscript 1 r subscript 2 over denominator r subscript 1 space plus space r subscript 2 end fraction
 

                                               OR


a) The outward electric flux due to the charge enclosed inside a surface is the number of electric field lines coming out of the surface. Outward flux is independent of the shape and size of the surface because:

i) Number of electric field lines coming out from a closed surface is dependent on charge which does not change with the shape and size of the conductor.

ii) Number of electric lines is independent of the position of the charge inside the closed surface.

b) Magnitude of electric field at any point on the axis of a uniformly charged loop is given by, 

E space equals space fraction numerator lambda over denominator 2 epsilon subscript o end fraction fraction numerator r R over denominator left parenthesis r squared plus R squared right parenthesis to the power of bevelled 3 over 2 end exponent end fraction 

 

Electric field at the centre of loop 1 due to charge present on loop 1 = 0

Electric field at the centre of loop1 due to charge present on loop 2 is given by, 

straight E '  equals straight space fraction numerator plus straight lambda over denominator 2 straight epsilon subscript straight o end fraction fraction numerator straight R square root of 3 straight R over denominator left parenthesis left parenthesis straight R square root of 3 right parenthesis squared plus straight R squared right parenthesis to the power of bevelled 3 over 2 end exponent end fraction 

rightwards double arrow space E space apostrophe space equals space fraction numerator 3 over denominator 16 epsilon subscript o end fraction lambda over R, is the required electric field.

2585 Views