﻿ 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? from Physics Current Electricity Class 12 CBSE

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 by

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(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,

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, 1 = 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, 2 = kl2                                                                        … (2)

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

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,

; which is the required condition of balance for Wheatstone bridge.
b)

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(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

Proof:

V = IR = I

i.e., V =

For unifrom current and cross- sectional area, we have

ii)

Given,

E = 2 V; R = 15  ; RAB = 10

Potential difference across the wire, =
Therefore, potential gradient = 0.8/1 = 0.8 V/ m

Potential difference across AO =

Therefore,

Length, AO =
= ; which is the required balance length of the wire.

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(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.

Electric field, E = -eE

Acceleration of each electron,          ... (2)
Here,

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

Drift velocity is given by,

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
Current density, J  =
From ohm's law, we have

J =
Here,  is the conductivity of the material through whic the current is flowing,

Thus,

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.

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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.

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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 -and + 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.

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.

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,

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

So,

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

Consider, I as the current flowing through the cell.

Therefore,

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

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,

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,

, is the required electric field.

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