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Current is defined as the rate of flow of charge.

I = Q/t

Its S.I. unit is Ampere.

Electric potential at a point is defined as the amount of work done in bringing a unit positive charge from infinity to that point. Its unit is the volt.

It is the property of a conductor to resist the flow of charges through it. It's S.I. unit is Ohm.

The potential difference between two points is equal to the work done in moving a unit positive charge from one point to the other. 

It's S.I. unit is Volt.

One volt is the potential difference between two points in an electric circuit when 1 joule of work is done to move charge of 1 coulomb from one point to other.

If a body is free to fall, on releasing it from a height, it falls downwards towards the earth's surface. For, this one end has to be at higher level and other at lower level, so that gravity could effect on this difference and body could freely fall. Same way to make flow of the charge through a conductor, the gravity of course has no role of play; there should be difference of electric potential. This difference gives the flow of charge in a conductor.

If a body is free to fall, on releasing it from a height, it falls downwards towards the earth's surface. For, this one end has to be at higher level and other at lower level, so that gravity could effect on this difference and body could freely fall. Same way to make flow of the charge through a conductor, the gravity of course has no role of play; there should be difference of electric potential. This difference gives the flow of charge in a conductor.

It is the property of a conductor to resist the flow of charges through it. It's S.I. unit is Ohm.

In a metal, the charges responsible for the flow of current are the free electrons. The direction of flow of current is conventionally taken opposite to the direction of motion of electrons.

(a), (c)

(a) The meter bridge can have no other neutral point for this set of resistances.

(c) When the jockey contacts a point on the meter wire to the right of D, current flows from B to the wire through galvanometer.  

Resistance of a wire is inversely proportional to the area of cross-section of the wire.

R ∝ 1/A

R ∝ 1/πr² 

This means if a wire of same length, but of double radius is taken, its resistance is found to be one-fourth.

Resistance of a wire is directly proportional to the length of the wire.

R ∝ I

The resistance of a conductor depends on the number of collisions which the electrons suffer with the fixed positive ions while moving from one end to the other end of the conductor. Obviously the number of collisions will be more in a longer conductor as compared to a shorter conductor. Therefore, a longer conductor offers more resistance.

(b), (c)

(b) Errors of measurement do depend on the accuracy with which R₂ and R₁can be measured. 

(c) If the student uses large values of R₂ and R₁, the currents through the arms will be feeble. This will make determination of null point accurately more difficult. 

With the increase in temperature of conductor, both the random motion of electrons and the amplitude of vibration of fixed positive ions increase. As a result, the number of collisions increases. Hence, the resistance of a conductor increases with the increase in its temperature. The resistance of filament of a bulb is more when it is glowing (i.e., when it is at a high temperature) as compared to when it is not glowing (i.e., when it is cold).

In the balanced condition,

\(\frac{P}{Q}=\frac{R}{S}\)

∴ \(\frac{4+4}{1}=\frac{4+x}{2}\)

\(\Rightarrow\) \(x=12\Omega\)

(D) the potential difference across the galvanometer is zero.

Iron wire will have more resistance than copper wire of the same length and same radius because resistivity of iron is more than that of copper.

(i) Resistance of a wire is directly proportional to the length of the wire means with the increase in length resistance also increases.

R ∝ I

(ii) Resistance of a wire is inversely proportional to the area of cross-section of the wire. If area of cross-section of the wire is more, then resistance will be less and vice versa.

R ∝ 1

(iii) Resistance increases with the increase in temperature since with increase in temperature the number of collisions increases.

(iv) Resistance depends on the nature of conductor because different substances have different concentration of free electrons. Substances such as silver, copper etc. offer less resistance and are called good conductors; but substances such as rubber, glass etc. offer very high resistance and are called insulators.

Although Ohm’s law has been found valid over a large class of materials, there do exist materials and devices used in electric circuits where the proportionality of V and I does not hold. The deviations broadly are one or more of the following types: 

(a) V ceases to be proportional to I. 

(b) The relation between V and I depends on the sign of V. In other words, if I is the current for a certain V, then reversing the direction of V keeping its magnitude fixed, does not produce a current of the same magnitude as I in the opposite direction. 

(c) The relation between V and I is not unique, i.e., there is more than one value of V for the same current I. A material exhibiting such behaviour is GaAs.

Limitations of Ohm’s law: 

1. Ohm’s law applicable only for good conductors. 

2. Ohm’s law applicable only, when the physical conditions like temperature, pressure and tension remains constant. 

3. Ohm’s law is not applicable at very low temperature and very high temperature. 

4. Ohm’s law is not applicable for semiconductors, thermistors, vacuum tubes, discharge tubes. 

Electric energy: The total work done by the source of emf in maintaining an electric current in a circuit for a given time is called electric energy consumed in the circuit. Its SI unit is joule. 

Electric power : The rate at which work is done by a source of emf in maintaining an electric current through a circuit is called electric power of the circuit. Its SI unit is watt. 

The average velocity of free electrons in a metal at room temperature is zero.

Terminal potential difference across a cell

V = ε - Ir

i. In series arrangement, current, \(I_S=\frac{E}{R_1+R_2+r}\)

ii. In parallel arrangement, current, \(I_P=\frac{E}{\frac{R_1R_2}{R_1+R_2}+r}\) 

Obviously, IP > IS, so V_P < VS

That is series arrangement will have higher terminal potential difference.

It is defined as the average velocity gained by the free electrons of a conductor in the opposite of the externally applied electric field. 

The reciprocal of electrical resistivity of material of a conductor is called conductivity. i.e. σ = (1/ρ)

Answer is 2 ohms

Let ' R be their equivalent resistance of the 4Ω and 6Ω resistors connected in parallel.

Then, 1/R = 1/4 + 1/6 = 3 + 2/12 = (5/12)Ω 

Or, R' = 12/5 = 2.4 Ω 

Emf is the potential difference between the positive and negative electrodes in an open circuit. i.e. when no current flowing through the cell.  

R_eq = R₁ + R₂ + R₃ + R₄ +…………… 

R_eq = R + R + R +………………….P times = P(R)  

The effective resistance when two or more resistors are connected in series increases and is greater than the greatest of individual resistance. 

The effective resistance when two or more resistors are connected in parallel decreases and is smaller than the smallest of individual resistance. 

When the wire of 15Ω resistance is stretched to double its original length, then its resistance becomes

R’ = n² × 15 = 2² × 15 = 60Ω 

When it cut into two equal parts, then resistance of each part becomes

R' = R'/2 = 60/2 = 30Ω 

These parts are connected in parallel, then net resistance of their combination is

R = R'/2 = 30/2 = 15Ω 

So, the current drawn from the battery

l = V/R = 3/15 = 1/5 

or l = 0.2 A.

The resistance changes is 4 times.

A solar cell is a device that converts solar energy into electrical energy.

The working of an electric bell is based on the magnetic effect of electric current.

The Ni-Cd cell is rechargeable.

The potential difference delivered by the Ni-Cd cell is (about) 1.2 V.

[Note: In this cell, nickel oxide is the positive electrode and cadmium is the negative electrode. It is often used as a dry cell and it can be recharged. These cells are used in some portable machines that run on electricity like a drilling machine or a gardening tool.]

When a glass rod is rubbed on a silk cloth, the glass rod acquires positive charge.

Correct option is (d) 2 V

Correct option is (b) 1.2 V

(i) Charge carriers in silver foil are free electrons.

(ii) Charge carriers in hydrogen discharge tube are electrons (e–) and positive hydrogen ions (H+).

(iii) Charge carriers in germanium semiconductor are electrons (e–) and holes (o+)

(iv) Charge carriers in nichrome wire are electrons.

(v) Charge carriers in superconductor are electrons.

In first case, resistance R is in parallel with cell ε, so p.d. across R = ε

i.e., ε = RI ...(i)

In second case, X is in parallel with cell so p.d. across X = ε

i.e., ε = XI ...(ii)

Let k be the potential gradient of potentiometer wire. If l1 and l2 are the balancing length corresponding to resistance respectively, then

ε = kl₁ ...(iii)

ε = kl₂ ...(iv)

From (i) and (iii) RI = kl₁ ...(v)

From (ii) and (iv) XI = kl₂ ...(vi) 

Dividing (vi) by (v), we get

\(\frac{X}{R}=\frac{l_2}{l_1}\) \(\Rightarrow X=\frac{l_2}{l_1}R\)

Here, R = 10.0 Ω, l1 = 58.3 cm, l2 = 68.5 cm

∴ \(X=\frac{68.5}{58.3}\times10.0=11.75\Omega\)

If we fail to find the balance point with the given cell ε, then we shall take the driver battery (B1) of higher emf than given emf (ε).

Drift velocity is defined as the average velocity acquired by the free electrons in a conductor under the influence of an electric field applied across the conductor. It is denoted by vd.

Current, I = neA. vd

Drift velocity : It is the average velocity acquired by the free electrons superimposed over the random motion in the direction opposite to electric field and along the length of the metallic conductor.

Let n = number of free electrons per unit volume, v_d = Drift velocity of electrons Total number of free electrons passing through a cross section in unit time

N/t = Anv_d

So, total charge passing through a cross section in unit time

i.e., current, l = Q/t = N/t = Anev_d.

(a) Drift velocity alone.

(b), (d)

(b) conservation of charge. 

(d) the fact that there is no accumulation of charges at a junction.

(a), (d)

(a) Potential drop across AB is nearly constant as R′ is varied. 

(d) I ≥V/r+R always.

The resistance of a wire is inversely proportional to its cross-sectional area. Thus, a thick wire has less resistance, and a thin wire has more resistance.

No current will flow through 2Ω resistor, because in a closed loop, total p.d. must be zero. So

10 V – 5I₁ = 0 . . . (1)

20 V – 10I₂ = 0 . . . (2)

Equation (1) and (2) have no solutions and resistor 2Ωis not part of any loop ABCD and EFGH.