*Q. A wire is stretched to double its length. What happens to its resistance and resistivity?*

*Q. A wire is stretched to double its length. What happens to its resistance and resistivity?*

**Ans:** Since the resistivity of the wire depends upon the nature of the material of the conductor, therefore the resistivity remains same for the same conductor. Let us consider a wire of length *l*, resistance R, resistivity and area of cross section A. Then resistance

R = $\frac{\rho l}{A}$

Or, R = $\frac{\rho {{l}^{2}}}{Al}$

Or, R = $\frac{\rho {{l}^{2}}}{V}$ …………. (i), where A*l* = V is the volume of the conductor

If the wire is stretched to double its length, then the new length is

*l’* = 2*l *and the new resistance R’ can be written as

R = $\frac{\rho l’ {^{2}}}{V’}$ = $\frac{\rho {{(2l)}^{2}}}{V}$ ……………(ii), where V = V’ because volume remains constant.

Also $\rho $ = $\rho’$, because the material of the conductor is same

Dividing (ii) by (i)

R’ = R 2^{2} = 4 R

Thus, when a wire is stretched to double its length, its resistance becomes 4 times of the original resistance.

*Q. You are given n wires each of resistance ‘R’ . What is the ratio of maximum to minimum resistance that can be obtained from these wires?*

*Q. You are given n wires each of resistance ‘R’ . What is the ratio of maximum to minimum resistance that can be obtained from these wires?*

**Ans:** Since in series combination effective resistance increases, therefore, when n wires each of resistance R connected in series the maximum resistance is obtained is given by

R_{s} = R + R + …. ….upto n terms

Or, R_{s} = n R ………………(i)

In parallel combination effective resistance decreases, so minimum resistance is given by

$\frac{1}{{{R}_{p}}}$ = $\frac{1}{R}$ + $\frac{1}{R}$ + …. upto n terms = $\frac{n}{R}$

Or, $\frac{1}{{{R}_{p}}}$ = $\frac{n}{R}$

Or, R_{p }= $\frac{R}{n}$…….(ii)

Where R_{s} and R_{p} be the effective resistance in series and parallel combination.

Dividing Eq. (i) by Eq. (ii), we get

$\frac{{{R}_{s}}}{{{R}_{p}}}$ = $\frac{nR}{R/n}$ = n^{2}

Thus required ratio, R_{max }: R_{min}= n^{2}:1

*Q. You are given 2 wires each of resistance ‘R’ . What is the ratio of maximum to minimum resistance that can be obtained from these wires?*

*Q. You are given 2 wires each of resistance ‘R’ . What is the ratio of maximum to minimum resistance that can be obtained from these wires?*

**Ans:** Since in series combination effective resistance increases, therefore, when 2 wires each of resistance R connected in series the maximum resistance is obtained is given by

R_{s} = R + R

Or R_{s} = 2R ……………… (i)

In parallel combination effective resistance decreases, so minimum resistance is given by

$\frac{1}{{{R}_{p}}}$ = $\frac{1}{R}$ + $\frac{1}{R}$

Or, $\frac{1}{{{R}_{p}}}$ = $\frac{2}{R}$

Or, R_{p }= $\frac{R}{2}$ ……………(ii)

Where R_{s} and R_{p} be the effective resistance in series and parallel combination.

Dividing Eq^{n}. (i) by Eq. (ii), we get

$\frac{{{R}_{s}}}{{{R}_{p}}}$ = $\frac{2R}{R/2}$ = 2^{2 }= 4

Thus required ratio, R_{max }: R_{min} = 4:1

*Q. Why ammeter is always connected in series?*

*Q. Why ammeter is always connected in series?*

**Ans:** Ammeter is an instrument which is to measure current flowing in a wire. It is necessary to connect the ammeter in series so that the current to be measured passes through it. Ideally, the resistance of an ammeter is zero so that the equivalent resistance of the dc circuit does not change and hence doesn’t change the current in the circuit. In practice, the resistance is not zero but can be very low as compared to the other resistors.

*If ammeter is connected in parallel, the equivalent resistance decreases drastically which increases the current in the dc circuit.*

*Q. Why a voltmeter is always connected in parallel with the load resistance?*

*Q. Why a voltmeter is always connected in parallel with the load resistance?*

**Ans:** A voltmeter is a high resistance instrument which is used to measure the potential difference. When it is connected in parallel across any element of a circuit, it draws a very small amount of current from the main circuit and hence current in the circuit is altered insignificantly due to the connection of voltmeter. Most of the current passes through the element across which it is connected. Hence, p.d across the element is not affected.

*If the voltmeter is connected in series, the resistance of the dc circuit becomes high. The current decreases considerably and voltmeter will not read the required p.d.*

*Q. Explain the difference between resistance and resistivity of a wire.*

*Q. Explain the difference between resistance and resistivity of a wire.*

**Ans:** Resistance is the property of a wire to oppose the flow of charge through it. Its unit is $\Omega $(ohm).

From Ohm’s law, V = IR

Or, R = $\frac{V}{I}$

**Resistivity:** From the relation, R = $\rho $$\frac{l}{A}$

If *l* = 1m, A = 1m^{2} then R = $\rho $

Hence, the resistivity of the material of a conductor is defined as the resistance of unit cross – sectional area and unit length. Its unit is Ohm-meter (m). The value of resistivity of the wire is independent of its dimension.

*Q. What are the factors on which the resistance of a conductor depends?*

*Q. What are the factors on which the resistance of a conductor depends?*

**Ans:** The resistance of a conductor depends upon (i) nature of material (i.e. resistivity,$\rho $ ), (ii) length, (iii) area of cross-section and (iv) temperature.

We know that

R = $\rho $$\frac{l}{A}$ where R is the resistance of a wire of length *l* and cross-sectional area ‘A’ and ” $\rho $is the resistivity of the material of the wire. From the above relation it can be concluded that

(i) R $\propto $ *l*

(ii) R$\propto $ $\frac{1}{A}$

The resistance of a conductor also varies with temperature. If ${{R}_{\theta }}$ and R_{0} are the resistances of the wire at $\theta $^{º}C and 0^{º}C respectively, then the relation between them is given by ${{R}_{\theta }}$ = R_{0 }[1 +$\alpha $$\theta $ ] where $\alpha $ is the temperature coefficient of the material of the wire.

*Q. Why do electrons acquire steady drift velocity?*

*Q. Why do electrons acquire steady drift velocity?*

**Ans:** When a battery is connected across the ends of conductor, an electric field is setup. This electric field accelerates the electrons in the conductor. Electrons gain Kinetic energy depending on the amount of the field. During movement, electrons collide with atomic lattice and lose their energy. So, they can’t be accelerated continuously but they soon acquire a steady average drift velocity.

*Q. Why do we use connecting wires made of copper?*

*Q. Why do we use connecting wires made of copper?*

**Ans:** The resistance of a wire is given by,

R = $\frac{\rho l}{A}$

For the same dimensions ( *l* and A), the copper wire has low resistance than other due to low resistivity and therefore easily conducts electricity . Moreover, it is diamagnetic and so does not get magnetized due to the passes of current through it. Thus it does it does not disturb the current flowing in the circuit. Also copper is easily available and cheaper as well as compare to other good conductors.

*Q. Two copper wires of different diameters are joined end to end. If a current flows in a wire combination, what happens to the drift velocity of electrons when move from the larger diameter wire to the smaller diameter?*

*Q. Two copper wires of different diameters are joined end to end. If a current flows in a wire combination, what happens to the drift velocity of electrons when move from the larger diameter wire to the smaller diameter?*

**Ans: **

We have the relation between current and drift velocity

I = V_{d }e n A

For the same value of I and n we can write

V_{d} $\propto $$\frac{1}{A}$ Since A = $\frac{\pi {{d}^{2}}}{4}$

So, V_{d} $\propto $$\frac{1}{{{d}^{2}}}$

So when diameter decreases, the drift velocity increases.

*Q. Will the drift velocity of electrons change if the diameter of a connecting wire is halved? Why? *

*Q. Will the drift velocity of electrons change if the diameter of a connecting wire is halved? Why?*

**Ans:** We have the relation between current and drift velocity

I = V_{d }e n A

Case:1 I = V_{1 }e n A_{1}

I = V_{1 }e n $\frac{\pi {{d}_{1}}^{2}}{4}$……………(1)

Case:2 when diameter is halved i.e. d = d/2

I = V_{2 }e n A_{2}

I = V_{2 }e n $\frac{\pi {{d}_{2}}^{2}}{4}$……………(2)

Dividing eqn (2) by (1),

$\frac{{{V}_{2}}{{(d/2)}^{2}}}{{{V}_{1}}{{d}^{2}}}$= 1

V_{2} = 4 V_{1 }

Thus when diameter is halved the drift velocity increases to 4 times the initial drift velocity.

*Q. Two wires, of one of copper and another of iron, have the same diameter and carry the same current. In which wire the drift velocity of electrons will be more? *

*Q. Two wires, of one of copper and another of iron, have the same diameter and carry the same current. In which wire the drift velocity of electrons will be more?*

**Ans:**

We have the relation between current and drift velocity

I = V_{d }e n A

For same value of area of cross-section (or diameter) and current,

we can write V_{d} $\propto $$\frac{1}{n}$

Thus the drift velocity of electrons will be more in the wire in which less number of free electrons per unit volume.

Furthermore, the electronic configuration of Cu is

Cu (29): *1S*^{2}* 2S*^{2 }*2P*^{6 }*3S*^{2}* 3P** ^{6}* 4S

^{1}3d

^{10}i.e. one Cu-atom contributes 1 free electron.

and the electronic configuration of Fe is

Fe (26): *1S*^{2}* 2S*^{2 }*2P*^{6 }*3S*^{2}* 3P** ^{6}* 4S

^{2}3d

^{6}i.e. one Fe-atom contributes 2 free electrons.

Thus we can conclude that n_{Fe} >$$n_{Cu} and hence V_{Cu} > V_{Fe }i.e drift velocity of electrons is greater in copper than in iron.

*Q. Why is it essential that the resistance of a voltmeter be very high?*

*Q. Why is it essential that the resistance of a voltmeter be very high?*

**Ans:**

A voltmeter is a device which is used to measure the potential difference across a resister. It is always connected in parallel with the resistor. The resistance of voltmeter is essential to be very high because when a very high value resistor is connected in parallel, the equivalent resistance of the circuit doesn’t decrease significantly and hence doesn’t increase the current significantly in the circuit.

*Q. The resistance of an ammeter must essentially be very small. Why?*

*Q. The resistance of an ammeter must essentially be very small. Why?*

**Ans:**

An ammeter is a device which is used to measure the current in the circuit. It is always connected in series with the resistor. The resistance of ammeter is essential to be very low because when a very low value resistor is connected in series, the equivalent resistance of the circuit doesn’t increase significantly and hence doesn’t decrease the current significantly in the circuit.

*Q. A cylindrical rod has resistance R. If we double its length and diameter, what is its resistance in terms of R’?*

*Q. A cylindrical rod has resistance R. If we double its length and diameter, what is its resistance in terms of R’?*

**Ans:**

The resistance of a wire is given by,

R = $\frac{\rho l}{A}$ = $\frac{\rho l}{\pi d{{{}_{1}}^{2}}/4}$ ………… (i)

When the length and diameter is doubled i.e. new length = 2*l* and diameter = 2d then new resistance R’ becomes R’ = $\frac{\rho \times 2l}{\pi {{(2d)}^{2}}/4}$ …………. (ii)

Dividing (ii) by (i) $\frac{R’}{R}$ = $\frac{1}{2}$

R’ = $\frac{1}{2}$ R.

Thus if we double the length and diameter of the wire, the resistance becomes half than that of initial resistance.

*Q. Resistors R*_{1} and R_{2} are connected in parallel to an emf source that has negligible internal resistance. What happens to the current through R_{1} when R_{2} is removed from the circuit?

*Q. Resistors R*

_{1}and R_{2}are connected in parallel to an emf source that has negligible internal resistance. What happens to the current through R_{1}when R_{2}is removed from the circuit?**Ans:**

When R_{1} and R_{2} are connected in parallel

The current through R_{1} is I_{1} = $\frac{{{V}_{1}}}{{{R}_{1}}}$ since V = IR

Or, I_{1} = $\frac{E}{{{R}_{1}}}$since in parallel combination V_{1}, V_{2}, E are equal.

When R_{2} is removed

The current through R_{1 }is I_{1} = $\frac{E}{{{R}_{1}}}$ i.e. I_{1} = I_{1}

The current through R_{1} remains same when R_{2} is removed dc circuit. But total current decreases.

Thus, we can conclude that in parallel combination, the current through a resistor remains same if we remove or add other resistor(s) parallel to the given resistor.

*Q. Resistors R*_{1} and R_{2} are connected in series to an emf source that has negligible internal resistance. What happens to the current through R_{1} when a third resistor R_{3} is connected in parallel with R_{2}?

*Q. Resistors R*

_{1}and R_{2}are connected in series to an emf source that has negligible internal resistance. What happens to the current through R_{1}when a third resistor R_{3}is connected in parallel with R_{2}?**Ans:**

When a third resister R_{3} is connected in parallel with R_{2} then equivalent resistance of the circuit decreases and hence the current in the circuit (i.e. the current through R_{1}) increases.

*Q. A large number of electrons are present in metals. Why is there no current in the absence of electric field across it?*

*Q. A large number of electrons are present in metals. Why is there no current in the absence of electric field across it?*

**Ans:**

The electric current is due flow of charge in particular direction. But this does not happen in the above case. Though a large number of electrons are present in metals, in the absence of electric field, these free electrons have random motion. During such motion, they collide with positive ions of the metal again and again. After each collision, their directions change. So these electrons have no net flow in the particular direction and hence no current flows.

*Q. Why are alloys constantan and manganin used to make standard resistors?*

*Q. Why are alloys constantan and manganin used to make standard resistors?*

**Ans:**

The resistance of a conductor also varies with temperature as

${{R}_{\theta }}$ = R_{0 }[1 + $\alpha \theta$ ] where $\alpha$ is the temperature coefficient of the resistance of the wire.

The alloys constantan and manganin are used to make standard resistors because the value of their temperature coefficient of resistance ( $\alpha$) is negligible. i.e. with change in temperature the resistance of such substance has negligible effect.

*Q. Give an example of non-ohmic conductor and present its current voltage characteristics graph.*

*Q. Give an example of non-ohmic conductor and present its current voltage characteristics graph.*

**Ans:**

A conductor which does not obey Ohm’s law is known as non-ohmic conductor. For example: electrolyte, junction diode etc.

If a graph is plotted between current and voltage for non ohmic conductor, a straight line passing through origin will not be obtained.

I-V graph of a junction diode (a non- ohmic conductor) is shown in figure.

*Q. How can a galvanometer be converted into ammeter? Explain.*

*Q. How can a galvanometer be converted into ammeter? Explain.*

**Ans:**

A galvanometer can be converted into ammeter by connecting a suitable shunt (a low value resistance) in parallel with the galvanometer.

If the galvanometer has resistance ‘G’ and shows full scale deflection when an amount of current I_{g} flows through it. The required value of shunt to convert the galvanometer into an ammeter to measure current upto I ampere is

S = $\frac{{{I}_{g}}}{I-{{I}_{g}}}$.G

*Q. How can a galvanometer be converted into voltmeter? Explain.*

*Q. How can a galvanometer be converted into voltmeter? Explain.*

**Ans: **

A galvanometer can be converted into voltmeter by connecting a high value resistance in series with the galvanometer.

If the galvanometer has resistance ‘G’ and shows full scale deflection when an amount of current I_{g} flows through it. The required value of high value resistance to convert the galvanometer into a voltmeter to measure potential difference upto V volt is

R = $\frac{V}{{{I}_{g}}}$– G