Solution:

Ans. (i) $V=\operatorname{Ir}$ (without voltmeter)

(ii) Percentage error $=\left(\frac{V-V^{\prime}}{V}\right) \times 100 \quad 1 / 2$

$$ =\left(\frac{r}{r+R_{v}}\right) \times 100 $$

(iii) $R_{v} \rightarrow \infty, V^{\prime}=I r=V$

[CBSE Marking Scheme, 2016]

[I] Q. 17. In the two electric circuits shown in the figure, determine the readings of ideal ammeter $(A)$ and the ideal voltmeter $(V)$.

(b)

A [Delhi I, II, III 2015]

Ans. In circuit (a),

(a)

Total Resistance $=2 \Omega$

$$ \text { Current } i=(15 / 2) \mathrm{A}=7.5 \mathrm{~A} \quad 1 / 2 $$

Potential difference between the terminals of $6 \mathrm{~V}$ battery

In circuit (b),

$$ \begin{aligned} V & =E-i r \ & =[6-(7.5 \times 1)] \mathrm{V} \ V & =-1.5 \mathrm{~V} \end{aligned} $$

$$ \begin{aligned} \text { Effective emf } & =(9-6) \mathrm{V} \ & =3 \mathrm{~V} \end{aligned} $$

Total resistance $=2 \Omega$

Current,

$$ i=\left(\frac{3}{2}\right) \mathrm{A}=1.5 \mathrm{~A} $$

Potential difference across $6 \mathrm{~V}$ cell

$$ \begin{aligned} V & =E+i r \ & =6+1.5 \times 1 \ & =7.5 \mathrm{~V} \end{aligned} $$

(AI Q. 1. (i) Derive an expression for drift velocity of electrons in a conductor. Hence deduce Ohm’s law.

(ii) A wire whose cross-sectional area is increasing linearly from its one end to the other, is connected across a battery of $\mathrm{V}$ volts. Which of the following quantities remain constant in the wire?

(a) drift speed

(b) current density (c) electric current

(d) electric field

Justify your answer.

A [Delhi I, II, III 2017]

Ans. (i) Derivation of the expression for drift velocity 2

Deduction of Ohm’s law

2

(ii) Name of quantity and justification $\quad 1 / 2+1 / 2$

(i) Let an electric field $E$ be applied to the conductor. Acceleration of each electron is

$$ a=-\frac{e E}{m} $$

Velocity gained by the electron

$$ v=-\frac{e E}{m} t $$

Let the conductor contain $n$ electrons per unit volume. The average value of time ’ $t$ ‘, between their successive collisions, is the relaxation time, ’ $\tau$ ‘.

Hence average drift velocity, $v_{d}=\frac{-e E}{m} \tau$

The amount of charge, crossing area $A$, in time $\Delta t$, is $=n e A v_{d} \Delta t=I \Delta t$

Substituting the value of $v_{d}$, we get

$$ \begin{aligned} I \Delta t & =n e A\left(\frac{e E \tau}{m}\right) \Delta t \ \therefore \quad I & =\left(\frac{e^{2} A \tau n}{m}\right) E \ \therefore \quad \frac{I}{A} & =\left(\frac{e^{2} \tau n}{m}\right) E \ & =\sigma E \quad\left(\sigma=\frac{e^{2} \tau n}{m} \text { is the conductivity }\right) \end{aligned} $$

But $I=J A$, where, $J$ is the current density,

$\Rightarrow \quad \mathrm{J}=\left(\frac{e^{2} \tau n}{m}\right) E$

$\Rightarrow \quad \mathrm{J}=\sigma E$

This is Ohm’s law

[Note : Credit should be given if the student derives the alternative form of Ohm’s law by substituting $E=\frac{V}{l}$ ]

(ii) (b) Current density will remain constant in the wire.

All other quantities, depend on the cross sectiona area of the wire.

[CBSE Marking Scheme 2017]

Detailed Answer :

(ii) Out of these, current density remains constant in a wire whose cross-sectional area increases linearly from its one end to other as current density is :

$$ \begin{aligned} J & =\frac{I}{A} \text { and } I=\frac{V}{R} \ \therefore \quad & J=\frac{V}{R A} \text { and } R=\rho \frac{l}{A} \end{aligned} $$

$$ \therefore \quad=\frac{V A}{\rho l A}=\frac{V}{\rho l}=\text { constant } $$

It is current per unit area that depends on area of cross-section.

$1 / 2$

Drift speed is given as :

$$ v_{d}=\frac{I}{\text { Ane }} $$

Electric field $=\frac{J}{\sigma}$; where $\sigma$ is electrical conductivity

$1 / 2$