Question: Q. 16. The potential difference across a resistor ’ $r$ ’ carrying current ’ $I$ ’ is Ir.
(i) Now if the potential difference across ’ $r$ ’ is measured using a voltmeter of resistance ’ $R_{v}{ }^{\prime}$ ’ show that the reading of voltmeter is less than the true value.
(ii) Find the percentage error in measuring the potential difference by a voltmeter.
(iii) At what value of $R_{v}$, does the voltmetermeasures the true potential difference?
U] [CBSESQP 2015-16]
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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
$1 / 2$