SATHEE: current-electricity Question 32

current-electricity Question 32

Question: Q. 8. Two material bars $A$ and $B$ of equal area of crosssection, are connected in series to a DC supply. A is made of usual resistance wire and $B$ of an $n$ -type semiconductor.

(i) In which bar is drift speed of free electrons greater?

(ii) If the same constant current continues to flow for a long time, how will the voltage drop across $A$ and $B$ be affected? Justify each error. A&E [CBSE Comptt. 2018]

Sol. (a) Drift speed in $B$ ( $n$ -type semiconductor) is higher

Reason : $I = n e A v_{d}$ is same for both $n$ is much lower in semiconductors.

(b) Voltage drop across $A$ will increase as the resistance of $A$ increases with increase in temperature. $1 / 2 + 1 / 2$ Voltage drop across $B$ will decrease as resistance of $B$ will decrease with increase in temperature.

$1 / 2 + 1 / 2$

[CBSE Marking Scheme 2018]

AI Q. 9. (a) Define the term ‘conductivity’ of a metallic wire. Write its SI unit.

(b) Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field $E$ .

R&U [Delhi & O.D. 2018]

Show Answer

Solution:

Ans. (a) Definition and SI unit of conductivity $1 / 2 + 1 / 2$

(b) Derivation of the expression for conductivity $1 \frac{1}{2}$ Relation between current density and electric field

(a) The conductivity of a material equals the reciprocal of the resistance of its wire of unit length and unit area of cross-section.

[Alternatively :

The conductivity $(σ)$ of a material is the reciprocalof its resistivity $(ρ)]$

(Also accept $σ = \frac{1}{ρ}$ )

Its SI unit is

$(\frac{1}{ohm-metre}) /$ ohm $^{- 1} m^{- 1} /$ (mhe $m^{- 1}$ ) / siemen $m^{- 1}$

(b) The acceleration, $a = - \frac{e}{m} \vec{E}$

The average drift velocity, $v_{d}$ , is given by

$v_{d} = - \frac{e E}{m} τ$

( $τ =$ average time between collisions/relaxation time)

If $n$ is the number of free electrons per unit volume, the current $I$ is given by

$\begin{aligned} I & = n e A | v_{d} | & = \frac{e^{2} A}{m} τ n | E | \end{aligned}$

But

We, therefore, get

$I = | j | A (j = current density)$

$| j | = \frac{n e^{2}}{m} τ | E |$

The term $\frac{n e^{2}}{m} τ$ is conductivity

$\begin{array}{ll} ∴ & σ = \frac{n e^{2} τ}{m} \Rightarrow & J = σ E \end{array}$

[CBSE Marking Scheme 2018]

Current Electricity

current-electricity Question 32

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Current Electricity

current-electricity Question 32

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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