Current Electricity — CBSE Class 12 · 2017 · Delhi Board Question with Solution

(i) Derive an expression for the drift velocity of electrons in a conductor and hence deduce Ohm's law. (ii) A wire whose cross-sectional area increases linearly from one end to the other is connected across a battery of

V

volts. Which of the following remains constant along the wire: drift speed, current density, electric current, electric field? \\ \textbf{OR} \\ (i) State the two Kirchhoff's laws and explain how they are justified. (ii) A current is drawn from a cell of EMF

E

and internal resistance

r

connected to a symmetric network of resistors, each of resistance

r

. Obtain (a) the current drawn from the cell and (b) the power consumed in the network.

Solution

(i) Under field

E

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

, and averaging over the relaxation time gives

v_{d} = - \frac{e E}{m} τ

. Then

I = n e A v_{d} = \frac{n e ^{2} A τ}{m} E

; writing

E = V / ℓ

gives

I = \frac{n e ^{2} A τ}{m ℓ} V

, i.e.

V = I R

with

R = \frac{m ℓ}{n e ^{2} A τ}

constant — Ohm's law (

J = σ E

). (ii) By continuity the same charge per second crosses every section, so the \textbf{current} is constant; since

J = I / A

v_{d} = J / n e

and

E = J / σ

, all of these change where

A

changes. \textbf{OR:} (i) \emph{Junction rule} — at any node

\sum I = 0

(conservation of charge); \emph{loop rule} — around any closed loop

\sum Δ V = 0

(conservation of energy). (ii) The symmetric network reduces to

R_{e q} = \frac{r}{3}

; the current from the cell is

I = \frac{E}{r + r /3} = \frac{3 E}{4 r}

, and the power dissipated in the network is

P = I^{2} R_{e q} = (\frac{3 E}{4 r})^{2} \frac{r}{3} = \frac{3 E ^{2}}{16 r} .

CBSE Class 12 Board · 2017 · Delhi · Subjective

CBSE Class 12 · 2017 · Delhi · Current ElectricityLong Answer

Solution

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