Q26 of 26 Page 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 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.


OR


(i) State the two Kirchhoff’s laws. Explain briefly how these rules are justified.



(ii) The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r as shown in the figure. Obtain the expression for (i) the current draw from the cell and (ii) the power consumed in the network.

i) Let us consider a metallic of conductor of cross-sectional area A, electron density n, and let e be the charge of an electron. Consider vd to be the drift velocity of the of the electrons. Consider an electric field E is applied to the conductor, and so the acceleration of the electron will be defined as,




Where me is the mass of the electron. So, velocity gained by the electron in infinitesimally small time Δt, is,



Where vi is the initial velocity of the electron, averaging this term for all the electrons present, we get the drift velocity vd,



Where is defined as the relaxation time (time between two successive collisions of electrons)


Let there be a steady state current I in the conductor. Now consider a segment of conductor of length Δx, the net charge flown through this volume of the conductor in time Δt, is I Δt, which can be written in terms of drift velocity as,


Net charge flown in time Δt = n e A |vd| Δt


Equating both the expressions we get,


I Δt = n e A |vd| Δt


Thus, we get,



Equating the equations (a) and (b), we get,



As we know that E=V/L.


We gwt




If physical condition remain unchanged then right hand side will be constant


We get,



Thus, we get ohms law verified.


ii) A wire whose cross-sectional area is increasing linearly from its one end to the other, is connected across a battery of V volts.


A. drift speed of the electrons is decreasing as it is inversely proportional to the area.



B. current density is also decreasing as the area is increasing as it is defined as the current per unit area of cross section of the conductor



C. electric current remains constant as it does not depend upon the area of cross section rather the resistance of the conductor and the potential difference applied determines how much current will flow.


D. electric field will also decrease as it depends on current density which is also decreasing,



OR


i) The Kirchhoff’s Rules: -


a) Junction rule/ Current rule: This rule states that at any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction.


Σi=0


b) Voltage rule/ loop rule: - Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero.


ΣΔV=0


The justification for current rule is that when the, circuit reaches the steady state after the switch is turned on, no charges are accumulating at any given point in the circuit and thus charges from leaving any point is equal to charges reaching at that point, which implies that the summation of all the currents at the given node is always zero, where incoming currents towards the node are taken to be positive and outgoing currents from the node are taken to be negative.


The justification for the Voltage rule comes from the law of conservation of energy, i.e., if we are moving in a closed loop and reach the point where we started then the net loss of potential must be zero or the voltage drops occurred in the closed loop must be equal to the source voltages occurred in the that very closed loop.


ii)


The resistors between the points A and C are in parallel, same is for resistors between points C and B, so equivalent resistance between the points AC and CB is,



Now, the two resistors of r/2 are in series, so the net equivalent resistance of the branch ACB will be.



Now the circuit is reduced to three resistors of resistance r in parallel and a battery of emf E and internal resistance r, the net resistance of the circuit is,


The equivalent resistance AB for the above question is three resistance in parallel combination.



We get


RAB= R/3



Where r/3 is resistance of the 3 resistors in parallel and r is the internal resistance of the cell.



The net resistance is,



So, the net current drawn from the battery is,



Also, the power consumed in the circuit can be given by the formula,


P = V I


For the circuit, V = emf of the cell, and I is the Inet so,




Conclusion: -


Net current in the circuit,



Net power consumed by the circuit,



More from this chapter

All 26 →
22

Describe the working principle of a moving coil galvanometer. Why is it necessary to use (i) a radial magnetic field and (ii) a cylindrical soft iron core in a galvanometer? Write the expression for current sensitivity of the galvanometer.

Can a galvanometer as such be used for measuring the current? Explain.


OR


A. Define the term ‘self-inductance’ and write its S.I. unit.


B. Obtain the expression for the mutual inductance of two long co-axial solenoids and would one over the other, each of length L and radii and and and number of turns per unit length, when a current I is set up in the outer solenoid

23

Mrs. Rashmi Singh broke her reading glasses. When she went to the shopkeeper to order new specs, he suggested that she should get spectacles with plastic lenses instead of glass lenses. On getting the new spectacles, she found that the new ones were thicker than the earlier ones. She asked this question to the shopkeeper but he could not offer satisfactory explanation for this. At home, Mrs. Singh raised the same question to her daughter Anuja who explained why plastic lenses were thicker.

A. Write two qualities displayed each by Anuja and her mother.


B. How do you explain this fact using lens maker’s formula?

 24

A. Draw a labeled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil.


B. A circular coil of cross-sectional area and 20 turns is rotated about the vertical diameter with angular speed of in uniform magnetic field of magnitude Calculate the maximum value of the current in the coil.


OR


A. Draw a labeled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of number of turns and currents in the two coils.


B. A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 3000 turns. Find the number of turns in the secondary to get the power output at 220 V.

 25

A. Distinguish between unpolarized light and linearly polarized light. How does one get linearly polarised light with the help of a polaroid?

B. A narrow beam of unpolarized light of intensity is incident on a polaroid The light transmitted by it is then incident on a second polaroid with its pass axis making angle of 60o relative to the pass axis of Find the intensity of the light transmitted by .


OR


A. Explain two features to distinguish between the interference pattern in young’s double slit experiment with the diffraction pattern obtained due to a single slit.


B. A monochromatic light of wavelength 500 mm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen.


Estimate the number of fringes obtained in Young’s double slit experiment with fringe width 0.5 mm, which can be accommodated within the region of total angular spread of the central maximum due to single slit. (the wavelength of the light is given in mm but all the calculations are done using it as nm)