Study guide H C Verma JEE Physics Ch. 32 ELECTRIC CURRENT IN CONDUCTORS

JEE 2008 syllabus

Electric current;
Ohm’s law;
Series and parallel arrangements of resistances and cells;
Kirchhoff’s laws and simple applications;
Heating effect of current.
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Verma - Topics

32.1 Electric current and current density
32.2 Drift speed
32.3 Ohm's law
32.4 Temperature dependence of resistivity
32.5 Battery and EMF
32.6 Energy transfer in an electric current
32.7 Kirchhoff's Laws
32.8 Combination of resistors in series and parallel
32.9 Grouping of batteries
32.10 Wheatstone bridge
32.11 Ammeter and Voltmeter
32.12 Stretched wire potentiometer
32.13 Chargin and discharging of capactiros
32.14 Atmospheric electricity
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Study Plan

Day 1
32.1 Electric current and current density
32.2 Drift speed
32.3 Ohm's law

Day 2
32.4 Temperature dependence of resistivity
32.5 Battery and EMF
Worked out examples 1 to 3

Day 3
32.6 Energy transfer in an electric current
32.7 Kirchhoff's Laws
32.8 Combination of resistors in series and parallel

Day 4
WOE 11 to 15
Exercises 1 to 5

Day 5
32.9 Grouping of batteries
32.10 Wheatstone bridge
WOE 4 to 10

Day 6
32.11 Ammeter and Voltmeter
32.12 Stretched wire potentiometer
WOE 16 to 20

Day 7
32.13 Chargin and discharging of capactiros
32.14 Atmospheric electricity
WOE 21 to 25

Day 8
WOe 26 to 33
Exercises 6 to 10

Day 9
Exercises 11 to 20

Day 10
Exercises 21 to 30

Day 11
Exercises 31 to 35


Day 12
Exercises 36 to 40


Day 13
Exercises 41 to 45


Day 14
Exercises 46 to 50


Day 15
Exercises 51 to 55


Day 16
Exercises 56 to 60


Day 17
Exercises 61 to 65


Day 18
Exercises 66 to 70

Day 19
Exercises 71 to 75


Day 20
Exercises 76 to 80






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Kirchoff's laws

The junction law: The sum of all currents directed toward a point in a circuit is equal to the sum of all the currents directed away from the point.

The loop law: The algebraic sum of all the potential differences along a closed loop in a circuit is zero.

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Heating effect of Current (Work done by electric current)
Joule's law


During the flow of current, the energy spent to overcome the resistance in the conductor is converted into heat energy and the resistor (conductor) gets heated. This is known as the heating effect of the electric current.

Joule's law
The quantity of heat (H) produced in a resistor of resistance R when a current I flows throug it for a time t, is directly proportional to:
i) square of the current
ii0 resistance of the resistor and
iii) time for which the current flows,

H α I²Rt

If H is measured in calories, the above relation can be written as

H = I²Rt/J

where J is the mechanical equivalent of heat.







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Effect and causes

In metallic conductors, the electric current is caused by the motion of negatively charged electrons. During their motion, the electrons collide with the vibrating atoms of the conductor. Due to these collisions, the motion of electrons is opposed. This opposition gives rise to the resistance of the conductor.

If the temperature of the metal is increased, the atoms of the metal vibrate with larger amplitudes and therefore the number of collisions between the atoms and electrons increaes. This results into an increase in the resistance of the metal as temperature increases





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Audiovisual lectures

Lesson 35: Current, Resistance, Power
www.curriki.org/nroc/Introductory_Physics_2/lesson35/Container.html

Lesson 36: DC Circuits with Batteries and Resistors
www.curriki.org/nroc/Introductory_Physics_2/lesson36/Container.html

Lesson 37: Capacitors in Circuits
www.curriki.org/nroc/Introductory_Physics_2/lesson37/Container.html





http://www.colorado.edu/physics/phys1120/phys1120_fa07/notes/notes/Knight28_current_lect.pdf
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JEE question 2007 paper I

A resistance of 2 Ω is connected across one gap of a metre-bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2 Ω, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is
(A) 3 Ω (B) 4 Ω (C) 5 Ω (D) 6 Ω

Solution: A

assume the length of the circuit having 2Ω resistance is l.

2/x = (l/(100-l))

When the load is reversed x is in the circuit having l+20. and 2Ω resistance is in the circuit having 100-l-20 = 80-l

x/2 = (l+20)/(80-l)

2/x = (80-l)/(l+20) = (l/(100-l)

(80-l)*(100-l) = l*(l+20)

8000 -180l +l^2 = l^2+20l

8000 = 200l
l = 40
x = 2*60/40 = 3Ω
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