Tiêu đề Med-RM_Phy_SP-3_Ch-18_Current Electricity.pdf ✅

Chapter Contents Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Electric Current Ohm’s Law Drift of Electrons and the Origin of Resistivity Limitations of Ohm’s Law Resistor Colour Codes Temperature Dependence of Resistivity Electrical Energy, Power Combination of Resistors – Series and Parallel Cells, emf, Internal Resistance Cells in Series and Parallel Kirchhoff’s Rules Wheatstone Bridge R-C Circuit Meter Bridge Potentiometer Thermoelectricity ELECTRIC CURRENT The time rate of flow of charge across any cross-section is called current. I = dq dt Its unit is ampere and dimension [M0L0T0A] 1 ampere = 1 coulomb/second (1 C/s) Steady Current When current does not vary with time.  Q  t. But currents are not always steady. In that case we define current in two ways 1. I av =   Q t . 2. in 0 lim t Q dQ I   t dt     . Conventional direction of current is taken as direction of motion of positively charged particles and opposite to the direction of motion of negatively charged particles. Example 1 : 106 positrons are flowing normally through an area in forward direction and same amount of electrons are flowing in backward direction in the interval of 10 ms. Find the current through the area. Solution : q+ = 106 × 1.6 × 10–19 C = 1.6 × 10–13 C q– = 106 × (–1.6 × 10–19 C) = –1.6 × 10–13 C Chapter 18 Current Electricity
190 Current Electricity NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Net charge through the area q = q+ – q–= 1.6 × 10–13 C – (–1.6 × 10–13 C) = 3.2 × 10–13 C Current through the area 13 11 3 3.2 10 C 3.2 10 A 10 10 s         q I t Example 2 : Charge through a cross-section of a conductor is given by Q = (2t2 + 5t)C. Find the current through the conductor at the instant t = 2 s. Solution : Charge Q = (2t 2 + 5t) C Instantaneous current 2   (2 5 ) dQ d I tt dt dt = 4t + 5 At t = 2 s, I = (4 × 2) + 5 = 13 A Example 3 : In neon gas discharge tube 2.9 × 1018 Ne+ ions move to the right through a cross-section of the tube each second, while 1.2 × 1018 electrons move to the left in this time. The electronic charge is 1.6 × 10–19 coulomb. What is the net electric current in the tube? Solution : Here since positive and negative charges are moving in opposite directions their equivalent current is in the same direction i.e. in the direction in which the positive ions are moving. So the total current will be the sum of the currents due to individual positive and negative charges. Hence net current = ( ) n ne p e t  = 18 18 –19 (2.9 10 1.2 10 ) 1.6 10 1 s    = 0.66 A to the right i.e. the direction in which the positive charge is moving. Example 4 : In the Bohr's model of hydrogen atom, the electron is assumed to rotate in a circular orbit of radius 5 × 10–11 m, at a speed of 2.2 × 106 m/s. What is the current associated with electron motion? Solution : I = q t = 2 qv R distance 2 speed R t v          ∵ = –19 6 –3 –11 1.6 10 C × 2.2 × 10 m/s 1.12 10 A 2 5 10 m      = 1.12 mA Example 5 : The current in a wire varies with time according to the relation i = 2.0A + (0.6A/s2)t2. (a) How many coulomb of charge pass a cross-section of the wire in the time interval between t = 0 and t = 10s? (b) What constant current would transport the same charge in the same time interval? Solution : (a)    t s idtq dtt 10 0 22 0 ])As6.0(A0.2[ = s10 0 3 2 3 )As6.0()A0.2(         t t = (2.0 A) (10s) + (0.6 As–2) 3 )s10( 3 = 20 As + 200 As = 220 As = 220 coulomb (b) Constant current A22 s10 C220  t q i
NEET Current Electricity 191 Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Note : Total charge in the interval t1 to t2 Q = 2 1 t t Idt  = Area below I versus t graph in the interval t1 to t2 as shown in figure. I t t2 t1 Average current in the interval t1 to t2 2 1 21 21 t t av Idt Q Area below I versus t graph I t t t t Time interval     OHM’S LAW If V be the potential difference between the ends of the conductor through which a current I is flowing, then Ohm’s law states that V  I or V = RI where R is the proportionality constant known as Resistance of the conductor, SI unit of resistance are V A–1 or ohm (). 2 ml R ne A   2 m ne    R A   where  is resistivity of material, A is area of cross section of conductor. l is length of conductor. Resistance of the conductor depends on 1. Dimensions of conductor 2. Material of conductor 3. Temperature of conductor Current Density and Electric Field Current density : It is defined as current per unit normal area. It is denoted as  I j A SI unit of j is A m–2 Vectorially     A I j A j dA Also, Vectorially E j   or j  E This represents the equivalence of Ohm’s law. where, 1    is called conductivity of the material of the conductor.
192 Current Electricity NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Dependence of resistance on length and area of cross-section. Note : Keeping volume/mass of the conductor constant, its resistance 2 2 L Ld m R A m A d      1. If length of conductor is made n times, then resistance of the conductor becomes n2 times the initial value. (R  L2) 2. If area of cross-section is made n times, then resistance of conductor becomes 2 1 n times the initial value. 2 1 R A        3. If length is increased by x% (x < 5), then resistance will increase by 2x%. (R  l2) 4. If area is increased by x% (x < 5), then resistance will decrease by 2x%. 2 1 R A        5. If radius of cross-section is increased by x% (x < 5), then resistance will decrease by 4x%. 4 1 R r        Example 6 : Resistance of a conductor of length l and area of cross-section A is R. If its length is doubled and area of cross-section is halved, then find its new resistance. Solution : Initial length = l, Area = A So, initial resistance R = A  Final length l = 2l, Area A = 2 A New resistance R = 2 4 4 2 l R A A A              Example 7 : If j and E are current density and electric field inside a current carrying conductor at an instant, then j E is (1) Positive (2) Negative (3) Zero (4) May be positive or negative Solution : Current density j is directed along E . So the angle between j E and is 0°  j E = jEcos0° = jE i.e., positive So, the option (1) is correct. DRIFT OF ELECTRONS AND THE ORIGIN OF RESISTIVITY In the absence of electric field a free electron suffers collision with heavy fixed ions and after collision it emerges with same speed but in random directions. If we consider N free electrons with velocity of i th electron ui (i = 1, 2, 3, ....., N) at a given cross-section, then average velocity of these free electrons through a cross-section = 1 1 0 N i i u N    When the electric field E is present inside the conductor.