Current Electricity XII Physics PDF

CURRENT ELECTRICITY Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune ...

CURRENT ELECTRICITY Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  CARRIERS OF CURRENT The charged particles which by flowing in a definite direction set up an electric current are called current carriers. Different types of current carriers In solids: In liquids: In gases: In metallic In electrolytic liquids, the In ionised gases, conductors, charge carriers are positive and electrons are the positively and negatively negative ions and charge carriers. The charged ions. For example, electrons are the electric current is CuS04 solution has Cu2+ charge carriers due to the drift of and SO- ions, which act as electrons. the charge carriers. Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  DRIFT VELOCITY AND RELAXATION TIME Metals have a large number of free electrons, nearly 1028 per cubic metre. In the absence of any electric field, these electrons are in a state of continuous random motion due to thermal energy. At room temperature, they move with velocities of the order of 105 m/s. However, these velocities are distributed randomly in all directions. There is no preferred direction of motion. On the average, the number of electrons travelling in any direction will be equal to number of electrons travelling in the opposite direction. If u1, u2,...., uN are the random velocities of N free electrons, then average velocity of electrons will be u1 + u2 +... + u N = 0 ---------------------- (1) Thus, there is no net flow of charge in any direction. In the presence of an external field E, each electron experiences a force, F = – eE , in the opposite direction of E (since an electron has negative charge). The electron undergoes an acceleration ‘a’ given by a = Force/mass = -eE/m where, m is the mass of electron As the electrons accelerate, they frequently collide with other electrons of the metal. Between two successive collisions an electron gains a velocity component in a direction opposite to E. However the gain in velocity lasts for a short time and is lost in the next collision. At each collision, the electron starts fresh with thermal velocity. Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  DRIFT VELOCITY AND RELAXATION TIME If electron which has random velocity u1 accelerates for time 1 then it will attain velocity v 1 = u 1 + a1, Similarly the velocities of other electrons will be v1 = u1 + a  1, v2 = u2 + a  2 --------- vN = uN + a  N Then the average velocity will be vd will be v1 + v2 + --------- + vN vd = N u1 + a  1+ u2 + a  2 --------- + uN + a  N ( u1 + u2 ------- + uN ) + a (  1 +  2 --------- +  N ) vd = = N N ( u1 + u2 ------- + uN ) (  1+  2 --------- +  N ) vd = + a ---------------- (2) N N From equation (1) This term is the average time between two successive collisions and this term is equal is known as relaxation time (  ) and is defined as the average time to 0. that elapses between two successive collisions of an electron.  vd = a  The term vd is called drift velocity and it is defined as the average velocity gained by the free electron of a conductor in the direction opposite to that of the externally applied electric field. Force -e E Acceleration, a = = -e E mass m Drift velocity = vd =  -e E m Drift velocity = vd = a  =  m Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  RELATION BETWEEN ELECTRIC CURRENT AND DRIFT VELOCITY Consider that a potential difference V is applied across a conductor of length l and of uniform cross-section A. The electric field E set up inside the conductor is given by E= V/ l V is the potential difference) Under the influence of field E, the free electrons begin to drift in the opposite direction of E with an average drift velocity vd Let the number of electrons per unit volume or electron density = n Number of electrons in length l of the conductor = n x volume of the conductor = n Al Charge on an electron = e  Total charge contained in length I of the conductor is q = en Al All the electrons which enter the conductor at the right end will pass through the conductor at the left end in time, Distance l  Therefore, the current, I = ne Avd t= = Velocity vd  Current density , which is current per unit area Charge ne Avd Therefore, the current, I =  j= I = time A A en Al  current density , j =ne vd  Therefore, the current, I = l/vd Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  DEDUCTION OF OHM’S LAW When a potential difference V is applied across a conductor of length l, the drift velocity in terms of V is given by: vd = acceleration x time vd = eE/ m vd = eV  /m l If the area of cross-section of the conductor is A and the number of electrons per unit volume or the electron density of the conductor is n, then the current through the conductor will be  I = ne Avd eV  I = ne A x ml V ml  = I ne2  A At a fixed temperature, the quantities m, l, n, e,  and A have constant values for a given conductor. Therefore, V = a constant, R x I This proves Ohm's law for a conductor and the resistance of the conductor is given by ml  R = ne2  A Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  RESISTIVITY IN TERMS OF ELECTRON DENSITY Consider a conductor ‘l ’ and area of cross-section ‘A’. If  is the resistivity of the conductor then l  R = ------------------ (1) A The resistance of the conductor is also given by ml  R = ------------------ (2) ne2  A From (1) and (2) m   = ne2  Thus, the resistivity 1. is independent of the dimensions of the conductor. 2. depends upon the number of free electrons per unit volume or electron density of the conductor. 3. depends upon the relaxation time , the average time between two successive collisions of an electron. Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  SOME POINTS TO REMEMBER  Collisions are the basic cause of resistance. When a potential difference is applied across a conductor, its free electrons get accelerated. On their way, they frequently collide with the positive metal ions i.e., their motion is opposed and this opposition to the flow of electrons is called resistance.  Larger the number of collisions per second, smaller is the relaxation time , and larger will be the Resistivity The number of collisions that the electrons make with the atoms/ions depends on the arrangement of atoms or ions in a conductor. So the resistance depends on the nature of the material (copper, silver etc.) of the conductor. The resistance of a conductor depends on its length. A long wire offers more resistance than short wire because there will be more collisions in the longer wire. The resistance of conductor depends on its area of cross section. A thick wire offers less resistance than a thin wire because in a thick wire, more area of cross-section is available for the flow of electrons. Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  MOBILITY OF CHARGE CARRIER The conductivity of any material is due to its mobile charge carriers. These may be electrons in metals, positive and negative ions in electrolytes; and electrons and holes in semiconductors. The mobility of a charge carrier is the drift velocity acquired by it in a unit electric field. It is denoted by .  = vd/ E  = qE / mE  = q  / m For electron :  = e  / m Unit of mobility = m2/V/s RELATION BETWEEN ELECTRIC CURRENT AND MOBILITY FOR A CONDUCTOR In a metallic conductor, the electric current is due to its free electrons and is given by I= neAvd But vd =  E  I = neAE This is the relation between electric current and electron mobility. Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune 1. A copper wire has a resistance of 10 and an area of cross-section 1 mm2. A potential difference of 10 V exists across the wire. Calculate the drift speed of electrons if the number of electrons per cubic metre in copper is 8 x 1028 electrons. 1. Drift velocity = 0.078 mm/s 2. A copper wire of diameter 1 mm carries a current of 0.2A. Copper has 8.4x1028 atoms per cubic meter. Find the drift velocity of electrons assuming that one charge carrier of 1.6 x 10-19 C is associated with each atom of the metal. 2. Drift velocity = 1.895 x 10-5 m/s 3. A current of 2 A is flowing through a wire of length 4 m and cross sectional area 1 mm2. If each cubic meter of the wire contains 1029 free electrons, find the average time taken by an electron to cross the length of the wire. 3. Time taken, 3.2 x 104 seconds) 4. A potential difference of 6 V is applied across a conductor of length 0.12 m. Calculate the drift velocity of electrons if the electron mobility is 5.6x10-6 m 2/V/s 4.Drift velocity = 2.8 x 10-4 m/s 5.The number density of electron in copper is 8.5x1028per m3. Determine the current flowing through a copper wire of length 0.2 m, area of cross section 1 mm2, when connected to a battery of 3 V. Given the electron mobility 4.5x10-6 m 2/V/s and charge on electron is 1.6 x 10-19C Current flowing through the wire = 0.918 A Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  INTERNAL RESISTANCE When the terminals of a cell are connected by a wire, an electric current flows in the wire from positive terminal of the cell towards the negative terminal. But inside the electrolyte of the cell, the positive ions flow from the lower to the higher potential (or negative ions from the higher to the lower potential). So the electrolyte offers some resistance to the flow of current inside the cell. The resistance offered by the electrolyte of a cell to the flow of current between its electrodes is called internal resistance of the cell. It is denoted by ‘r’. The internal resistance of a cell depends on following factors: 1. Nature of the electrolyte. 2. It is directly proportional to the concentration of the electrolyte. 3. It is directly proportional to the distance between the two electrodes. 4. It varies inversely as the common area of the electrodes immersed in the electrolyte. 5. It increases with the decrease in temperature of the electrolyte. The internal resistance of a freshly prepared cell is usually low but its value increases as we draw more and more current from it. Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune RELATION BETWEEN EMF, POTENTIAL DIFFERENCE AND INTERNAL RESISTANCE The EMF E = work done by the cell in carrying a unit charge along the closed circuit E = Work done in carrying a unit + charge from B to charge work done in carrying A against the a unit from A to B against the internal external resistance R resistance r  E = V + V’ By Ohm’s law V = IR and V’ = Ir E = IR + Ir E = I(R + r) E Hence the current in the circuit will be I = (R + r) Thus to determine the current in the circuit, the internal resistance r combines in series with external resistance R. The terminal p.d. of the cell that sends current I through the external resistance R is given by V = IR ER V= = E - Ir Terminal p.d = emf - potential drop across R+r the internal resistance The internal resistance, r = (E – V )/ I r = (E - V )R V Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune RELATION BETWEEN EMF, POTENTIAL DIFFERENCE AND INTERNAL RESISTANCE The potential difference, V = E - Ir Special Cases: (i) When cell is on open circuit, i.e., I =0, we have V open = E Thus the potential difference across the terminals of the cell is equal to its emf when no current is being drawn from the cell. (ii) A real cell has always some internal resistance r, so when current is being drawn from cell, V< E Thus the potential difference across the terminals of the cell in a closed circuit is always less than its emf Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  CELLS CONNECTED IN SERIES When the negative terminal of one cell is connected to the positive terminal of the other cell and so on, the cells are said to be connected in series. Consider two cells of emfs E1 and E2 and internal resistances r1 and r2 are connected in series between points A and C. Let I be the current flowing through the series combination Let VA,VB and VC be the potentials at points A, B and C respectively. The potential differences across the terminals of the two cells will be VAB = VA - VB = E1 - I r1 ------------------ (1) VBC = VB - VC = E2 - I r2 ------------------ (2) Thus the potential difference between the terminals A and C of the series combination is VAC = VA – VC = (VA - VB) + (VB – VC) = ( E1 - I r1) + (E2 - I r2) VAC = (E1 + E2 ) - I(r1 + r2) If we wish to replace the series combination by a single cell of emf Eeq and internal resistance req then VAC = Eeq – I req Comparing the last two equations, when cells are connected in series, Eeq = E1 + E2 and req = r1 + r2 Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  CELLS CONNECTED IN SERIES When two or more cells are connected in series, 1. The equivalent emf of a series combination of n cells is equal to the sum of their individual emfs. Eeq = E1 + E2 + E3 + ------------ + En 2. The equivalent internal resistance of a series combination of n cells is equal to the sum of their individual internal resistances. req = r1 + r2 + r3 + ----------- + rn 3. The above expression for Eeq is valid when the n cells assist each other i.e., the current leaves each cell from the positive terminal. However, if one cell of emf E2, say, is turned around in opposition' to other cells, then Eeq = E1 - E2 + E3 + ------------ + En Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  CELLS CONNECTED IN PARALLEL When the positive terminals of all cells are connected to one point and all the negative terminals to another point, the cells are said to be connected in parallel. Consider two cells of emfs E1 and E2 and internal resistances r1and r2 are connected in parallel between two points A and C. The currents I1 and I2 from the positive terminals of the two cells flow towards the junction B and current I flows out. As the two cells are connected in parallel, the potential difference V across both cells must be same and the current, I = I1 + I2 The potential difference between the terminals The potential difference between the terminals of of first cell is V = E1 - I1 r1 second cell is V = E2 – I2 r2 E1 - V E2 - V  I1 =  I2 = r1 r2 E1 - V E2 - V The current I = I1 + I2 = + r1 r2 E1 E2 1 + 1 The current I = r1 + r2 -V r1 r2 E1 E2 r1 + r2 I = r1 + r2 -V r1 r2 Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  CELLS CONNECTED IN PARALLEL E1 E2 r1 + r2 I = r1 + r2 -V r1 r2 r1 + r2 E1 r2 + E2 r1 V = - I r1 r2 r1 r2 E1 r2 + E2 r1 r1 r2 V = - I r1 + r 2 r1 + r2 If we wish to replace the parallel combination by a single cell of emf Eeq and internal resistance req, then V = Eeq – req Comparing the last two equations, we get E1 r2 + E2 r1 r1 r2 Eeq = And req = r1 + r 2 r1 + r2 Without limiting the rights under copyright, no part of this document may be reproduced, or transmitted in any form or by any means, or for any purpose, without the express written permission of Sanskriti Group of Schools, Pune  CONDITION FOR OBTAINING MAXIMUM CURRENT THROUGH AN EXTERNAL RESISTANCE WHEN THE EXTERNAL RESISTANCE IS CONNECTED ACROSS A SERIES COMBINATION OF CELLS When n cells of equal emf E are connected in series Total emf = sum of the emf of all the cells = nE Total internal resistance of n cells in series = r + r + r +...... n terms = nr Total resistance in the circuit = R + nr Total emf The current in the circuit = I = Total resistance nE I= R + nr Special Cases: (i) If the external resistance is much larger than the total internal resistance, R >> nr, then nE E I= =n = n times the current that can be drawn from one cell. R R (ii) If the external resistance is less than the total internal resistance R