THEORYCURRENT ELECTRICITY SYLLABUS : Electric current, Drift velocity, Ohm's law, Electrical resistance, Resistances of different materials, V-I characteristics of Ohmic and nonohmic conductors, Electrical energy and power, Electrical resistivity, Colour code for resistor, Series and parallel combinations of resistors; Temperature dependence of resistance. Electric Cell and its internal resistance, potential difference and emf of a cell, combination of cells in series and in parallel. Kirchhoff's laws and their applications. Wheatstone bridge, Metre bridge. Potentiometer-principle and its applications. ELECTRIC CURRENT : (a) When a charge flows in a conductor from one place to the other, then the rate of flow of charge is called electric current(i) (b) The electric current in measured by 'rate of flow of charge'. or (c) Charge flowing per second from any cross section of the conductor is called electric current, Ch arg e dq q Current i = = , if flow is uniform i= Time dt t (d) Unit : Ampere (A) 1 ampere = 1 coulomb/second. i.e. if 1 coulomb of charge flows per second then 1 ampere of current is said to be flowing. (e) Dimension : (M0L0T0A1) ne (f) If n electrons pass through any cross section in every t seconds then i = t where e = 1.6 ◊ 10ñ19 coulomb. (g) 1 ampere of current means the flow of 6.25 ◊ 1018 electrons per second through any cross section of conductor (h) Direction of flow of current is taken to be opposite to the direction of flow of electrons. (i) Value of the current is same throughout the conductor, irrespective of the cross section of conductor at different points. (j) Net charge in a current carrying conductor is zero at any instant of time. (k) Electric field outside a current carrying conductor is zero, but it is non zero inside the conductor v and is given by e = ñ l (l) Electric current is a scalar quality Although in diagrams, we represent current in a wire by an arrow but the arrow simply indicate the direction of flow of positive charges in the wire. CURRENT DENSITY : (a) The current density at a point in a conductor is the ratio of the current at that point in the conductor to the area of crossñsection of the conductor of that point. i (b) It is denoted by j i.e. j = A i = Electric current A = Area of cross section. Note : Area 'A' is normal to current 'I'. If A is not normal to I, but makes an angle with the normal to current, then I P A I 1 j= A = normal A cos I = j A cos = j . A 6.25 J = 2 2 10 6 4 2 ◊ 106 A/m2 200 A/cm2 (c) It is a VECTOR quantity. It's direction is the direction of motion of the positive charges at that point. (d) Units : ampere / meter2 (A/m2) (e) Dimension : [M0Lñ2T0A] (f) If, n = number of free electrons per unit volume of conductor. A = cross sectional area of conductor vd = Drift. velocity. then I = neA vd and J = ne vd (g) Drift velocity : An applied potential difference does not give an accelerated motion to electrons but simply gives them a small constant velocity ( 10ñ4 m/s) along the length of wire towards the end at higher potential. This is called Drift velocity of the electrons. Note : The speed of random motion of electrons is determined by temperature and is given by 1 3 3kT mv2 = kT v = 2 2 m where m is mass of electron, T is absolute temp. and k is Boltz mann's constant. (h) Electrons collide with the ions of metal while moving. The average timeñinterval between two successive collisions is called relaxationñtime, denoted by . The relations between relaxation time () and drift velocity (vd) is given vd = ñ e E m OHM'S LAW : (a) If there is no change in the physical state of a conductor (Such as temperature) then the ratio of the potential difference applied at it's ends and the current flowing through it is constant i.e. I V VI or V=IR; where, R is a constant. This is called 'Electrical resistance' of the conductor. T0 and = thermal coefficient of resistance so. (l) Inverse of resistivity is called conductance of wire denoted by 1 = (m) Units of conductance : Mho Important points : (a) If a conductor is stretched to n times of it's original length. then R = 2 l 22 R1 l 24 (ii) If the radius of wire is changed. It does not depend upon length area etc. (e) Dimensions of resistance : [M1L2Tñ3A ñ2] (f) If. as T increases R increases. I I I Semi I conductor Diode Torch (a) (b) (c) (d) Bulb V V V V (d) Units of resistance : ohm() 1 ohm = 1 volt / 1 ampere. transistors. L = length of conductor R = resistance of conductor A = cross sectional area of conductor perpendicular to current 1 L Then. Note : Effect of stretching a wire on its resistance R1 l 12 (i) If the length of wire is changed. is called 'Resistivity' or 'Specific resistance'. of the conductor. (i) Resistivity is also defined as the ratio of the intensity of the electric field E at any point within the conductor and the current density j at that point E = j or j E (j) Resistivity is characteristic property of the material of the conductor. R L . semiñconductors etc. it's new resistance will be n2 times older one . (c) The law is not applicable for ionized gases. Although it depends on temperature. then = R2 f14 (g) Units of = Ohmñmeter Dimensions of = [M1 L3 Tñ3 A2] (h) If T = Temperature in kelvin R = R0 (1 + (T ñ T0) where R0 = Resistance at temp. R R= A A This constant of proportionality.(b) This is true for metallic conductors only which have free electrons. It increases with increase in temperature (k) Value of resistivity is least for conductors and most for insulators. .... = 0 for alloys.. COMBINATION OF RESISTANCES : (a) Series Combination R1 R2 R3 I V (i) Same current passes through each resistance... I =R Where..... (c) Resistance of the conductor decreases linearly with temperature and becomes zero at a specific temperature..... This temperature is called critical or transition temperature. (d) There is no loss of energy in a circuit formed by super conductors.. Current passed in loop formed by superconductor will continue flowing for infinite time if there is no resistance in the loop.. (ii) Voltage across each resistance is directly proportional to it's value.. = ñve for semi conductors and insulators. This is true for x < 5 only. R = equivalent resistance.(b) if x% of change is brought in length of a wire. V = R1 + R2 + R3 + . conductor becomes a super conductor at this temperature.... V = V1 + V2 + V3 + . V2 = IR2 (iii) Sum of the voltages across resistances is equal to the voltage applied across the circuit i.... it's resistance will change by 2x%.. R0 = Resistance at 00 C t = change in temperature. . V1 = IR1. (c) If a conductor is stretched such that it's radius is reduced to 1/nth of it's original values. then resistance will increases n4 times similarly resistance will decrease n4 times if radius is increased n times by contraction ñ Effect of temperature on resistance : Rt (a) R0 o t∞C Rt = R0 (1 + t) where.......e... V = IR1 + IR2 + IR3 + ...... (b) R2 = R1 [1 + (t2 ñ t1)]. = Temperature coefficient of resistance at 0∫ C = +ve for metals.. This formula gives an approximate value.... Rt = Resistance at t0 C.. .e. R (b) The equivalent resistance of parallel combination is lower than the value of lowest resistance in the combination. V V V i = R + R + R + .. i3 = R etc.. 1 Note : (a) You are asked to find R and not in the question...... R1 i1 i1 i2 R2 V V(R1 R2 ) (i) i = i1 + i2 = R1R2 .. (c) For a parallel combination of two resistances . V 1R 2 3 where R = equivalent resistance. 1 2 3 (iii) Current flowing in the circuit is sum of the currents in individual resistances i. 1 2 3 i 1 1 1 1 = = R + R + R + .... (ii) Current in each resistance is inversely proportional to the value of resistance i. so be careful. i = i1 + i2 + i3 .Note : If n resistance (each R) are connected in series there resultant will be nR (iv) For a series combination of two resistances R1 R2 V (A) equivalent resistance R = R1 + R2 (B) I = V / (R1 + R2) R1V (C) V1 (voltage across R1) = IR1 = R R 1 2 R2V (D) V2 (voltage across R2) = IR2 = R R 1 2 (b) PARALLEL COMBINATION : i1 R1 i1 R2 A B i1 R3 V (i) There is same drop of potential across each resistance. i2 = R ... V V V i1 = R ..e.. In other words. (b) This law is based on 'law of conservation of energy'. their resultant will be R/n (ii) If n resistance are connected in series and parallel respectively the ratio of their resultant will be nR : R/n = n2 . i. E= Q (b) Terminal voltage : (i) The resistance offered by the electrolyte of the cell to the flow of current through it is called internal resistance of the cell. and the emf is taken as positive when we traverse from the negative to the positive electrode of the cell through the electrolyte.m. when we traverse in the direction of current then the product of the currrent and the corresponding resistance is taken as positive. These laws are as followsñ (a) First law : In an electrical circuit. CELLS : (a) Electro Motive Force (EMF) : The potential difference across the terminals of a cell when it is not giving any current is called EMF of the cell. KIRCHHOFF'S LAWS : Kirchoff in 1842 gave two laws for solving complicated electrical circuits. iR = E Important notes (a) In applying this law.f. the algebraic sum of the current meeting at any junction in the circuit is zero. (ii) When current is drawn through the cell or current is supplied to cell then. when a steady current flows in a circuit then their is neither accumulation of charge at point in the circuit nor any charge is removed from there. .'s in that mesh.Note : (i) If n resistances (each R) are connected in parallel. OR Sum of the currents entering the junction is equal to sum of the currents leaving the junction i = 0 i1 ñ i2 ñ i3 ñ i4 + i5 = 0 or i1 + i5 = i2 + i3 + i4 Note : This law is based on law of conservation of charge. the potential difference across its terminals is called terminal voltage. or The energy given by the cell in the flow of unit charge in the whole circuit (including the cell) is W called the EMF of the cell. (b) Second law : In a 'closed' mesh of a circuit the algebric sum of the products of the current and the resistance in each part of the mesh is equal to the algebric sum of the e.e. . r3 r1 r2 r3 ri E= = E3 r4 1 1 1 1 .... Combinations of cells : E1 r1 E2 r2 E3 r3 i1 R (a) Series Combination : (i) Equivalent emf E = E1 + E2 + E3 .. (ii) Equivalent internal resistance r is given by r = r1 + r2 + r3 .. then for series combinations of n such cells.. i = Rr E R R .. R = r i.... then terminal voltage is less than it's emf E.. I = r R (c) Cells are employed in series only when internal resistance is less than the load resistance.. r = internal resistance of battery (iv) When current is supplied to the cell.. the terminal voltage is greater than the emf E i. I I I r is + r + r + ......e.. V=Eñir E r Where V = terminal voltage.. I = (b) If nr << R..... E Ei (iii) Current.. Note : Direction of emf is taken into consideration... r 1 2 (ii) Equivalent emf E1 r1 E1 E2 E3 Ei i1 E2 r2 . I = R nr E nE Cases : (a) if nr >> R........ (b) Parallel Combination : (i) Equivalent internal resistance.. The load resistance must be equal to the equivalent internal resistance.. E4 r1 r2 r3 ri R E (iii) Current.(iii) When i current is drawn from cell. i = = r R r R i Imp : (iv) For maximum current......... nE (v) If all emf are equal (E)...e.. V=E+ir (v) Units of both emf and terminal voltage are volt..... I = R nE (b) If r >> nR. (iv) When all 'n' cells with emf E and internal resistance r each. For maximum current mR nr Internal resistance = External resistance R nr i. Q S (g) Meter bridge and post office box work on this principle. equivalent internal resistance = n E nE (v) In this (5) case I = R r = n nR r Cases : E (a) If r << nR. r then equivalent emf = E.e. E n (c) Mixed combination : m mnE i= .e. . then bridge is said to be balanced. (b) If ig = 0 i. P R (c) For ig = 0 (i) VD = VB (ii) = Q S (P Q) (R S) (d) Equivalent resistance in balanced condition = PQRS P R (e) If < then VB > VD and current will flow from B to D. are connected in parallel. R = m WHEAT STONE BRIDGE : (a) The configuration in the adjacent figure is called wheat stone bridge. Q S P R (f) If > . current in galvanometer is zero. I = r (c) This combination is used only when load resistance is lower than internal resistance. the VB < VD and current will flow from D to B. 2 Electrical power : The rate of loss of energy in an electrical circuit is called electrical power. V2 dW = VIdt = I2Rdt = dt = Vdq R This energy is equal to work done by battery or heat produced in the wire. horse power 1 watt = 1 joule/sec. It is denoted by'P' dW V2 P= = I2 R = IV = dt R units of power = joule/sec. If energy is to be written in calorieñ dW Then dW = cal = 24 dW cal When dW is energy in Joules. The battery constantly provide energy to continue the motion of electron and hence electric current in the circuit. It means that if 220V is applied across this instrument then 40W of power will be generated. Energy dissipated dW = Vdq = VIdt. V = IR. kilowatt hour 1 kilowatt hour (kwh) = 36 ◊ 105 Joule Combination of electrical instrumentsñ (a) If 220V and 40W is written on an electrical instrument then this is called it's standard Ratings. 1 HP = 746 watt unit of electrical energy = watt second. 4. This effect is also called 'Heating Effect of Current'. current starts flowing in it. If R = Resistance of wire I = Current in wire V = Potential difference across wire. This energy is given to ions of the metal during collision and thus temperature of wire rises.ELECTRIC ENERGY AND POWER : Electric energy : When a potential difference is applied across a wire. The free electrons collide with the positive ions of the metal and lose energy. Thus energy taken from the battery is dissipated. Thus. This energy is called electrical energy. energy taken from the battery gets transferred in to heat. Thus V2 (220 ) 2 the resistance will be given by R= = ohm P 40 (b) Series combination : . Flow of charge in 'dt' time = Idt. watt. P 1 2 3 Where P'1 s are standard powers of instrument (ii) In this combination. If standard rating is (V/P) then it resistance is R = V2/P (c) Find the currents and voltages in different branches using kirchoff's first and second laws. (d) If rating of a bulb is changed form V1/P1 to V2/P2 then V12 V22 V22 = =R or P2 = P1 P1 P2 V12 . (c) Parallel combination : (i) Net power dissipation P = P1 + P2 + P3 (ii) Bulb with least power will glow least or the bulb in which maximum current is flowing will glow brightest and viceñversa. Note : (a) These formulae are applicable only if the voltage ratings of all the instruments are equal along with the power source. (b) Replace the instrument by its's equivalent resistance. If voltage ratings are different then circuit is solved by considering equivalent resistances of the instruments as follows. (i) If total power dissipatted if P. the bulb with least power will glow most and bulb with highest power will glow least or we can say that bulb with highest R will glow brightest and bulb with least R will glow least. 1 1 1 1 then = P + P + P . EXERCISE-I Q.6 Net resistance between X and Y is ñ (A) 5 (B) 10 (C) 15 (D) 60 .2 When the resistance wire is passed through a die the crossñsection area decreases by 1%. The arm of B is double that of A.5 Net resitance between X and Y is ñ R (A) R (B) 2R (C) (D) 4R 2 Q.3 When the resistance of copper wire is 0.1 A current (I) flows through a uniform wire of diameter (d) when the mean drift velocity is V. The same current will flow through a wire of diameter d/2 made of the same material if the mean drift velocity of the electron is (A) V/4 (B) V/2 (C) 4V (D) 2V Q. Compare these resistances and find out the value of RA/RB is ñ 1 (A) 1 (B) 2 (C) (D) 4 2 Q.1 and the radius is 1 mm. the change in resistance of the wire is (A) 1% decrease (B) 1% increase (C) 2% decrease (D) 2% increase Q.4 In the following diagram two parallelopipedA and B are of the same thickness.14 ◊ 10ñ8 ohm ◊ m) ñ (A) 10 cm (B) 10 m (C) 100 m (D) 100 cm Q. then the length of the wire is (specific resistance of copper is 3. 8 For the following circuits.7 Net resistance between X and Y is ñ (A) 4 (B) 4. the potential difference between X and Y in volt is ñ 2 4 8 5 (A) (B) (C) (D) 3 3 9 3 Q. The equivalent resistance between point (A) and (B) is ñ (A) R (B) 5R (C) R/5 (D) 2R/5 .ammeter in ampere for the following circuit is ñ (A) 1 (B) 2 (C) 3 (D) 4 Q.9 Reading of ideal .Q.55 (C) 2 (D) 20 Q.10 Five identical resistance are connected as shown in fig. m.15 Two electric bulbs whose resistances are in the ratio of 1 : 2 are connected in parallel to a constant voltage source.f (E) and internal resistance (r) is connected in series with an external resistance (nr.14 A cell of e. To get maximum power in the external circuit. The equivalent resistance between points (A) and (B) is R1.11 Four identical resistances are joined as shown in fig.13 Two cells of same emf E and internal resistance r are connected in parallel with a resistance of R.) then the ratio of the terminal p.Q.F is (A) 1/n (B) 1/(n+1) (C) n/(n+1) (D) (n+1)/n Q. The power's dissipated in them have the ratio (A) 1 : 2 (B) 1 : 1 (C) 2 : 1 (D) 1 : 4 . to E.12 In the following circuit the resultant emf between AB is ñ (A) E1 + E2 + E3 + E4 (B) E1 + E2 + 2E3 + E4 (C) E1 + E2 + (E3/2) + E4 (D) E1 + E2 + (E3/4) + E4 Q. The equivalent resistance between points A and C is R2 then ratio of R1/R2 is ñ (A) 1 : 1 (B) 4 : 3 (C) 3 : 4 (D) 1 : 2 Q.M.d. the value of R is ñ E + ñ r Eñ + r R r (A) R (B) R = r (C) R = 2 r (D) R = 4r 2 Q. The same wire is connected across a diameter AB as shown in fig.m.2 The current (I) and voltage (V) graphs for a given metallic wire at two different temperature (T1) and (T2) are shown in fig. The effective resistance between two diagonal ends will be (A) 5/6 R (B) 6/5 R (C) 3R (D) 12 R Q.f (E) and internal resistance (r) are connected in series. The equivalent resistance is - (A) ohm (B) ( 2) ohm (C) ( 4) ohm (D) ( + 1) ohm Q.m. The ratio of the radii of two wires is 1 : 2.5 m-1 is bent into a circle of radius 1m.then the equivalent e. EXERCISE-II Q.5 Twelve wires of equal resistance (R) are connected to form a cube. The ratio of the specific resistance of the two materials is ñ (A) 1 : 1 (B) 1 : 2 (C) 1 : 4 (D) 4 : 1 Q.4 Two wires of equal lengths and of material (x) and (y) have same resistance.7 In fig the equivalent resistance between points (x) and (y) (A) 16 (B) 14 (C) 11 (D) 18 .1 A wire of resistance 0.3 A wire of resistance 2 is redrawn so that its length becomes four times.6 Five cells each of e. The resistance of the redrawn wire is ñ (A) 2 (B) 8 (C) 16 (D) 32 Q. It is concluded that (A) T1 > T2 (B) T1 < T2 (C) T1 = T2 (D) T1 = 2T2 Q. If due to oversight one cell is connected wrongly .f and internal resistance of the combination is (A) 5E and 5r (B) 3E and 3r (C) 3E and 5r (D) 5E and 4r Q. the reading of voltmeter is ñ (A) 1.6V Q.33 V (B) 0.8 In the circuit shown in fig. Then the relation between R . r1 .9 If fig. and r2 is - r1. the reading of an ideal voltmeter (v) is zero.5 A (D) 9A Q.r2 (C) R = r1 + r2 (D) R = r r 1 2 .10 In fig the current through resistance (R) is (A) 3A (B) 13A (C) 6.8 V (C) 2.r2 (A) R = r2 .33V Q.r1 (B) R = r1 . the difference of potential between (B) and (D) is - (A) + 0.67V (B) ñ0.Q.67V (C) 2V (D) 1.11 In the adjoining figure.0 V (D) 1. Statement 2 : Because effective internal resistance of cells decreases.33 A .33 A (C) 6A . 3A (D) 3A .R 3 (C) R R (D) R R R1 R 3 1 3 1 3 Q.12 In fig . and Statement-2 is the correct explanation of State- ment-1. the steady state voltage drop across capacitor (C) is VR1 VR 3 VR1 (A) V (B) R 3 R1.067A (B) 3. (D) Statement-1 is false but Statement-2 is true.15 Statement 1 : The total resistance in series combination of resistors increases and in parallel combination of resistors decreases. (C) Statement-1 is true but Statement-2 is false. .14 Statement 1 : When cells are connected in parallel to the external load. Q.f increases. (B) Both Statement-1 and Statement-2 are true but Statement-2 is not the correct explanation of Statement-1. (B) Both Statement-1 and Statement-2 are true but Statement-2 is not the correct explanation of Statement-1. (D) Statement-1 is false but Statement-2 is true. 7. 6A Q. (A) Both Statement-1 and Statement-2 are true. 3. Statement 2 : In series combination of resistors. the effective e.m.13 In fig the current in 3 and 6 resistance is respectively- (A) 7.Q. (C) Statement-1 is true but Statement-2 is false. and Statement-2 is the correct explanation of State- ment-1. the area of cross-section of the resistors increases.67A . the effective length of resistors increases and in parallel combination of resistors. (A) Both Statement-1 and Statement-2 are true. 5 A (B) 2 A (C) 1/3 A (D) 1 A Q.5 A 3 volt battery with negligible internal resistance is connected in a circuit as shown in the figure. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. The power consumed will be ñ [AIEEE-2003] (A) 500 watt (B) 250 watt (C) 1000 watt (D) 750 watt Q.8 An electric current is passed through a circut containing two wires of the same material. connected in parallel. Due to the consequent decrease in diameter the change in the resistance of the wire will be ñ [AIEEE-2003] (A) 100% (B) 50% (C) 300% (D) 200% Q. The current I in the circuit will be ñ [AIEEE-2003] (A) 1. 1000 watt bulb is connected across a 110 volt mains supply.4 A 220 volt. then the ratio of the currents passing through the wires will be ñ [AIEEE-2004] (A) 3 (B) 1/3 (C) 8/9 (D) 2 . The P2 : P1 is ñ [AIEEE-2002] (A) 1 (B) 4 (C) 2 (D) 3 Q.6 The total current supplied to the circuit by the battery is ñ [AIEEE-2004] (A) 1 A (B) 2 A (C) 4 A (D) 6 A Q.7 The resistance of the series combination of two resistance is S.2 A wire when connected to 220 V mains supply has power dissipation P1. If S = n P then the minimum possible value of n is ñ [AIEEE-2004] (A) 4 (B) 3 (C) 2 (D) 1 Q. Power dissipation in this case is P2.3 The length of a given cylindrical wire is increased by 100%.1 If energy consumption of this circuit is 150 watt then find the value of resistance ñ [AIEEE-2002] (A) 2 (B) 4 (C) 6 (D) 8 Q. EXERCISE-III Q. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3. When they are joined in parallel the total resistance is P. 10 Time taken by a 836 W heater to heat one litre of water from 10∞C to 40∞C is ñ [AIEEE-2004] (A) 50 s (B) 100 s (C) 150 s (D) 200 s Q. the ratio B /A of their respective lengths must be ñ [AIEEE 2006] 1 1 (A) (B) 2 (C) 1 (D) 4 2 Q.16 The Kirchhoff's first law (i = 0) and second law (iR = E). then ñ [AIEEE-2005] (A) R = R2 ◊ (R1 + R2)/(R2 ñ R1) (B) R = R2 ñ R1 (C) R = R1R2/(R1 + R2) (D) R = R1R2/(R2 ñ R1) Q.12 In the circuit. are respectively based on ñ [AIEEE 2006] (A) conservation of momentum.14 An energy source will supply a constant current into the load if its internal resistance is ñ [AIEEE-2005] (A) equal to the resistance of the load (B) very large as compared to the load resistance (C) zero (D) non-zero but less than the resistance of the load Q. conservation of momentum (D) conservation of energy.13 Two sources of equal emf are connected to an external resistance R. the resistance in ohms needed to be connected in series with the coil will be ñ [AIEEE-2005] (A) 103 (B) 105 (C) 99995 (D) 9995 Q. If the potential difference across the source having internal resistance R2 is zero. The internal resistances of the two sources are R1 and R2(R2 > R1). conservation of charge (B) conservation of charge. then for the two wires to have the same resistance. the value of the resistor R will be ñ AIEEE-2005] (A) 200 (B) 100 (C) 500 (D) 1000 Q. where the symbols have usual meanings.11 A moving coil galvanometer has 150 equal divisions.15 A material 'B' has twice the specific resistance of 'A'. conscrvation of charge .Q. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt.9 The thermistors are usually made of ñ [AIEEE-2004] (A) Metals with low temperature coefficient of resistivity (B) Metals with high temperature coefficent of resistivity (C) metal oxides with high temperature coefficient of resistivity (D) Semiconducting meterials having low temperature coefficient of resistivity Q. A circular wire made of 'B' has twice the diameter of a wire made of 'A'. In order that each division reads 1 volt. If the batteries A and B have negligible internal resistance. the galvanometer G shows zero deflection. conservation of energy (C) conservation of charge. three resistances P. Current ëIí enters at ëAí and leaves from ëDí.Q.03 A P2 to P1 (C) 0. obtain the potential V(r) at r. (iv) Repeat (i).21 A 5 V battery with internal resistance 2 and a 2V battery with internal resistance 1 are connected to a 10 resistor as shown in the figure. (iii) From the ërí dependence of E(r). We apply superposition principle to find voltage ëVí developed between ëBí and ëCí.5 A Q.33 A (D) 0. . The power consumed by it when operated on 110 volt will be ñ [AIEEE 2006] (A) 25 watt (B) 50 watt (C) 75 watt (D) 40 watt Q. where j is the current per unit area at ërí. (2 Questions) Consider a block of conducting material of resistivity ëí shown in the figure. The resistance of the wire at 0∫C will be [AIEEE 2007] (A) 2 ohm (B) 1 ohm (C) 4 ohm (D) 3 ohm Q. (ii) Calculate field E(r) at distance ërí from A by using Ohmís law E = j.18 In a Wheatstone's bridge. The calculation is done in the following steps : (i) Take current ëIí entering from ëAí and assume it to spread over a hemispherical surface in the block.17 The current I drawn from the 5 volt source will be ñ [AIEEE 2006] (A) 0.67 A (B) 0. 22 and 23 are based on the following paragraph.27 A P2 to P1 Directions : Questions No.17 A (C) 0.03 A P1 to P2 (B) 0. (ii) and (iii) for current ëIí leaving ëDí and superpose results for ëAí and ëDí. the condition for the bridge to be balance will be ñ [AIEEE 2006] P R(S1 S 2 ) P R P 2R P R(S1 S 2 ) (A) = 2S S (B) = S S (C) = S S (D) = SS Q 1 2 Q 1 2 Q 1 2 Q 1 2 Q.19 An electric bulb is rated 220 volt ñ 100 watt.27 A P1 to P2 (D) 0. [AIEEE-2008] P2 5V 2V 10 2 1 P1 The current in the 10 resistor is - (A) 0.20 The resistance of wire is 5 ohm at 50∫C and 6 ohm at 100∫C. Q and R are connected in the three arms and the fourth arm is formed by two resistances S1 and S2 connected in parallel. 24 Statement-1 : The temperature dependence of resistance is usually given as R = R0(1 + t).05% (B) increase by 0. This implies that = 2.1% longer. Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is true.2% (C) decrease by 0. The respective temperature coefficients of their series and parallel combinations are nearly 1 2 1 2 1 2 (A) 1 + 2.5 ◊ 10ñ3/∫C. The resistance of a wire changes from 100 to 150 when its temperature is increased from 27∫C to 227∫C.23 V measured between B and C is . I V I a b a A B C D Q. Statement-2 is not a correct explanation for Statement-1 (C) Statement-1 is true.2% (D) decrease by 0. Statement-2 is false. [AIEEE-2010] 1 2 2 2 1 2 1 2 (C) . 2 2 Q.25 Two conductors have the same resistance at 0∞C but their temperature coefficients of resistance are 1 and 2. Statement-2 : R = R0(1 + t) is valid only when the change in the temperature T is small and R = (R ñ R0) << R0.22 For current entering at A. (B) . (D) Statement-1 is false. its resistance will : [AIEEE-2011] (A) increase by 0.26 If a wire is stretched to make it 0. [AIEEE-2008] I I I I I I I (A) ñ ( a b) (B) ñ 2 (a b ) (C) 2(a b) (D) ñ (a b) a 2a a Q. [AIEEE-2008] I I I I (A) (B) (C) (D) r 2 2r 2 4r 2 8r 2 Q. Statement-2 is true. the electric field at a distance ërí from A is .05% . 1 2 (D) 1 2 . Statement-2 is true. Statement-2 is true. Q. [AIEEE-2009] (A) Statement-1 is true.
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