9-ELECTRIC CURRENT-01-Theory

ELECTRIC CURRENT Electric current across an area held perpendicular to the direction of flow of charge is defined to be the amount of charge flowing across the area per unit time. If charge ΔQ passes through the area in time interval Δt , at uniform rate the current I is defined by I  ΔQ Δt ...(i) If rate of flow of charge is not steady then instantaneous current is given by I  lim Q  dQ ...(ii) t0 t dt The unit of current is ampere. Ampere (A) is the S.I. base unit, defined in terms of its magnetic effect. Smaller currents are more conveniently expressed in milliampere (1mA = 10-3 A) or microampere 1μA  106 A. . Illustration: Two boys A and B are sitting at two points in a field. Both boys are sitting near assemblence of charged balls each carrying charge +3e. A throws 100 balls per second towards B while B throws 50 balls per second towards A. Find the current at the mid point of A and B. Solution: Let mid point be C as shown 100e A C B 50e Charge moving to the right per unit time = 100 × 3e = 300e Charge moving to the left per unit time = 50 × 3e = 150e Movement of charge per unit time is 300e –150e = 150e towards right I = 150e = 150  1.6  10–19 A. Illustration: Flow of charge through a surface is given as Q  4t2  2t (for 0 to 10 sec.) (a) Find the current through the surface at t = 5sec. (b) Find the average current for (0 – 10 sec) Solution: (a) Instantaneous current I  dQ  d (4t2  2t)  8t  2 dt dt at t = 5 sec; I = 8  5  2  42 Amp (b) Average current Q Q 4  (10)2  2 10 420 I  t  t  10   42 Amp. 10 ELECTROMOTIVE FORCE(EMF) To maintain a steady electric current, the conductor cannot be isolated; it must be part of a closed circuit that includes an external agency or device (figure). This device is required to transport the positive charge from B back to A, i.e., from lower to higher potential and thus maintain the potential difference between A and B. The external device will need to do work for transporting positive charge from lower to higher potential. Such a device is the source known as electromotive force abbreviated as emf. It is the analogue of the pump in the water flow circuit. I The external source, as said above, does work on taking a positive charge from lower to the higher potential. A natural way of characterizing the external source of energy is in terms of the work that it needs to do per unit positive charge in transporting it from lower to higher potential. This is known as electromotive force or emf of the device, denoted by ε . ε  Vopen The emf of a source is thus the potential difference between its two terminals in open circuit i.e. when no external resistances are connected. RESISTANCES AND RESISTIVITY OHM’S LAW Consider a closed circuit having a source of emf and a conductor in the external circuit. Let the voltage drop across the ends of the conductor be V and let I be the steady current flowing through the conductor. The quantity V/I is a measure of the resistance offered by the conductor for the steady flow of charge through it. The more the resistance, the less is the current I for a given voltage difference, V. For many conductors, it is found experimentally that the ratio V/I is a constant at constant temperature, the constant is called the resistance of the conductor and is denoted by R. V  R or V  IR I ...(iii) The constancy of R implies that V and I are linearly related – a graph between measured values of V and I is a straight line. The unit of resistance is ohm 1Ω  1VA1 . Ohm’s law is only an empirical law that holds approximately for many substances over certain ranges of V and I. RESISTIVITY The resistance of a resistor (an element in a circuit with some resistance R) depends on its geometrical factors (length, cross-sectional area) as also on the nature of the substance of which the resistor is made. It is convenient to separate out the ‘size’ factors from the resistance R so that we can define a quantity that is characteristic of the material and is independent of the size or shape. Consider a rectangular slab of length l and area of cross section A. For a fixed current I, if the length of the slab is doubled, the potential drop across the slab also doubles. (It is the electric field that drives the current in the conductor and potential difference is electric field times the distance). This means that resistance of the slabs doubles with the doubling of its length. That is, R  l. Next, imagine the slab as being made of two parallel slabs, each of area A 2 . If for a given voltage V, the current I flows across the full slab, it is clear that through each half-slab, the current flowing is I 2 . Thus, the resistance of each half-slab is twice that of the full slab. That is, R  1 . A Combining the two dependences, we get R  l A ...(iv) or R  ρl A ...(v) where ρ is a constant of proportionality called resistivity. It depends only on the nature of the material of the resistor and its physical conditions such as temperature and pressure. The unit of resistivity is ohm m ( Ωm ). The inverse of ρ is called conductivity, and is denoted by σ . The unit of σ is Ωm1 or mho m-1 or siemen m-1. A perfect conductor would have zero resistivity and a perfect insulator would have infinite resistivity. Though these are ideal limits, the electrical resistivity of substances has a very wide range. Metals have low resistivity of 10-8 Ωm to 10-6 Ωm , while insulators like glass or rubber have resistivity, some 1018 times (or even more) greater. Generally, good electrical conductors like metals are also good conductors of heat, while insulators like ceramic or plastic materials are also poor thermal conductors. ORIGIN OF RESISTIVITY The motion of charge carriers (electrons) in a conductor is very different from that of charges in empty space. In the latter case, under an external electric field, the charge carriers would accelerate. In a conductor, on the other hand, when the current is steady, the charge carriers move with a certain average drift velocity. At any temperature, the electrons in a metal have a certain distribution of velocities. When there is no external field, all directions are equally likely, and there is no overall drift. In the presence of an external eE field, each electron experiences an acceleration of m opposite to the field direction. But this acceleration is momentary, since electrons are continually making random collisions with vibrating atoms or ions or other electrons of the metal. After a collision, each electron makes a fresh start, accelerates only to be deflected randomly again. We next define a physical quantity called current density vector, denoted by j. The direction of j is the direction of flow of positive charge (or opposite to the direction of drift of electrons in a metal). The magnitude of j is the amount of charge flowing per unit cross sectional area per second. Thus, if δS is an area element, j. δS is the amount of charge flowing across the area element per second. If A is taken to be the cross-sectional area of a wire (with the direction of A along the conventional current), j. A is nothing but the current through the wire. In this case, j is parallel to A, so → dI  j.ds → I   j.ds I = jA If the drift speed of electrons is υd' the amount of charge flowing across a unit cross-sectional area in unit time is contained in a cylinder of base of unit area and height υd i.e., in a volume 1 × υd = υd (Fig.). If n is the number density of electrons in the metal, i.e., the number of electrons per unit volume, the total magnitude of charge contained in the cylinder of volume υd is n e υd . Therefore, j  n e υd and I  n e d A Here, e is the magnitude of electronic charge. Then we can rewrite j  E  σE ρ The more familiar form V = IR is in terms of directly measurable quantities like V and I, while the form j = σ → relates the basic vector quantities of the problem, namely, current density vector and the electric field vector. Illustration: Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 × 10-7 m2 carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0 × 103 kg/m3, and its atomic mass is 63.5 amu Solution: The direction of drift velocity of conduction electrons is opposite to the electric field direc- tion, i.e., electrons drift in the direction of increasing potential. The drift speed υd is given by υd  I / neA . Now, e = 1.6 × 10-19 C, A = 1.0 × 10-7 m2, I = 1.5 A. The density of conduction electrons, n is equal to the number of atoms per cubic metre (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one). A cubic metre of copper has a mass of 9.0 × 103 kg. Since 6.0 × 1023 copper atoms have a mass of 63.5 g. 6.0 1023 n  63.5  9.0 106 = 8.5 × 1028 m-3 which gives υd  1.5 8.5 1028 1.6 1019 1.0 1.07 = 1.1 × 10-3 m s-1 . Illustration: A potential difference of 100 V is applied to the ends of a copper wire one metre long. Calculate the average drift velocity of the electrons. Compare it with thermal velocity at 270C. (use the results of Previous Illustration). given   5.81107 1m1 Solution: Since ΔV  100V,l  1m .  electric field  ΔV  100  100Vm1 l 1 Also, conductivity σ  5.81107 Ω1m1 N = 8.5 × 1028 m-3    5.81107 100  d en E  1.6 1019  8.5 1028 =0.43 m s-1 υrms  3k BT m  = 1.17 × 105 m s-1 TEMPERATURE DEPENDENCE OF RESISTIVITY The resistivity of all metallic conductors increases with temperature. Over a limited temperature range that is not too large, the resistivity of metallic conductor can often be represented approximately by a linear relation: where ρT  ρ0 1  αT  T0  ρ0 is the resistivity at a reference temperature (i) T0 and ρT its value at temperature T. The factor α is called the temperature coefficient of resistivity and has dimensions of per degree Celsius. RESISTORS IN SERIES AND IN PARALLEL Figure illustrates four different ways in which three resistors having resistances, R1, R2 and R3 might be connected between points x and y. Figure(a) shows the resistances in series that means one after another. The current is the same in each element. R2 x R1 I a R2 (a) R1 b R3 y I (c) R2 R3 (b) (d) The resistors in Figure(b) are said to be in parallel between the points x and y. Each resistor provides an alternative path between the points. The potential difference is the same across each element. In Figure(c) resistors, R2 and R3 are in parallel with each other and this combination is in series with resistance R1. In Figure(d), R2 and R3 are in series and this combination is in parallel with R1. It is always possible to find a single resistor that could replace a combination of resistors in any given circuit and leave unchanged the potential differences between the terminals of the combination and the current in the rest of the circuit. The resistance of this single resistor is called equivalent resistance of the combination. If any one of the networks were replaced by its equivalent resistance R, we could write Vxy  IR or R  Vxy I where Vxy is the potential difference between the terminals x and y of the network and I is the current at point x or y. Hence, the method of computing an equivalent resistance is to assume a potential difference Vxy across the actual network, compute the corresponding current I, and take the ratio Vxy /I. If the resistors are in series as in Figure(a), current in each one must be the same and equal to the line current I. Hence, Vxa  IR1 , Vab  IR 2andVby  IR3 Vxy  Vxa  Vab  Vby = I (R1 + R2 + R3) Vxy Vxy  R I 1  R 2  R 3 But I is, by definition, the equivalent resistance R. Therefore, R  R1  R 2  R 3 The equivalent resistance of any number of resistors in series equals the sum of their individual resistances. If the resistors are in parallel as in Figure (b), the potential difference between the terminals of each must be same and equal to Vxy . If the current in each are denoted by I1, I2 and I3 respectively, then I  Vxy , I R1  Vxy , I R 2  Vxy R 3 since charge is not accumulated at x, it follows that  1 1 1  I  I1  I2  I3  Vxy   or I  1  1  1  1 R 2   R 3  Vxy R1 R 2 R 3 But I Vxy  1 , so that 1  1  1  1 R R1 R 2 R 3 for any number of resistors in parallel, the reciprocal of the equivalent resistance equals the sum of the reciprocal of their individual resistances. For special case of two resistors in parallel, 1  1  1 R R1 R 2  R 2  R1 R1R 2 and R  R1R 2 R1  R 2 also, Since Vxy  I1R1  I2R 2 I1  R 2 I2 R1 and the current carried by two resistors in parallel are inversely proportional to their resistances. The equivalent resistance of the networks in Figure(c) and Figure(d) could be found by the same general method, but it is simpler to consider them as combination of series and parallel arrangements. Thus, in figure (c) the combination of R2 and R3 in parallel is first replaced by its equivalent resistance which then forms a simple series combinations with R1. In Figure(d), the combination of R2 and R3 in series forms a simple parallel combination with R1. Not all networks, however, can be reduced to simple series-parallel combinations and special methods must be used for handling such networks. Illustration: Find the effective resistance between the point A and B. Solution: Resistors AF and FE are in series with each other. There- fore, network AEF reduces to a parallel combination of E 3 D 3 6 F 6 6 C two resistors of 6 each. Req.  6  6  3 . 6  6 3 3 A 3 B Similarly, the resistance between A and D is given 6  6  3 . 6  6 Now, resistor AC is in parallel with AD. Therefore, the resistance between A and C is 6  3  2 . 9 Now AC and BC are in series with Req  5 and this R is in parallel with AB Req.  5  3  15  8 8 ELECTRIC CIRCUITSAND KIRCHHOFF’S RULES Most electrical circuits consist not merely a single source and a single external resistor, but comprise of a number of sources, resistors or other elements such as capacitors, motors, etc. interconnected in a complicated manner. The general term applied to such a circuit is called network. INTERNAL RESISTANCE Suppose for flow of unit charge in the circuit. potential difference across the load is V By energy conservation eq  vq  w' with q = 1 unit then ε  V  w ' where w is the part of the total work done by source of emf per unit charge ε used up in the source itself . In a cell, this happens because in moving the positive ions from lower to higher potential (or negative ions from the higher to lower potential), the other ions and atoms of the electrolyte offer resistance. Now, by Ohm’s law V = IR. Assuming that Ohm’s law is also valid for the flow of current in the source, we can assign an internal resistance r to the source and write W' = Ir, ε  V  Ir  IR  Ir  IR  r this can also be written as V  ε  Ir showing that the external voltage is less than the emf of the source by the quantity Ir. It is as if an internal resistance r combines in series with the external resistance R to determine the current in the circuit for a given source of emf. Clearly, If I = 0, i.e., the circuit is open, V = Vopen. KIRCHHOFF’S RULES Using the fact that there is no net current at the junction i.e., the incoming current equals the outgoing current and if we complete the circuit via a path the total potential change is zero. These facts, called Kirchhoff’s rules, are very useful for many electrical circuit problems; they are discussed in detail below : (a) Junction rule: This rule is based on the fact that charge cannot accumulate at any point in a conductor in a steady situation. It states that ‘at any junction of several circuit elements, the sum of currents entering the junction must equal the sum of currents leaving it’. Unless this happens, net positive or negative charge will accumulate at the junction at a rate equal to the net electrical current at the junction. Consider for example, the circuit in figure given below. At the junction ‘a’, I3 flows in and I1 and I2 flow out. We must then have I3  I1  I2 ...(i) Applying this to another junction of this circuit leads to nothing new equation. For further progress in analyzing this circuit, we need another rule, given below: (b) Loop rule : ‘The algebraic sum of changes in potential around any closed resistor loop must be zero’. Otherwise, one can continuously gain energy by circulating charge around a closed loop in a particular direction. So, this rule is based on energy conservation. Now consider the loop ‘ahdcba’ in Figure. we have, from Kirchhoff’s second rules, -30I1, -41I3 + 45 = 0 ...(ii) For the second loop, the circuit ‘ahdefga’ is taken. We have -30I1, +21I2 – 80 = 0 [iii] Calculate Using Eq. (i), (ii) & (iii) we I1 = 0.86A, I2 = 2.59A and I3 = 1.73A, respectively. 30 h I1 a d I2 20 80 V SIGN CONVENTION IN APPLYING KIRCHHOFF’S RULES First, all quantities, known as unknown, should be labelled carefully, including an assumed sense of directions for each unknown current or emf. The solution is then carried out using the assumed directions, and if the actual direction of a particular quantity is opposite to the assumed direction, the value of the quantity will emerge from the analysis with a negative sign. Hence, Kirchhoff’s rules, correctly used, give the direction and magnitude of unknown currents and emf’s. Usually in labelling currents it is advantageous to use the point rule immediately to express the currents in terms of as few quantities as possible. For example, Figure(a) shows a circuit correctly labelled, and Figure(b) shows the same circuit, relabelled by applying the point rule to point a to eliminate I3. The following guidelines will help with the problems of signs: 1. Choose any closed loop in the network, and designate a direction (clockwise or counter clockwise) to transverse the loop in applying the loop rule. 2. Go around the loop in the designated direction, adding emf’s and potential differences. An emf is counted as positive when it is traversed from (-) to (+) and negative when transformed from (+) to (-). An IR term is counted negative if the resistor is traversed in the same direction of the assumed current, and positive if in the opposite direction. 1,r1 2,r2 R1 R2 R1 a R Figure (a) Figure (b) 3. Equate the sum of step (2) to zero. 4. If necessary, choose another loop to obtain different relations between the unknowns, and continue until there are as many equations and unknowns or until every circuit element has been included in at least one of the chosen loops. Illustration: Find the potential differnce VA  VB shown in fig. and the rate of production of heat in R1 in the network 2 i3 G R3 2 i1 1 D 1 2 F H R4 1 C with 1 =12 V and 2 = 3 V A 2 i2 R2 2 B Solution: At the steady state, the capacitor branch CDGH acts as an open branch, i.e., of infinite resistance with zero current in it. If i1 ,i2 and i3 be the currents in the ED, AB and GF branches respectively, then from the Kirchhoff’s first law, we get at point D i1  i2  i3  0 or i3  i1  i2 . Assuming the sources of negligible resistance and applying Kirchhoff’s second law to the meshes ABCDEA and EDGFE, we get, i2 R 2  i1R1  2  1 and i1R1  i1  i2  R3  1  i2 R3  i1 R1  R3   1 solving these equations for i2 , and i1 , we get 122  2  3 2 and i1  i2  4  4  4 32  2 12  2 4  4  4  3.5 amp  1 amp  VA  VB  3  1 2  5 volts. Rate of production of heat in R1  i12 R1  3.52  2  24.5 joules. ENERGY, POWER AND HEATING EFFECT When a current I flows for time t from a source of emf E, then the amount of charge that flows in time t is Q = I t. Electrical energy delivered W = Q. V = V I t Thus, Power given to the circuit, = W/t =VI or V2/R or I2R In the circuit E. I = I2R + I2r, where E I is the rate at which chemical energy is converted to electrical energy, I2R is power supplied to the external resistance R and I2r is the power dissipated in the internal resistance of the battery. An electrical current flowing through conductor produces heat in it. This is known as Joule’s effect. The heat developed in Joules is given by H = I2.R.t I R Illustration : An electric bulb rated 220V and 60 W is connected in series with another electric bulb rated 220V and 40 W. The combination is connected across 220 volt source of emf which bulb will glow more ? Solution : V2 P V2  Resistance of first bulb is R1  1 V2 And resistance of the second bulb is R 2  2 In series same current will pass through each bulb 2 V2  Power developed across first is 2 V 2 P'1  I 1 P' P and that across second is P'2  I  1  2 P2 P'2 P1 as P2  P1  P2  1 P1  P'1  1  P'2 P'1  P'2 The bulb rated 220 V & 40 W will glow more. MAXIMUM POWER TRANSFER THEOREM In a circuit, for what value of the external resistance the maximum power drawn from a battery? For the shown network power developed in resistance R equals E2 .R E 2 P  R  r  ( I  R  r and P  I R) I Now, for dP/dR = 0 (for P to be maximum dP  0 ) dR  E2. R  r2  2RR  r R  r4  0  R  r  2R or R = r The power output is maximum, when the external resistance equals the internal resistance. R = r Illustration : A copper wire having a cross-sectional area of 0.5 mm2 and a length of 0.1 m is initially at 250C and is thermally insulated from the surroundings. If a current of 10 A is set up in this wire (a) Find the time in which the wire starts melting. The change of resistance of the wire with temperature may be neglected. (b) What will this time be, if the length of the wire is doubled ? Density of Cu = 9 × 103 Kg m-3 specific heat of Cu = 9 × 10-2 Cal Kg-1 0C-1 , M.P. (Cu) 10750C and specific resistance = 1.6 × 10-8 Ωm . Solution : (a) Mass of Cu = Volume × density = 0.5 × 10-6 × 0.1 × 9 × 103 = 45 × 10-5 Kg. Rise in temperature = θ = 1075-25 = 1050 0C. Specific heat = 9 × 10-2 Kg-1 0C × 4.2 J  I2Rt  mSθ  t  mSθ I2.R but R  ρL A  1.6 108  0.1 0.5 106  3.2 10 3.Ω  t  45  4.2 105 1050  0.09 10 10  3.2 103  558s (b) When the length of wire is doubled, R is doubled, but correspondingly mass is also doubled. Therefore, wire will start melting in the same time. BATTERY COMBINATION : (i) ies Combination :- eq = n req = nr I = n nr  R ε L eq n (a) RL >>> nr  I = R . This combination is useful as current across load resistance can be varied by varying nunmber of cells in the circuit.  (b) RL <<< nr  I = r . This combination is not prefered (c) In general, equivalent emf of the battery is worked out using Thevenin theorem (i) VAB = 1 + 2 RAB = r1 + r2  RL (ii)  VAB = |1 – 2| RAB = r1 + r2 (ii) PARALLEL COMBINATION :-  eq =  Req = r n ,  I =  r I = R n n r  nR εeq Illustration: Find equivalent εeq of the battery as shown in figure and VA  VB ? (i) (ii) Solution: (i) Current in the circuit i  1  2 r1  r2 ...(i) eq.  1  2 and VA – 2 – r2 1  2 r  r = VB  VA – VB = 1 2 1r2  2r1 r  r 1 2 (ii) ent i  1  2 r1  r2 ...(i)  1  2  and V –  + r  r  r  = V A 1 1  1 2  B VA – VB = 1r2  2r1 r  r . 1 2 (iii) MIXED COMBINATION  nr req = m , I = n R  nr m mn  I = nr  mR EFFECTIVE GROUPING OF CELLS For effective grouping current should be maximum  mR + nr should minimum Now, mR + nr =  mR 2   nr 2   mR  nr 2  2 The second term is nonzero the term in the parenthesis   nr 2  0  mR = nr R  n r m Illustration: Length of AB = 6 m Find the position of Null pt ? Given RAB  15r Soln. V  iR    15r .x  ...(i) AJ AJ 16r  6    also for J to be null point   VAJ  2 ...(ii) equating (i) and (ii) x  320cm Illustration : There are 27 cells with an internal resistance 0.4 Ω and an external resistance 2.4 Ω . What is the most effective way of grouping them ? Solution : Let there be n cells series of m parallel branches mn = 27 ...(i) mR = nr ...(ii) 1.2m = n (0.4)  3m = n ...(iii) from equation (i) and (iii), we get m = 3 & n = 9 SUPERPOSITION PRINCIPLE Concepts : Whenever a circuit has more than one cell or battery, the superposition principle may be used to find current and voltages. This principle is based on the fact that every cell or battery acts independent of the presence of others. According to this principle, the total current I in the circuit equals the algebraic sum of current I1, I2, In produced by each source (cell or battery) taken one at a time. Mathematically, I= I1 + I2 + + In The superposition splits the original two - source problem into two one - source problems. Instead of solving a difficult two source problem, we can solve to simple problems. Determine the current I in the 2Ω resistor 6 I = I1 + I2 = 2 4 + 2 = 3 + 2 = 5 A. AMMETER It is an instrument used to measure currents. It is put in series with the branch in which current is to be measured. An ideal Ammeter has zero resistance. A galvanometer with resistance G and current rating ig can be converted into an ammeter of rating I by connected a suitable resistance S in parallel to it. (The resistance connected in parallel to the ammeter is called a shunt.) Thus S(I – ig) = igG  S  ig G I  ig Illustration : A galvanometer having a coil resistance of 100 Ω gives a full scale deflection when a current of 1 mA is passed through it. What is the value of the resistance which can convert this galvanometer into ammeter giving full scale deflection for a current of 10A ? Solution : S  ig .G i  ig 103 A100Ω  10 103 A  0.1 9.99  S  1 99.99 Ω  102 Ω VOLTMETER It is an instrument to find the potential difference across two points in a circuit. It is essential that the resistance Rv of a voltmeter be very large compared to the resistance of any circuit element with which the voltmeter is connected. Otherwise, the voltmeter itself becomes an important circuit element and alters the potential difference that is measured. Rv >> R For an ideal voltmeter Rv =  . ig (G  R)  V  R  V  G i g Illustration : A galvanometer having a coil resistance 100 Ω gives a full scale deflection when a current of 1 mA is passed through it. What is the value of the resistance which can convert this galvanometer into a meter giving full scale deflection for a potential difference of 10V ? Solution : V = lg [G+Rv] 10=(10-3) (100 + R ) R   10  100  9,900Ω  9.9KΩ  v  103    WHEATSTONE BRIDGE Figure shows the fundamental diagram of wheatstone bridge. The bridge has four resistive arms, together with a source of emf (a battery) and a galvanometer. The current through the galvanometer depends on the potential difference between the point c and d. The bridge is said to be balanced when the potential difference across the galvanometer is 0V so that there is no current through the galvanometer. This condition occurs when the potential difference from point c to point a, equals the potential difference from point d to point a; or by referring to the other battery terminal, when the voltage from other point c to point b equals the voltage from point d to point b. Hence, the bridge is balanced when I1R1  I2 R 2 …(i) if the galvanometer current is zero, the following conditions also exist: I1  I3  ε R1  R 3 …(ii) and I2  I4  ε R 2  R 4 …(iii) a  Combining Eqs. (i), (ii) and (iii) and simplifying, we obtain R1 R1  R 3  R 2 R 2  R 4 …(iv) from which we get R1R 4  R 2R 3 …(v) Equation (v) is the well known expression for balance of the wheatstone bridge. If three of the resistances have known values, the fourth may be determined from Equation (v). Hence, if R4 is the unknown resistor, its resistance can be expressed in terms of remaining resistors R 4  R 3 2 1 …(vi) Resistance R3 is called the standard arm of the bridge and resistors R2 and R1 are called the ratio arms. Illustration: The four arms of a wheatstone bridge (figure) have the following resistances : AB = 100 Ω , BC = 10 Ω , CD = 5 Ω and DA = 60 Ω . The galvanometer of 15 Ω resistance is connected across BD. Calculate the current through the galvanometer when at potential difference of 10 V is maintained across AC. Solution: Considering the mesh BADB, we have 100 I1 + 15 Ig – 60 I2 = 0 or 20 I1 + 3 Ig – 12 I2 = 0 …(i) Considering the mesh BCDB, we have – 10 (I1 – Ig) + 15 Ig + 5 (I2 + Ig) = 0  10 I1 – 30 Ig – 5 I2 = 0 2 I1 – 6 Ig – I2 = 0 …(ii) Considering the mesh ADCEA, – 60 I2 – 5 (I2 + Ig) + 10 = 0  65 I2 + 5 Ig = 10 13 I2 + Ig = 2 …(iii) A I Multiplying Equation (ii) by 10 20 I1 – 60 Ig – 10 I2 = 0 …(iv) From (iv) and (i) we have 63 Ig – 2 I2 = 0 I  63 I 2 2 g  31.5Ig …(v) Substituting the value of I2 into Equation (iii), we get 13 (31.5 Ig) + Ig = 2 410.5 Ig = 2 Ig = 4.87 mA. METRE BRIDGE – Special case of Wheatstone Bridge This is the simplest form of wheatstone bridge and is specially useful for comparing resistances more accurately. The construction of the metre bridge is shown in the Figure. It consists of one metre resistance wire clamped between two metallic strips bent at right angles and it has two points for connection. There are two gaps; in one of them a known resistance and in second an unknown resistance whose value is to be determined is connected. The galvanometer is connected with the help of jockey across BD and the cell is connected across AC. After making connections, the jockey is moved along the wire and the null point is from two resistances of the wheatstone bridge, wire used is of uniform material and cross-section. The resistance can be found with the help of the following relation: R  ℓ1 S 100  ℓ1  R  ℓ1 S 100  ℓ1 R  S l1 100  l1 . . . (1) where σ is the resistance per unit length of the wire and l1 is the length of the wire from one end where null point is obtained. The bridge is most sensitive when null point is somewhere near the middle point of the wire. This is due to end resistances. END CORRECTION : Sometimes at the end points of the wire, some length is found under the metallic strips and as a result, in addition to length l1 or (100 –l1) some additional length should be added for accurate measurements. The resistance due to this additional length is called end resistance. If the end resistance is small it can be determined by first introducing known resistances P and Q in the gap and obtaining the null point reading l1, then interchanging P and Q and obtaining the null point reading l2. Let  and  be the lengths on the respective end under the metallic strips, then we have P  Q Q  l1  α 100  l1  β l2   ...(i) …(ii) P 100  l2   Solving Equations. (i) and (ii) for α and β , we have α  Ql1  Pl2 P  Q …(iii) β  Pl1  Ql2 100 P  Q …(iv) hence the value of are required. α,β can be calculated and suitably accounted for when accurate measurements Illustration: Figure (a) shows a metre bridge (which is nothing but a practical Wheatstone bridge) con- sisting of two resistors X and Y together in parallel with a metre long constantan wire of uniform cross section. With the help of a movable contact D, one can change the ratio of the resistances of the two segments of the wire, until a sensitive galvanometer G connected across B and D shows no deflection. The null point is found to be at a distance of 33.7 cm from the end A. the resistance Y is shunted by a resistance Y of 12.0 Ω [Figure (b)] and the null point is found to shift by a distance of 18.2 cm. Determine the resistance of X and Y. (a) (b) Solution: Since the wire is of uniform cross-section, the resistances of the two segments of the wire AD and DC are in the ratio of the lengths of AD and DC. Using the null-point conditions of a wheatstone bridge, we have (X/Y) = (33.7/66.3) (1) When Y is shunted by a resistance of 12.0 Ω , net resistance changes Y’ = 12 Y (Y + 12) Since Y’ is less than Y, the ratio X/Y’ is greater than X/Y’. Thus the null point must shift towards the end C, i.e., (X/Y’) = (51.9/48.1) or X (Y + 12) / 12 Y = (51.9 / 48.1) Y  12   51.9   66.3 i.e.   12 48.1 33.7   which give Y = 13.5 Ω and X = 6.86 Ω , using (1) POTENTIOMETER This instrument is identical to the meter bridge but more sensitive than meter bridge because resistance wire is of more than meter in length. A B It obeys the rule R  ℓ (for constant cross section area of wire and tem- perature) Let wire AB has total resistance R0  Current I in the circuit = R0  R p.d. across AB V   R  0 R  R 0 p.d. per unit length    R0  1    R  AB  It is used in several process like comparing emfs of two batteries and to calculate resistnace of unknown resistor. Let 1 is connected by connecting 1  3 and a null point is observed at ‘C’ on AB A B   R0  AC  then 1 R  R  AB  ...(i) and similarly C is null point for 2 in circuit    R0  AC then    0 R  AB ...(ii) A B Using (i) and (ii) 1  AC 2 AC Illustration: A resistance of R Ω is powered from a poten- tiometer of resistance R 0Ω (figure). A volt- age V is supplied to the potentiometer. Derive A C an expression for the voltage fed into the branch BC when the slide is in the middle of the potentiometer. Solution: While the slide is in the middle of the potentiometer only half of its resistance R 0 / 2 will be between the points A and B. Hence, the total resistance between A and B, say, R1, will be given by the following expression : 1  1   1  R1 R R 0 / 2 R1  R 0R R 0  2R The total resistance between A and C will be sum of resistance between A and B and B and C, i.e., R1  R 0 / 2  The current flowing through the potentiometer will be I  V  R1  R 0 / 2 2V 2R1  R 0 The voltage V1 taken from the potentiometer will be the product of current I and resistance R1, IR0  2V  R0 V1  2    2R1   R 0  Substituting for R1, we have V1  2V  R  R   R 0 2 2 0   R 0 Brain Teaser: Would you prefer a voltmeter or a potentiometer to measure the emf of a battery? RC-CIRCUIT Charging: Let us assume that the capacitor in the shown network is uncharged for t < 0. The switch is connected to position 1 at t = 0. Now, ‘C’ is getting charged. If the charge on capacitor at time ‘t’ is q. Writing the loop rule, q  1 S R IR  E  0 C E C  R dq  E  q dt c  RC dq  EC  q I dt  dq  1 dt EC  q RC q dq  t Integrating 0 EC  q RC dt - ln | EC - q|q  1 .t RC  ln EC  q EC   t RC qmax q = EC  q  EC[1 et / RC ] q = 0.63 EC  At t = 0, q = 0 and at t =  , q = E C (the maximum charge.) thus, q  q max 1  et / RC   i  dq  qmax et / RC  E et / RC dt RC R i  i max et / RC where imax  R imax i = E/R t TIME CONSTANT τ It is the time during which the charging would have been completed, had the growth rate been as it began initially. Numerically it is equal to RC. DISCHARGING Consider the same arrangement as we had in previous case with one difference that the capacitor has charge q0 for t<0 and switch is connected to position 2 at t = 0. If the charge on capacitor is q at any later moment t then the loop equation given as Flip the switch to 2 q  IR  0 1 C  R dq  q dt C  C dq  q 1 dt RC R 1 Integrating, at t = 0, q = q0 t  t, q  q  q dq   1 t q0 q  ln RC 0   t or RC q  q0 .et / RC i  q0 et / RC RC i   EC et / RC RC i  i0 e t / RC ‘-ve’ sign indicates that the discharging current flows in a direction opposite to the charging current. q i O -imax =/R t * Just as the flow of charge starts to the plates of capacitor with no initial charge, potential difference across the capacitor is zero. But at that moment current is not zero. * When the capacitor is fully charged that state is known as steady state. Flow of charge i.e. Current through capacitor is zero in steady state but potential difference across capacitor is non zero. Illustration: Corresponding to the shown circuit find the current through battery (a) at t = 0 just after closing the switch. (b) After a long time find the current through the battery and charge at capacitor. R Solution:(a) Just after closing the switch potential difference across capacitor is zero. Let current be as shown. Applying kirchhoff’s loop law(KLL) to circuit ABEFA C I2 R R B I1  D C I1 2 E S F A   RI1  0 I1+ I2 …(i) Applying KLL to circuit ACDFA   RI2  0 …(ii) Current through battery I  I      2 1 2 R R R (b) After long time, capacitor is in steady state fully charged, cur- C I2 R rent through it is zero only current is through CD. I1  0 R Applying KLL to ACDFA B 1  D +q -q E S F ε  RI2  0 A I1+ I2   I2  R Current through battery I  I  0     1 2 R R Let charge at capacitor be q Applying KLL to ABEFA   R  I1  q  0 C  q  C .