Intermediate Physics "current Electricity" By Abubakkar Marwat

F.Sc-PHY (Part-II)

13-CURRENT ELECTRICITY

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Current electricity: The study of charges in motion is called current electricity. 13.1 ELECTRIC CURRENT: The flow of charges in a conductor per unit time is called electric current, denoted by I. if ∆Q = charge passing through any cross-section of conductor ∆t = time the charge flows, then

I = ∆Q/∆t ------- {13.1} Thus, current can also be defined as “rate of flow of charges” Unit: the S.I unit of current is “ampere” denoted by A. from equ. 13.1 We have, 1A = 1C/1s Ampere: “one ampere is the current when charge is flowing at the rate of one coulomb per second through the conductor”. In metallic conductors, charge carriers are protons (+ve) and electrons (-ve), in electrolytes, the charge carriers are positive and negative ions & in gases, carriers are electrons and ions. 13.1.1 Current Direction: Early scientists suggested the current as flow of protons from positive to negative terminal of a battery. But later on it was found that “current is due to the flow of electrons from negative to the positive terminal of battery”. It has been found experimentally that positive charge in one direction is equivalent to a negative charge flowing in opposite direction. Conventional current a) Electronic current: current due to the flow of electrons from negative to positive terminal of the battery is called electronic current. b) Conventional current: current flowing in a conductor from positive to Electronic negative terminal of the battery is called conventional current. current 13.1.2 Current through a metallic conductor: The valence electrons in a metal are in constant random motion. These valence electrons are called free electrons. In any section of the wire, the rate of flow of electrons from left to right is equal to flow from right to left, i.e., net flow of electrons is zero and hence “no current”. Now if ends of this wire are connected to a battery, an electric field E (from +ve to –ve terminal) will be setup at every point of the wire. Free electrons will then move towards positive terminal i.e., against the electric field E. (Fig 13.1)

Fig 13.1

13.1.3 Drift velocity: While flowing at high velocities (several hundreds km/s) through the wire, the free electrons collide with the atoms and molecules of the wire and transfer energy to them. Thus their velocity is decreased and acquires a uniform velocity called drift velocity of free electrons in opposite direction of electric field (-E). Drift velocity is of the order of 10-3 m/s. EXAMPLE 13.1: 1x107 electrons pass through a conductor in 1 µs. Find current in ampere. Solution: No of electrons = 1.0x107 Charge on electron = 1.6x10-19 C Charge on 1.0x107 electrons ∆Q = 1.6x10-19 x 1.0x107 Time t = 1.0 µs = 1x10-6 s Current through the conductor I = ∆Q/∆t = 1.6x10-19 x 1.0x107/1x10-6 = 1.6 x 10-6 A 13.2 Source of current: When two conductors at different potentials are joined together by a metallic wire, current will flow from higher potential to lower potential until both are at same potential, and current will Abu Bakkar Marwat--B.Sc Textile Engineering (Spinning} NTU, Faisalabad. [email protected]

F.Sc-PHY (Part-II)

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decrease from maximum to zero. So in order to maintain a potential difference at ends of conductor and to flow current constantly, source of current (battery or cell) is used. The source of current converts its chemical energy (cell), mechanical energy (electric generators), heat energy (thermo-couples) or solar energy (solar cells) into electrical energy. 13.3 DIFFERENT EFFECTS OF CURRENT: 13.3.1 Heating effect of current: When current flows through a metallic wire, the free electrons collide with the atoms of the wire and transfer their energy to them. Thus vibrational energies of the atoms are increased which appears as heat. It is found that the heat H produce by a current I in the wire of resistance R during a time interval t is given as: H = I2Rt Use: this heating effect of current is used in electric heater, kettle, toaster and electric iron etc. 13.3.2 Magnetic effect of current: Whenever current flows through a conductor, a magnetic field is developed around the conductor. The strength of magnetic field depends on value of current and distance from the conductor. The pattern of field produced by a current carrying straight wire, a coil and a solenoid is shown in fig 13.3a (a,b,c). Use: the magnetic effect of current is used in electric motors.

Fig 13.3a

(a)

(b)

(c)

13.3.3 Chemical effect of current: When current is passed through certain liquids, chemical changes occur in the liquid and it decomposes into positive and negative ions; process is called electrolysis. “electrolysis is the process of decomposition of liquids into ions by passing current”. The chemical effect produced, depends on: i) Nature of the liquid ii) quantity of current passed through the liquid. 13.3.4 Electrolysis of copper sulphate solution: Technical terms: 1. Electrolyte: the liquids conducting current are known as electrolytes. 2. Electrode: the metallic wire or plate through which current is passed into the electrolyte. 3. Anode: the electrode by which current enters the solution and is connected to the positive terminal of the battery is called anode. Power supply 4. Cathode: the electrode by which the current leaves the solution and connected with negative terminal of the battery is called cathode. 5. Voltmeter: a device used to determine potential difference and contains two electrodes and liquid (CuSO4). Fig 13.3b

Cathode (Cu) Electrolysis of CuSO4: Voltmeter contains dilute solution of CuSO4. Anode and cathode both Anode (Cu) are copper plates (Fig 13.3b). When CuSO4 is dissolved in water, it splits into CuSO4 solution Cu+2 and SO4 -2 ions. When current is passed, Cu+2 moves towards cathode and the following reaction occur: Cu+2 + 2e-
Cu {copper atoms thus formed are deposited at cathode} -2 The SO4 ions move towards the anode. Copper ions from anode combine with SO4-2 ions to form Cu+2 + SO4-2
CuSO4 copper sulphate.

Abu Bakkar Marwat--B.Sc Textile Engineering (Spinning} NTU, Faisalabad. [email protected]

F.Sc-PHY (Part-II)

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The copper deposited on cathode and an equal amount is dissolved from anode into the solution. Thus the density of copper sulphate solution is not changed. 13.4 State and Explain Ohm’s Law. Give its graphical representation. Ohm’s Law: “current flowing through a conductor is directly proportional to the potential difference across its ends provided that the physical state (temperature etc.) of the conductor remains constant”. Therefore; I ∞ V or V = I R ------- {13.2} Where “R” is constant of proportionality and is called resistance of the conductor. Resistance: it is the opposition to the flow of electrons. From equ. 13.2, we have: R = V/I “resistance of a conductor is the ratio of potential difference across its ends to the current flowing through the conductor”. The resistance of a conductor depends upon: 1. Nature of the conductor (copper, iron etc.) 2. Dimensions (length, thickness etc.) of the conductor 3. Physical state (temperature, density, hardness etc.) of the conductor The S.I unit of resistance is Ohm, denoted by Ω. From equ. 13.3, 1 Ω = 1V/1A “resistance of a conductor is one ohm when potential difference of 1 V produces a current of 1 A through the conductor” Graphical Representation: A conductor is said to obey ohm’s law if its resistance R remains constant, i.e., the graph of V vs. I is exactly a straight line (a). The conductors obeying ohm’s law are called Ohmic conductors. The b c devices which do not obey Ohm’s law are called non Fig 13.4 a Ohmic (e.g., filament bulb, semiconductor diode). If resistance increases with temperature (filament I I I lamp), the graph is not a straight line (13.4b). V For semi-conductor diode, current-voltage plot is not V V a straight line (13.4c), showing that the semiconductor is also non Ohmic device. 13.4.1 Combination of Resistors: a) Series combination: If different resistances are joined (end to end) with each other such that there is only one path for the flow of electric current, then the combination of such resistances is called Series Combination. 1. Current through each resistor is same. 2. Potential difference across each resistor is different depending upon the value of resistance. 3. Equivalent resistance of circuit is equal to the sum of individual resistances, i.e., equivalent resistance is more than either of the resistances in the circuit. Re = R1 + R2 + R3 +…. +Rn b) Parallel combination: If different resistances are joined (to a common point) such that there are more than one path for the flow of current in a circuit then the combination of resistances is called Parallel Combination. 1. Current through each resistor is different. 2. Potential difference across each resistor is same. 3. Equivalent resistance of circuit is always less than either of the resistances in the circuit. 1/Re = 1/R1+1/R2+1/R3 + ….. + 1/Rn Abu Bakkar Marwat--B.Sc Textile Engineering (Spinning} NTU, Faisalabad. [email protected]

F.Sc-PHY (Part-II)

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13.5 Resistivity: The resistance of a conductor of unit length and unit area of cross section is called resistivity or specific resistance, denoted by ρ. It is found that the resistance R of a wire is directly proportional to its length L and inversely proportional to its area of cross section A. therefore, R ∞ L and R ∞ 1/A => R ∞ L/A R = ρ L/A ----- {13.3} If L = 1m, and A = 1m2 then, ρ = R “resistivity is the resistance of unit length & unit cross-sectional area of a material at certain temperature”

S.I unit of resistivity is ohm-m or Ωm. Resistivity depends on nature of the material (not on its dimensions) whereas, resistance depends on both nature and dimensions of the material. 13.5.1 Conductance: Conductance is the reciprocal of resistance. => Conductance = 1/R Unit of conductance is “mho” or “siemen”. Similarly, conductivity is the reciprocal of resistivity ρ => σ = 1/ ρ S.I unit of conductivity is “mhom-1”  Good conductor {higher value of conductivity}, insulators {higher value of resistivity} 13.5.2 Effect of Temperature on Resistance “R” and Resistivity “ρ”: Resistance of a conductor increases with an increase in temperature. As the flowing electrons collide with the atoms of the conductor and transfer their energy to atoms due to which they start vibrating. At higher temperatures, amplitudes of these vibrations increase and hence more resistance to the flow of electrons. If Rt = resistance of conductor at temperature ∆t, and Ro = resistance of conductor at 0°C, then (Rt – Ro) ∞ Ro and (Rt – Ro) ∞ ∆t (Rt – Ro) ∞ Ro∆t (Rt – Ro) = α Ro ∆t => Rt = Ro + α Ro ∆t (Rt – Ro) = Ro (1+α ∆t) {α = temperature coefficient} From the above equation: α = Rt – Ro/ Ro ∆t ----- {13.4} the above equation can also be written for resistivity: as Rt = ρtL/A and Ro = ρoL/A α = (ρt – ρo)/ ρo ∆t ----- {13.5} Temperature coefficient: “the increase in resistance per original resistance per degree rise of temperature is called temperature coefficient”. The unit of temperature coefficient α is (°C)-1 or k-1. The substances (e.g., germanium and silicon) whose resistance decreases with temperature have negative temperature coefficient. Iron and platinum have same resistivity but have different α. Example 13.2: 0.75A current flows through an iron wire when a battery of 1.5V is connected across its ends. The length of the wire is 5.0 m and its cross sectional area is 2.5 x 10-7 m2. Resistivity? Solution: Current I = 0.75 A, potential difference V = 1.5 V, length of wire L = 5.0 m, area of cross-section A=2.5x10-7 m2

Now

R = V/I = 1.5/0.75 = 2.0 Ω ρ = RA/L = 2x2.5x10-7/5 = 1.0 x 10-7 Ω m (Ans)

Example 13.3: a platinum wire has resistance 10 Ω at 0°C and 20 Ω at 273°C. find α ? Solution: Ro = 10 Ω, Rt = 20 Ω, to = 0°C (273 k), t = 273°C (546 k), ∆t = 546-273 = 273 k

So,

α = Rt – Ro/ Ro ∆t = (20 Ω-10 Ω)/(10 Ωx273 k) = 3.66 x 10-3 k-1

(Ans)

Abu Bakkar Marwat--B.Sc Textile Engineering (Spinning} NTU, Faisalabad. [email protected]

F.Sc-PHY (Part-II)

13-CURRENT ELECTRICITY

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13.6 COLOR CODE FOR CARBON RESISTORS: Carbon resistors consist of a high-grade ceramic rod on which, a thin resistive film of carbon is deposited. Due to their smaller size, they are color coded to mark their resistance in ohm. Dark colors correspond to low numbers as compared to light ones (darkest “black” has 0 value, while “lightest” white has value 9). Color code consists of 4-bands of different colors and are read as follows: 4th band indicates the tolerance in %age

Second band gives the second digit

3rd band indicates no. of zeros after the 1st two digits First band indicates the first digit in numerical value

Color Black Brown Red Orange Yellow Green Blue Violet Gray White

Value 0 1 2 3 4 5 6 7 8 9

Tolerance Golden = ±5 % Silver = ±10 % No band = ±20 %

Tolerance is the possible variation from the marked value. e.g., a 1000 Ω resistor with a tolerance of ±10% will have an actual resistance b/w 900 Ω and 1100 Ω.

13.6.1 RHEOSTAT: It is a variable resistor consisting of a manganin (alloy of copper, manganese & nickel) wire wound over an insulating cylinder. Ends of the wire are connected to two contacts which can be moved over the wire (Fig 13.6a). It can be used as variable resistor and a potential divider. A V

C

C B

VAC A

13.6c) Rheostat as a potential divider

B 13.6b) rheostat as variable resistor

13.6a) A Rheostat

a) Rheostat as variable resistor: While using rheostat as variable resistor, one of the fixed terminals A or B and sliding terminal C are used in the circuit (Fig 13.6b). Sliding away from terminal (say A), will increase the length and hence the resistance of the wire in the circuit increases. Sliding C towards A will decrease the resistance. b) Rheostat as Potential divider: While using rheostat as potential divider, terminals A & B are connected to a battery of V volts (Fig 13.6c). If R is the resistance of the wire AB, the current I passing through is given as: I = V/R The potential difference between the portion BC of the wire is given by: VBC = current x resistance = V/R x r where r is the resistance of portion BC of the wire. If C is at B then VBC = 0 and if C is at A, then VBC = V, i.e., output potential varies from 0-V. 13.6.2 THERMISTOR: A semi conducting, heat sensitive resistor is called thermistor which consists of a mixture of oxides of manganese, nickel, cobalt, copper, iron etc. The resistance of a thermistor decreases with increasing temperature (i.e. temperature coefficient of thermistor is negative). Thermistors with positive temperature coefficient are also available. Thermistors with high negative temperature coefficient (very low resistance) are very accurate for measuring low temperature, especially near 10 K. They are used as fire alarms and temperature sensors in which, changes of voltage are converted into electrical voltage. Abu Bakkar Marwat--B.Sc Textile Engineering (Spinning} NTU, Faisalabad. [email protected]

F.Sc-PHY (Part-II)

13-CURRENT ELECTRICITY

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13.7 ELECTRICAL POWER & POWER DISSIPATION IN RESISTORS: “the electrical energy supplied by the battery in unit time is called electrical power of battery”. I -

In the circuit (Fig 13.7), I is the current supplied by the battery. Terminal A is at higher potential than B, and potential difference b/w them is V. therefore, A battery is continuously ding work in moving charge Q from low potential to high V C potential, through the potential difference V. B Work done = ∆W = V x ∆Q This is the energy supplied by the battery. The rate of supplying energy is called 13.7a) electrical power. Electrical power = Energy supplied / time taken = V ∆Q/ ∆t = V I {∆Q/ ∆t = I} Electrical power P = V I ----- {a} {this power is dissipated in resistor R} Also P = IR x I P = I2 R ----- {b} Also; P = (V2/R2) R = V2/R ------ {c} Unit of power: The S.I unit of power is watt (W). from equ {a} 1 W = 1 V x 1 A “when a current of 1 A passes through a potential difference of one volt then the power dissipated is one watt”. watt is also equal to J/s 13.8 Electromotive force and Potential difference: (electro
electric means charges, motive
causes motion, force) i.e., it is a force that cause the charges to flow through the circuit. “electromotive force is the work done on unit charge which forces the charge to move throughout the circuit”. A source of electrical energy (cell or battery) is connected across a resistor, maintains a steady current through it. The charge ∆Q enters the cell from low potential (negative terminal) and leaves it at high potential (positive terminal) and for this, a constant energy ∆W is required. Therefore, electromotive force of the battery is: E = ∆W/ ∆Q ----- {a} Now, let “r” be the internal resistance of the cell and “R” be the external resistance connected across the cell. The current flowing through the circuit is given as: I = E /R+r E = I (R + r) ----- {b} E = IR + Ir E = V + Ir ----- {c} (close circuit) In case of open circuit, I = 0, and E = V (terminal voltage) Terminal voltage: • The potential difference across external resistance when circuit is closed is called terminal potential difference. V = E – Ir • The potential difference at the terminals of the source when circuit is open, is called terminal potential different. E = V (Ir = 0) Difference between EMF & Potential Difference: • emf is the “cause” while potential difference is the “effect” • emf is always present even when no current is drawn through the battery, but potential difference across the conductor when there is no current, is zero. Example 13.4: potential difference b/w the terminals of a battery in open circuit is 2.2 V. When it is connected across a resistance of 5 Ω, potential drops to 1.8 V. calculate current & internal resistance of the battery. Abu Bakkar Marwat--B.Sc Textile Engineering (Spinning} NTU, Faisalabad. [email protected]

F.Sc-PHY (Part-II)

Solution: emf, E = 2.2 V,

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resistance, R = 5 Ω, potential difference, V = 1.8 V current I = V/R = 1.8 / 5 = 0.36 A. E = V + Ir => r = (E – V)/I = (2.2 – 1.8)/0.36 = 1.11 Ω

13.8.1 Maximum Power: The loss of potential energy per second, during the flow of charges from higher potential to lower potential, appears as power delivered to the resistor. Pout = V I = I2R ----- {a} As I = E/(R+r), put in equ. {a} E2R E2R E2 R E2R Pout = = = = ( R + r ) 2 R 2 + r 2 + 2 Rr R 2 + r 2 + 2 Rr + 2 Rr − 2 Rr R 2 + r 2 − 2 Rr + 4 Rr E2R Pout = ( R − r )2 + 4 Rr When R = r, i.e. if internal resistance of the battery equals the load resistance, maximum power is delivered to the load (resistance). Value of this maximum output power is: Pout = E2/4R 13.9 KIRCHHOFF’S RULE: Kirchhoff’s rules are applied for the analysis of the circuits which have components that are not in series, in parallel or in series-parallel. a) Kirchhoff’s 1st rule (KCL-Kirchhoff’s Current Law): it states that: “sum of all the currents meeting at a point in the circuit is zero” i.e., Sum of all the currents flowing towards a point = Sum of all the currents flowing away from point Or ∑I=0 Current flowing towards a point is taken as positive and flowing away from a point is taken as negative. In Fig 13.9a, four wires meet at a point A and currents flowing into A are I1 and I2 and currents flowing away from the pit are I3 and I4. So I1 + I2 = I3 + I4 ----- {KCL} This law can also be stated as; “the sum of all the currents flowing towards a point is equal to the sum of all the currents flowing away form the point”. It is also known as Kirchhoff’s point rule which is a manifestation of law of conversion of charge. b) Kirchhoff’s 2nd rule (KVL-Kirchhoff’s Voltage Law): it states that: “the algebraic sum of voltage changes in a closed circuit or a loop must be equal to zero”. Consider a closed loop (Fig 13.9b). The direction of the current I depends on the cell having the greater emf. If E1>E2, then current flows in the counter clockwise direction. When a positive charge ∆Q due to the current I in the closed circuit, passes through the cell E1 form low (-ve) to high potential (+ve), it gains energy because work is done on it. When current passes through E2 it losses energy equal to –E2∆Q because charge passes from high to low potential. In going through the resistor R1 and R2, the charge ∆Q loses energy equal to –IR1∆Q and –IR2∆Q respectively. According R1 E1 to law of conservation of energy the total change in energy of system is zero; E2 E1∆Q-IR1∆Q-E2∆Q-IR2∆Q = 0 I Or E1-IR1-E2-IR2 = 0 ----- {KVL} R2 Fig 13.9b

Abu Bakkar Marwat--B.Sc Textile Engineering (Spinning} NTU, Faisalabad. [email protected]

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Following rules are followed for finding the potential changes: • The direction of the current I depends on the cell having the greater emf. If E1>E2, then current flows in the direction of E1 (from +ve to –ve terminal of E1) • When current enters the battery through –ve terminal, energy is gained (E is +ve) and when it enters through +ve terminal, energy is lost (E is –ve). • Energy is also lost across resistors so for Fig 13.5b, energy loss is taken as (-IR2-IR1). Example 13.6: calculate the currents in the three resistances of the circuit shown in Fig. Solution: First we select two loops abcda and ebcfe for which loop currents are I1 and I2 in clockwise direction, respectively. Now applying Kirchhoff’s law to loop abcda: -E1 – I1R1 – (I1 –I2) R2 + E2 = 0 Substituting values: - 40 – 10I1 – (I1 – I2)x 30 + 60 = 0 20 - 10I1 – 30I1+30I2 = 0 40I1 - 30I2 = 20 => 4I1 – 3I2 = 2 A ----- {a} e b a E3 E2 E1 Similarly, applying Kirchhoff’s law to loop ebcfe: - E2 – (I2 – I1) R2 – I2R3 + E3 = 0 R3 I2 R2 I1 R1 Substituting values: - 60 – (I2 – I1)x 30 – 15I2 + 50 = 0 f c d R1=10 Ω, R2=30 Ω, R3=15 Ω - 10 – 30I2 + 30I1 – 15I2 = 0 E1=40V, E2= 60 V, E3=50 V 30I1 – 45I2 = 10 A => 6I1 – 9I2 = 2 A ----- {b} From equ. {a}, we have: 4I1 = 2 A + 3I2 => I1 = (2A + 3I2)/ 4 Putting value of I1 in equ. {b}: 6 (2A + 3I2)/4 – 9I2 = 2A 3A + 9/2I2 – 9I2 = 2A => -9/2I2 = -1A I2 = 2/9 A ---- {c} Now putting the value of I2 in equ. {a} 4I1 – 3 (2/9)A = 2A => 4I1 = 2A + 2/3A => 4I1 = 8/3A I1 = 2/3 A ----- {d} Now B Current through R1 = I1 = 2/3 A = 0.66 A Current through R2 = (I1 – I2) = 2/3 – 2/9 = 4/9 A = 0.44 A C I1 I2 Current through R3 = I2 = 2/9 A = 0.22 A A 13.10 WHEATSTONE BRIDGE: S I3 D It is an electrical circuit used to measure an unknown resistance. Construction: It consists of four resistances R1, R2, R3, and R4 connected in such a way as to form a mesh ABCDA. A battery is connected between points A and C. a sensitive galvanometer is connected between points B and D. Working: If switch S is closed, current will flow through galvanometer. We have to setup a condition under which, no current pass through galvanometer even if switch is closed. For this applying KVL to the loops ABDA, BCDB & ADCA and assuming anti clockwise currents I1, I2 and I3 through the loops, respectively. Loop ABDA: - I1R1 – (I1 – I2) Rg – (I1 – I3) R3 = 0 ----- {a} Abu Bakkar Marwat--B.Sc Textile Engineering (Spinning} NTU, Faisalabad. [email protected]

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Loop BCDB: - I2R2 – (I2 – I3) R4 – (I2 – I1) Rg = 0 ----- {b} Current flowing through the galvanometer will be zero if, I1 – I2 = 0 or I1 = I2 Applying this condition to {a} and {b} {a} => - I1R1 = (I1 – I3) R3 ----- {c} {b} => - I1R2 = (I2 - I3) R4 ----- {d} Dividing {c} by {d}: R1/R2 = R3/R4 ----- {e} (Wheatstone bridge equation) To determine unknown resistance: An unknown resistance is connected at R4 in the circuit and equ. {e} is used to find it, if values of R1, R2 and R3 are known. 13.11 POTENTIOMETER: Usually, a voltmeter is used to measure potential difference between two points. It is connected between two points in parallel with the circuit. The resistance of the voltmeter must be very large as compared to the resistance of the circuit otherwise; the voltmeter would alter the potential difference between the two points by drawing current through itself. An ideal voltmeter would have an infinite resistance. Potentiometer is an electrical circuit which can measure and compare potential differences accurately, without drawing any current through itself. Construction and working: It consists of a resistor R in the form of a wire on which a terminal C can slide (Fig 13.11a). The resistance between A and C can be varied from 0 to R as the sliding contact is moved from A to B. If a battery of emf E is connected across R, the current flowing through it is I = E/R. If resistance between A and C is “r” then potential drop between A and C will be rI = r E/R. thus as C is moved from A to B, “r” varies from 0 to R and potential drop between A and C changes from 0 to E. Potential divider: It is a circuit that is used to measure the unknown emf of a source (Fig 13.11b). Here R is in the form of a straight wire of uniform area of cross section A. A source of emf Ex (whose emf is to be measured) is connected between A and the sliding contact C through a galvanometer G. Positive terminal of Ex and that of potential divider are connected to same point A. Now if in the loop AGCA, the point C and negative terminal of Ex are at the same potential then two terminals of the galvanometer will be at same potential and no current will flow through the galvanometer. Under this condition, the emf Ex of the cell is equal to the potential difference between A and C whose value is: Ex = Er/R. In case of a wire of uniform cross section, the resistance is proportional to the length of the wire. Therefore unknown emf is also given by Ex = Er/R = E l/L {L total length of wire AB, l = length from A to C} This method can be used to compare the emfs E1 and E2 of two cells. The balancing lengths l1 and l2 are found separately for the two cells. Then; E E1 = E l1/L and E2 = E l2/L A B Dividing these two equations; r C E1/E2 = l1/l2 Fig 13.11a i.e., the ratio of the emfs is equal to ratio of the balancing lengths. E

-:---------------x---------------:-

A Ex

B C Fig 13.11 b

Abu Bakkar Marwat--B.Sc Textile Engineering (Spinning} NTU, Faisalabad. [email protected]