UNIT 5: Electric Current and DirectCurrent Circuit (D.C.)


 Jonah Matthews
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1 UNT 5: Electric Current DirectCurrent Circuit (D.C.) SF07 5. Electric Current, Consider a simple closed circuit consists of wires, a battery a lamp as shown in figure 5.a. F r e E r rea, From the figure, Fig. 5.a Direction of electric field or electric current : Positive to negative terminal. Direction of electron flows : Negative to positive terminal. The electron accelerates because of the electric force acted on it. Definition is defined as the total (net) charge, Q flowing through the area per unit time, t. Mathematically, dq nstantaneous or current Q t SF07 dt
2 t is a basic scalar quantity. The S.. unit of the electric current is the ampere (). ts dimension is given by [ ] ampere of current is defined to be one coulomb of charge passing through the surface area in one second. 5. Current Density, J Definition is defined as the current flowing through a conductor per unit crosssectional sectional area. Mathematically, J t is a vector quantity. ts unit is amperes per square metre ( m  ) : electric current :cross sectional area of the conductor The direction of current density, J always in the same direction of the current. e.g. rea, r J 0 SF07 3 J r 5.3 Drift velocity of Charges in a Conductor, v d Consider a segment of two different currentcarrying material as shown in figure 5.3a 5.3b. E r J r v r L d Figure 5.3a shows the charges carrier is positive, the electric force is in the same direction as E the drift velocity v d is from left to right. Figure 5.3b shows the charges carrier is negative (electron), the electric force is opposite to E the drift velocity v d is from right to left. Consider a situation of figure 5.3a, a semiconductor with crosssectional area length L. Suppose there are n charged particles per unit volume N n SF07 4 v r d Fig. 5.3a : Semiconductor L v r d E r J r L v r d Fig. 5.3b : Metal
3 Then the number of charges, N along the conductor is given by Therefore the total charge Q that passes through the area along the conductor is The time required for the charge moving along the conductor is Since N nl Q Ne Q ( nl)e t L v d Q t ( nl) e nevd L v d then the drift velocity v d is given by v d ne SF07 5 or v d J ne n : number of charge per unit volume e : magnitude of the positive charge (electron) J Definition Density of the charge carrier Consider a situation of figure 5.3b, when an electric field exits in the wire, the electron feel a force initially begin to accelerate. The acceleration a due to electron is given by F e ee ma ee ee a m 5.4. Electrical conduction in the metal F e F ma E : electric field strength m : mass of the electron 5.4 Mechanism of Electrical Conduction n metal the charge carrier is free electrons a lot of free electrons are available in it. They move freely romly throughout the crystal lattice structure of the metal but frequently interact with the lattices. When the electric field is applied to the metal, the freely moving electron experience an electric force tend to drift towards a direction opposite to the direction of the field. Then an electric current is flowing in the opposite direction of the electron flows. SF07 6
4 5.4. Electrical conduction in the semiconductor n a pure semiconductor such as silicon, the charge carriers is free electrons free positive holes. When an electron moves from the valence b into the conduction b by increases the temperature of the semiconductor, it leaves behind a vacant site called a hole. n electron from a neighbouring atom can move into this hole leaving the neighbour with the hole. n this way, the hole can travel through the semiconductor as an additional charge carrier. n a pure or intrinsic semiconductor, valence b holes conduction b electrons are always present in equal numbers. When an electric field is applied, they move in opposite directions as shown in figure 5.4a. e Conduction electron Conduction b Fig. 5.4a hole Energy gap pplied E r field alence b SF07 7 h Thus a hole in the valence b behaves like a positively charged particle, even though the moving charges in that b are electrons. The drifting of electrons produce a current e while the drifting of holes produce a current h. Therefore the net current flowing in the semiconductor is given by e + h e > h When the temperature of a semiconductor increases, the number of free electrons holes increases. Hence the current flowing also increase Electrical conduction in the superconductor Superconductor is a class of metals compound whose resistance decreases to zero when they are below the critical temperature T c. Table below shows the critical temperature for various superconductors. Material Pb Hg Sn l Zn T c (K) SF07 8
5 The value of T c is sensitive to chemical composition, pressure molecular structure. The remarkable features of superconductors is that once a current is set up in them, it persists without any applied potential difference (because 0). Example : current of.0 flows through a copper wire. Calculate a. the amount of charge, b. the number of electrons flow through a crosssectional area of the copper wire in 30 s. (Given the charge of electron, e.60x09 C) Solution:.0, t30 s a. From the definition of electric current, thus the amount of charge is Q t Q 60 C b. The number of electrons flow is Q Ne 60 N. 6 x0 N 3. 75x0 electron 0 9 SF07 9 Example : silver wire.6 mm in diameter transfers a charge of 40 C in 80 min. Silver contains 5.8 x 0 8 free electrons per cubic metre. Determine a. the current in the wire. b. the magnitude of the drift velocity in the wire. c. the current density in the wire (Given the charge of electron, e.60x09 C) Solution: d.6 x 03 m, t80x s, n5.8 x0 8 m 3, Q40 C a. From the definition of electric current, thus b. The magnitude of the drift velocity is c. By applying the equation of the current density, thus J Q t v d ne 4 vd nπd e 4 πd 8. 75x0 πd 4 SF07 0 v d. 78x0 J. 65x0 4 6 m m s
6 Example 3 : high voltage transmission line with a diameter of.00 cm a length of 00 km carries a steady current of 000. f the conductor is copper wire with a free charge density of 8.49 x 0 8 electrons m 3, find the time taken by one electron to travel the full length of the line. (Given the charge of electron, e.60x09 C) (Serway & Jewett,pg.855,no.56) Solution: d.00 x 0  m, 000, n8.49 x0 8 m 3, L00x0 3 m By using the equation of the drift velocity for electron, v d ne 4 vd nπd e v. 34x0 4 d Therefore the time taken for one electron travels through the line is L v d t t 8. 55x0 πd 4 m s 8 SF07 s 5.5 esistance esistivity 5.5. esistance, Definition is defined as the ratio of the potential difference across an electrical component to the current passing through it. Mathematically, t is thus a measure of the component s opposition to the flow of the electric charge. t is a scalar quantity its unit is ohm (Ω ) or  n general, the resistance of a metallic conductor increases with temperature, as the resistance of a semiconductor decreases with temperature. Note that if the temperature of the metallic conductor is constant hence its resistance also constant esistivity, ρ Definition is defined as the resistance of a unit crosssectional sectional area per unit length of the material. Mathematically, ρ l : potential difference (voltage) : current l :length of the material :cross sectional area SF07
7 t is a scalar quantity its unit is ohm metres (Ω m) t is a measure of a material s ability to oppose the flow of an electric current. t also known as specific resistance. 5.6 Conductance Conductivity 5.6. Conductance, G Definition is defined as the reciprocal of electrical resistance in a directcurrent current circuit. Mathematically, G or G t is a scalar quantity its unit is per ohm (Ω  ) SF Conductivity, σ Definition is defined as the reciprocal of the resistivity of a material. Mathematically, σ ρ σ t is a scalar quantity its unit is Ω  m  or l ρ l Example 4 : What diameter must an aluminium wire have if its resistance is to be same as that of an equal length of copper wire with diameter.0 mm. (Given ρ(aluminium) is.75x08 Ω m ρ(copper) is.7x08 Ω m) Solution: Cu l, d Cu.0x03 m, l Cu l l earrange the equation of resistivity, thus the resistance is given by ρl SF07 4
8 Since l Cu ρlll ρculcu l Cu ρl ρcu πd π l dcu mm d Cu Example 5 : When 5 is applied across a wire that is 0 m long has a 0.30 mm radius, the current density is.4 x 0 4 m . Find the resistivity of the wire. (Halliday,esnick&walker,pg.63,no.3) Solution: 5, r0.30x03 m,j.4x0 4 m , l0 m From the equation of the resistance, ρl J then the diameter of the aluminium wire is πd 4 J SF07 5 ρ 8. x0 4 Ω m ρl 5.7 Ohm s Law States that the potential difference across a metallic conductor is proportional to the current flowing through it if its temperature e is constant. Mathematically, Then T constant : resistance a conductor Ohm s law also can be stated in term of electric field E current density J. Consider a uniform conductor of length l crosssectional area. potential difference maintained across the conductor sets up an electric field E this field produce a current that is proportional to the potential difference as shown in figure 5.7a. l Fig. 5.7a E r SF07 6
9 f the field is assumed to be uniform, the potential difference is related to the field through the relationship below : El From the Ohm s law, then El J E ρj or J σe J ρl ρ σ ρl E : magnitude of electric field J :current density ρ : resistivity of the conductor σ : conductivity of the conductor SF07 7 The potential difference against current graphs of various material can be shown in figure 5.7b, 5.7c, 5.7d 5.7e. Gradient M 0 Fig. 5.7b : Metal 0 Fig. 5.7c : Semiconductor 0 SF07 8 Fig. 5.7d : Carbon 0 Fig. 5.7e : Electrolyte
10 Example 6 : wire 4.00 m long 6.00 mm in diameter has a resistance of 5 mω. potential difference of 3.0 is applied between the end. Determine a. the current in the wire. b. the current density. c. the resistivity of the wire material. (Halliday,esnick&walker,pg.630,no.6) Solution: d6.00x03 m, l4.00 m, 5x03 Ω, 3.0 a. From the Ohm s law, thus the current is. 5x0 b. By applying the equation of the current density, thus πd J 4 4 J πd J 5.3x0 7 m  SF c. By applying the equation of the resistivity, thus πd ρ l πd ρ 4l ρ.x0 7 Ω m Example 7 : (exercise) The rod in figure below is made of two materials. The figure is not drawn to scale. Each conductor has a square cross section 3.00 mm on a side. The first material has a resistivity of Ω m is 5.0 cm long, while the second material has a resistivity of Ω m is 40.0 cm long. Find the resistance between the ends of the rod. (Serway & Jewett,pg.853,no.4) 4 ns. : 378 Ω SF07 0
11 5.8 Conductivity in terms of Microscopic Quantities. n the microscopic model to explain for electrical conduction in metals, there are 4 assumption we need to considered. Each metal atom contributes one free electron. The free electron are in constant rom motion, colliding with each other the metal ions in the crystal lattice. The mean velocity of the free electrons is zero. There is no net transfer of free electrons in any direction before a potential difference is applied across the metal. When the electric field E is applied to the metal, each electron experiences a electric force FeE. ccording to the Newton s second law, Fm e a, then the acceleration of the electron is given Consider for a free electron whose rom velocity immediately after a collision is v 0, its final velocity just before the next collision v is given by r r ee a m e v + v0 at SF07 Similarly for the other N free electrons in the metal, their respective velocities just before the next collision are v v0 + at, v 3 v03 + at3,..., v N v0 N + at N The drift velocity v d of the free electron is defined as the mean of v,v,v 3,,v N. Hence the drift velocity is given by v d < v, v, v3,..., vn > vd < v0i + ati > i,, 3,..., N vd < v0i > + a < ti > But < v 0 i > 0 because the mean of the rom motion velocities before applied the electric field is zero the mean time interval between successive collision is < t i > τ Therefore, the drift velocity is v d aτ ee vd τ m e Since the current density is given by J nev d SF07 J σe (Ohm s s law)
12 Then the conductivity of the metal is Note: σe nev d ee σe ne τ me ne τ σ m e n : number of free electrons per unit volume e :charge of the electron : mass of the electron m e From the formula of the conductivity in terms of microscopic quantities, we get σ n For metal such as copper, the number of free electron per m 3 is 0 9 then metals are good conductors of electricity. For Materials such as silicon carbon, the value of n is small compared to metal, hence semiconductors are poor conductors of electricity. From that formula, conductivity do not depend on the strength of electric field applied to the metal. SF ariation of esistance with Temperature The resistivity of a conductors varies approximately linearly with temperature according to the expression below Since ρ ρ0 + ( α T ) ρ :final resistivity ρ 0 :initial resistivity α : temperature coefficient of resistivity (unit : K T : temperature difference ( TT0 ) ρ then the expression above can be written as 0 + α T ( ) :final resistance :initial resistance Metal When the temperature increases, the number of free electrons per unit volume in metal remains unchanged. Metal atoms in the crystal lattice vibrate with greater amplitude cause the number of collisions between the free electrons metal atoms increase. Hence the resistance in the metal also increases. SF )
13 5.9. Semiconductor When the temperature increases, the semiconductor atoms acquire the extra energy cause the valence electron escapes from the covalent bond. Thus the number of free electrons per unit volume in the semiconductor increases cause its resistance decreases Superconductor Superconductor is a class of metals that have zero resistance at lower temperature (below critical temperature). When the temperature of the metal decreases, its resistance decreases to zero at critical temperature e.g. mercury acquire the zero resistance at temperature of 4 K esistance against Temperature T graph for various materials. a. Metal b. Semiconductor 0 T SF T c. Superconductor d. Carbon 0 T T c Example 8 : certain resistor has a resistance of.48 Ω at 0.0 C a resistance of.5 Ω at 34.0 C. Find the temperature coefficient of resistivity. (Young & Freedman,pg.974.no.5.6) Solution: 0.48 Ω, T C,.5 Ω, T34.0 C By applying the equation of resistance varies with temperature, thus 0 + α T T T α 0 T 0 3 o α.54x0 C SF ( ) [ + ( T )] T 0 T
14 Example 9 : 5.00 m length of.0 mm diameter wire carries a 750 m current when.0 m is applied to its end. f the drift velocity of the electron has been measured to be.7 x 05 m s , determine a. the resistance of the wire. b. the resisitivity of the wire. c. the current density. d. the electric field inside the wire. e. the number of free electrons per volume. f. the conductivity of the wire. g. the mean time interval between successive collision of the electron. (Given the charge of electron, e.60x09 C m e 9. x03 kg) Solution: l5.00 m, d.0x03 m, 750x03,.0x03, v d.7 x 05 m s  a. From the Ohm s law, thus the resistance is. 9x0 b. From the definition of the resistivity, thus πd ρ SF07 7 l Ω 4 πd ρ 4l ρ.8x0 Ω m c. By applying the equation of the current density, thus πd J 4 4 J πd J.4x0 5 m  d. From the relationship between E for uniform E, thus e. By applying the equation of drift velocity, then 8 E l E 4.4x0 J v d ne J n v e n 8.8x0 SF07 d m electrons m
15 f. From the definition of the conductivity, thus σ ρ σ 5.6 x0 g. By applying the equation of conductivity in terms of microscopic quantities, ne τ σ me meσ τ ne τ.3x0 Example 0 : (exercise).0 m length of wire is made by welding the end of a 0 cm long silver wire to the end of an 80 cm long copper wire. Each piece of wire is 0.60 mm in diameter. potential difference of 5.0 is maintained between the ends of the.0 m composite wire. Determine a. the current in the copper silver wire. b. the magnitude of the electric field in copper silver wire. c. the potential difference between the ends of the silver section of wire. (Given ρ(silver) is.47x08 Ω m ρ(copper) is.7x08 Ω m) (Young & Freedman,pg.976.no.5.56) SF07 ns. : 45,.76 m ,.33 m , Ω 4 m s Energy Electrical Power 5.0. Energy Consider a circuit consisting of a battery that is connected by wires to an electrical device (such as a lamp, motor or battery being charged) as shown in figure 5.0a the potential different across that electrical device is B. Electrical device B B current flows from the terminal to the terminal B, if it flows for time t, the charge Q which it carries from B to is given by Then the work done on this charge Q from B to is W B Q B W B Bt Fig. 5.0a This work represents electrical energy supplied to the electrical device. f the electrical device is passive resistor (device which convert all the electrical energy supplied into heat), the heat H dissipated is given by t H W t H t or H or SF07 30 Q t
16 5.0. Electrical Power, P Definition is defined as the energy liberated per unit time in the electrical device. The electrical power P supplied to the electrical device is given by W t P t t P When the electric current flows through wire or passive resistor, hence the potential difference across it is then the electrical power can be written as P or P : current : resistance of the resistor (wire) : potential difference (voltage) t is scalar quantity ts unit is watts (W). SF Electromotive Force (e.m.f.), Terminal Potential Difference nternal esistance 5.. Electromotive Force (e.m.f.), Terminal Potential Difference Consider a circuit consisting of a battery (cell) that is connected by wires to an external resistor as shown in figure 5.a. Battery (cell) r B Fig. 5.a Electromotive force (e.m.f.), is defined as the energy provided by the source (battery/cell) to each unit charge that flows from the source. Terminal potential difference (voltage), B is defined as the work done in bringing a unit (test) charge from point B to point. SF07 3 current flows from the terminal to the terminal B. For the current to flow continuously from terminal to B, a source of electromotive force (e.m.f.), is required such as battery to maintained the potential difference between point point B.
17 The unit for both e.m.f. potential difference is volt (). When the current flows naturally from the battery there is an internal drop in potential difference (voltage) equal to r. Thus the terminal potential difference (voltage), B is given by B r n general, the equation above can be written as t r t ( r) + : e.m.f. t : terminal potential difference (voltage) r : nternal drop in potential r : total external resistance r : nternal resistance of a cell (battery) then (5.) Note : Equation (5.) is valid if the battery (cell) supplied the current to the circuit t < SF07 33 For the charging of battery, becomes t + For the battery without internal resistance or if no current flows in the circuit (open circuit), then equation (5.) can be written as 5.. nternal esistance of a cell (battery), r Definition is defined as the resistance of the chemicals inside the cell (battery) between the poles is given by r r The value of internal resistance depends on the type of chemical material in the cell (battery). The symbol of e.m.f. internal resistance in the circuit can be shown in figure 5.b. Fig. 5.b SF07 34 r t t > then the equation (5.) when the cell (battery) is used. : potential difference across internal resistance : current in the circuit r or r
18 Example : battery of internal resistance 0.3 Ω is connected across a 5.0 Ω resistor. The terminal potential difference measured by the voltmeter is.5. Calculate the e.m.f. of the battery. Solution: r0.3 Ω, 5.0 Ω, t.5 The current flows in the circuit is given by t t By applying the equation of terminal potential difference, thus the e.m.f. is given by t r. 8 Example : When a 0 Ω resistor is connected across the terminals of a cell of e.m.f. internal resistance r a current of 0.0 flows through the resistor. f the 0 Ω resistor is replaced with a 3.0 Ω resistor the current increases to 0.4. Find r. SF07 35 Solution: nitially: Ω r By applying the equation of e.m.f., ( r). ( 0 0 r) () Finally : Ω r By applying the equation of e.m.f., By equating eq. () () then r.0 Ω Substituting for r in either eq. () or (). ( 3 0 r) () SF07 36
19 Example 3 : For the circuit shown below, given, r.0 Ω 4.0 Ω. r Calculate the ammeter voltmeter reading. Solution: By applying the equation of e.m.f., the current flows in the circuit is + r ( + r) Therefore the ammeter reading is.0. The voltmeter reading is given by t ( ). 0 SF t. 5. Combinations of Cells 5.. Cells in Series Consider two cells connected in series as shown in figure 5.a. r r The total e.m.f., the total internal resistance, r are given by + r r + r Note: f one cell, e.m.f. say, is turned round in opposition to the others, then but the total internal resistance remains unaltered. 5.. Cells in Parallel Consider two equal cells connected in parallel as shown in figure 5.b. r The total e.m.f., the total internal resistance, r are given by r + Fig. 5.b Fig. 5.a SF07 r 38 r r
20 Note: f different cells are connected in parallel, there is no simple formula for the total e.m.f. the total internal resistance Kirchhoff s laws have to be used. 5.3 Combinations of esistors The symbol of resistor in electrical circuit can be shown in figure 5.3a esistors in Series or Fig. 5.3a Consider three resistors are connected in series to the battery as shown in figure 5.3b. 3 3 Fig. 5.3b SF07 39 The properties of resistors in series are given below. The same current flows through each resistor 3 ssuming that the connecting wires have no resistance, the total potential difference, is given by + + (5.3a) 3 From the definition of resistance, ; ; 3 3 ; eq Substituting for,, 3 in eq. (5.3a) gives + + eq eq + + resistance : equivalent(effective) eq 3 3 SF07 40
21 5.3. esistors in Parallel Consider three resistors are connected in parallel to the battery as shown in figure 5.3c 5.3d Fig. 5.3d Fig. 5.3c The properties of resistors in parallel are given below. There is the same potential difference, across each resistor SF Charge is conserved, therefore the total current in the circuit is given by + + (5.3b) 3 From the definition of resistance, ; ; 3 ; 3 eq Substituting for,, 3 in eq. (5.3b) gives + + eq eq Example 4 : For the circuit shown below,.0 Ω SF Ω 4.0 Ω 6. 0
22 Calculate : a. the total resistance of the circuit. b. the total current in the circuit. c. the potential difference across 4.0 Ω resistor. Solution:.0 Ω, Ω, Ω, 6.0 a. connected in parallel with 3, then 3 3 connected in series with combination of resistors, 3, therefore the total resistance total in the circuit is given by + total total 5.0 Ω b. The total current is given by + total Ω SF07 43 c. The potential difference across.0 Ω is. 4 Therefore the potential difference across Ω is given by Example 5 : For the circuits shown below, calculate the equivalent resistance between points x y. a. b. (exercise) x.0 Ω.0 Ω.0 Ω.0 Ω 8.0 Ω 6.0 Ω 6.0 Ω 0.0 Ω 3.0 Ω x 9.0 Ω y 6.0 Ω 8.0 Ω SF07 44 ns. : 8.0 Ω y
23 Solution: a. x.0 Ω 5.0 Ω 3.0 Ω.0 Ω y connected in series with, thus x Ω x x Ω 5.0 Ω 3.0 Ω x 4.0 Ω y Ω SF07 45 x connected in parallel with 3, thus + y x 3 y 0.8 Ω x 5.0 Ω y 0.8 Ω y 4 y connected in series with 4, thus z y Ω x z Ω 5.0 Ω z 3.8 Ω y SF07 46
24 z connected in parallel with 5, thus the equivalent resistance is given by eq eq z Ω 5 Example 6 : (exercise) a. Find the equivalent resistance between points a b in figure below. b. potential difference of 34.0 is applied between points a b. Calculate the current in each resistor. (Serway & Jewett,pg.885,no.6) ns. :7. Ω,.99 for 4.00 Ω 9.00 Ω,.7 for 7.00 Ω, 0.88 for 0.0 Ω SF Kirchhoff s Laws The laws are useful in solving complex circuit problems. This laws consist of two statements. Kirchhoff s first law (junction/current law) states the algebraic sum of the currents entering any junctions in a circuit must equal the algebraic sum of the currents leaving that junction. or in out For example : a b Kirchhoff s second law (loop/voltage law) states in any closed loop, the algebraic sum of e.m.f.s is equal to the algebraic sum of the products of current resistance. or n any closed loop, SF07 48
25 Note : a. For e.m.f : Travel Travel  : b. For product of Travel + Travel c. Problem solving strategy (Kirchhoff( Kirchhoff s Laws): Choose labeling the current at each junction in the circuit given. Choose any one junction in the circuit apply the Kirchhoff s first law. Choose any two closed loops in the circuit designate a direction (clockwise or anticlockwise) to travel around the loop in applying the Kirchhoff s second law. Solving the simultaneous equation to determine the unknown currents unknown variables. SF07 49 For example : Consider a circuit shown in figure 5.4a. E D 3 C L3 3 3 Fig. 5.4a L t junction or D (applying the Kirchhoff s first law) : + () 3 For the closed loop (either clockwise or anticlockwise), apply the Kirchhoff s second law. SF07 50 L 3 3 in out F 3 B
26 From closed loop L FEDF E F L D + + () From closed loop L BCD SF D (3) C 3 L 3 3 B From closed loop L3 FECBF E F L3 3 C (4) B By solving equation () any two equations from the closed loop, hence each current in the circuit can be determined. SF07 5
27 Example 7 : For the circuits shown below. a. Calculate the currents., Ω 3 Ω 4,4 Ω 7 Ω b. Calculate the currents, 3. Neglect the internal resistance in Ω each battery Ω Ω SF Solution: a. 3 Ω, Ω L 4,4 Ω 7 Ω From closed loop L : pplying Krchhoff s nd law : b. 3 L L Ω 3 Ω 0.5 Ω 3. 0 By applying Kirchhoff s st law : in out + () 3 By applying Kirchhoff s nd law : From closed loop L : () SF
28 From closed loop L : (3) 3 By solving this three simultaneous equations, we get ; 4. 6 ; Example 8 : (exercise) For the circuit shown below. 3 Note : From the calculation, sometimes we get negative value of current. This negative sign indicates that the direction of the actual current is opposite to the direction of the current drawn. SF07 55 Given 8, Ω, 3 3 Ω, Ω 3. gnore the internal resistance in each battery. Calculate a. the currents. b. the e.m.f.. ns. :, 4, Electrical nstruments 5.5. Potential (voltage) Divider potential divider produces an output voltage that is a fraction of the supply voltage. This is done by connecting two resistors in series as shown in figure 5.5a. Since the current flowing through each resistor is the same, thus Fig. 5.5a Therefore, the potential difference (voltage) across is given by Similarly, + SF07 56 eq eq
29 esistance can be replaced by a uniform homogeneous wire as shown in figure 5.5b. a l Therefore, the potential difference (voltage) across the wire with length l is given by Similarly, c ac Fig. 5.5b l ρ l l l + The total resistance, ab in the wire is ρl ab ac + cb ρl ρl ρ ab + l + l Since the current flowing through the wire is the same, thus mportant From Ohm s law : l l l + SF07 57 b ( l + l ) ρl ab ρ l ( ) ( + ) l l ρl 5.5. Potentiometer Consider a potentiometer circuit is shown in figure 5.5c. +  (Driver cell accumulator) G C x (Unknown oltage) Fig. 5.5c Potentiometer can be used to Compare the e.m.f.s of two cells. Measure an unknown e.m.f. of a cell. Measure the internal resistance of a cell. Jockey B The potentiometer is balanced when the jockey (sliding contact) is at such a position on wire B that there is no current through the galvanometer. Thus Galvanometer reading 0 When the potentiometer in balanced, the unknown voltage (potential difference being measured) is equal to the voltage across C. x C SF07 58
30 Compare the e.m.f.s of two cells. n this case, a potentiometer is set up as illustrated in figure 5.5d, in which B is a wire of uniform resistance J is a sliding contact (jockey) onto the wire. n accumulator X maintains a steady X current through the wire B. l l C J () S () Fig. 5.5d D G nitially, a switch S is connected to the terminal () the jockey moved until the e.m.f. exactly balances the potential difference (p.d.) from the accumulator (galvanometer reading is zero) at point C. Hence fter that, the switch S is connected to the terminal () the jockey moved until the e.m.f. balances the p.d. from the accumulator at point D. Hence SF07 59 D B then C C C ρ l ρl C () D D ρ l By dividing eq. () with eq. () then then ρ l ρ l Measure an unknown e.m.f.. of a cell. ρl D By using the same circuit shown in figure 5.5d, the value of unknown e.m.f. can be determined if the cell is replaced with stard cell. stard cell is one which provides a constant accurately known e.m.f. Thus the e.m.f. can be calculated by using equation below. l l SF07 60 l l ()
31 Measure the internal resistance of a cell. Consider a potentiometer circuit as shown in figure 5.5e. S l 0 r C J Fig. 5.5e G n accumulator of e.m.f. maintains a steady current through the wire B. nitially, a switch S is opened the jockey J moved until the e.m.f. exactly balances the e.m.f. from the accumulator (galvanometer reading is zero) at point C. Hence then C C C ρ l 0 ρl 0 C fter the switch S is closed, the current flows through the resistance box the jockey J moved until the galvanometer reading is zero (balanced condition) at point D as shown in figure 5.5f. SF07 6 B () l r S Fig. 5.5f D J G Hence then t D D D ρ l From the equation of e.m.f., By substituting eq. () () into eq. (3), we get D 0 r (4) SF07 6 B l l l l l r 0 t (4) () t + r t r t r t t ρl (3)
32 The value of internal resistance, r is determined by plotting the graph of /l against /. earranging eq. (4) : Then compare with l Y M + 0 l 0 X + Therefore the graph is straight line as shown in figure 5.5g. l l 0 0 SF07 63 r l Fig. 5.5g C Gradient, M r l 0 Example 9: Cells B centrezero galvanometer G are connected to a uniform wire OS using jockeys X Y as shown in figure below. O B X G Y S The length of the uniform wire OS is.00 m its resistance is Ω. When OY is 75.0 cm, the galvanometer does not show any deflection when OX 50.0 cm. f Y touches the end S of the wire, OX 6.5 cm when the galvanometer is balanced. The e.m.f. of the cell B is.0. Calculate a. the potential difference across OY when OY 75.0 cm. b. the potential difference across OY when Y touches S the galvanometer is balanced. c. the internal resistance of the cell. d. the e.m.f. of cell. SF07 64
33 Solution: l OS 00 cm, OS Ω, B.0. a. a O a l OY () l OX () X a B a G 0 When G 0 (balance condition), thus B a B OX () OX () Y l OY() 75.0 cm, l OX() 50.0 cm Therefore the potential difference across OY is given by OY ( ) a OY ( ) OY ( ). 5 SF07 65 S Since wire OS is uniform hence a 6 OX ( ) a OX ( ) l OX ( ) OX ( ) los OX ( ) 6. 0 Ω loy ( ) OY ( ) los 9. Ω OY ( ) 0 OS OS b b b. b O l OX () b B l OY () X S G 0 When G 0 (balance condition), thus B OX () b B OX () Therefore the potential difference across OY is given by OY ( ) b OY ( ) OY ( ). 6 Y l OY() 00 cm, l OX() 6.5 cm Since wire OS is uniform hence l OX ( ) OX ( ) los OX ( ) 7. 5 Ω OY ( ) OS OY ( ) Ω OX ( ) b OX ( ) a 0. 3 OS SF07 66
34 c. From the equation of e.m.f., the e.m.f. of cell is given by ( r) + For case in question (a) : For case in question (b) :. + 6 b OY 0. 3 a ( OY ( ) + r By equating eq. () eq. (), hence the internal resistance of cell is 6 d. By substituting r.5 Ω into eq. (), thus ( 9 0 r) ( ( ) + r ( + r) ( r) 0. 3( + r) r.5 Ω ( ) ) ) () () SF07 67 Q Example 0: (exercise) n the potentiometer circuit shown below, PQ is a uniform wire of length.0 m resistance 0.0 Ω. S is an accumulator of e.m.f..0 negligible internal resistance. is a 5 Ω resistor is a 5.0 Ω resistor when S S open, galvanometer G is balanced when QT is 6.5 cm. When both S S are closed, the balance length is T 0.0 cm. Calculate P a. the e.m.f. of cell. b. the internal resistance of cell. c. the balance length QT when S G is opened S closed. d. the balance length QT when S is opened S closed. S ns. :0.50, 7.5 Ω, 5.0 cm, 5.0 cm SF07 68
35 5.5.3 Wheatstone Bridge t is used to measured the unknown resistance of the resistor. Figure 5.5h shows the Wheatstone bridge circuit consists of a cell of e.m.f. (accumulator), a galvanometer, know resistances (, 3 ) unknown resistance x. 3 C G D Fig. 5.5h Dividing gives 0 X SF07 69 B The Wheatstone bridge is said to be balanced when no current flows through the galvanometer. Hence C CB D DB Then Potential at C Potential at D Therefore C D BC BD Since thus 3 3 X 3 X X The application of the Wheatstone bridge is Metre Bridge. Figure 5.5h shows the Metre bridge circuit. (Unknown Thick copper X resistance) (resistance box) strip Wire of uniform resistance l The metre bridge is balanced when the jockey J is at such a position on wire B that there is no current through the galvanometer. Thus the current flows through the resistance X but current flows in the wire B. Let x : p.d. across x : p.d. across, t balance condition, 0 Fig. 5.5h SF07 70 X J G J ccumulator JB Jockey l B
36 By applying Ohm s law, thus X J Dividing gives JB X Example : n unknown length of platinum wire 0.90 mm in diameter is placed as the unknown resistance in a Wheatstone bridge as shown in figure below. ρl X J J JB ρl ρl X l l ρl JB esistors have resistance of 38.0 Ω 46.0 Ω respectively. Balance is achieved when the switch closed 3 is 3.48 Ω. Find the length of the platinum wire if its resistivity is 0.6 x 08 Ω m. (Giancoli,pg.683.no.70) SF07 7 Solution: d0.93x03 m, 38.0 Ω, 46.0 Ω, Ω, ρ0.6x08 Ω m t balance condition, the ammeter reading is zero thus the resistance of the platinum wire is given by 3 X X 4.Ω From the definition of resistivity, hence the length of the platinum wire is πd X ρ l 4 πd X l 4ρ 3 π( 0. 93x0 ) ( 4. ) l 8 4( 0. 6 x0 ) l 6. 4 m SF07 7
37 5.5.4 Ohmmeter t is used to measure the unknown resistance of the resistor. Figure 5.5 shows the internal connection of an Ohmmeter. 0 Ω M s M s X : meter (coil) resistance : variable resistance : unknown resistance P X Fig. 5.5 When nothing is connected to terminals P Q, so that the circuit is open (that is, when ), there is no current no deflection. When terminals P Q are short circuited (that is when 0), the ohmmeter deflects fullscale. For any value of X the meter deflection depends on the value of X. SF07 73 Q mmeter t is used to measure a current flows in the circuit. mmeter is connected in series with other elements in the circuit because the current to be measured must pass directly through the ammeter. n ammeter should have low internal resistance ( M ) so that the current in the circuit would not affected. The maximum reading from the ammeter is known as full scale deflection (fs). f the full scale current passing through the ammeter then the p.d. across that ammeter is given by fs fsm M : meter(coil) resistance fs :full scale current : full scale potential difference (p.d.) fs f the meter is used to measure currents that are larger than its full scale deflection ( > fs ), some modification has to be done. resistor has to be connected in parallel with the meter (coil) resistance M so that some of the current will bypasses the meter (coil) resistance. This parallel resistor is called a shunt denoted as S. SF07 74
38 Figure 5.5J shows the internal connection of an ammeter with a shunt in parallel. 0 max fs S M S Since shunt is connected in parallel with the meter (coil) resistance then fs M S M S S ( ) S Fig. 5.5J fs M fs fs Therefore the shunt resistance is given by S M fs SF07 75 S fs oltmeter t is used to measure a potential difference (voltage) across electrical elements in the circuit. oltmeter is connected in parallel with other elements in the circuit therefore its resistance must be large than the resistance of the element so that a very small amount of current only can flows through it. n ideal voltmeter has infinite resistance so that no current exist in it. To measure a potential difference that are larger than its full scale deflection ( > fs ), the voltmeter has to be modified. resistor has to be connected in series with the meter (coil) resistance M so that only a fraction of the total p.d. appears across the M the remainder appears across the serial resistor. This serial resistor is called a multiplier or bobbin denoted as B. Figure 5.5K shows the internal connection of a voltmeter with a multiplier in series. SF07 76 fs Fig. 5.5K 0 max B M Electrical element
39 Since multiplier is connected in series with the meter (coil) resistance then the current through them are the same, fs. The p.d. (voltage) across the electrical element is given by + B M Hence the multiplier resistance is + B fs B fs fs Note : To convert a galvanometer to ammeter, a shunt (parallel resistor) is used. To convert a galvanometer to voltmeter, a multiplier (serial resistor) is used. fs M M SF07 77 Example : milliammeter with a full scale deflection of 0 m an internal (coil/metre) resistance of 40 Ω is to be used as an ammeter with a full scale deflection of 500 m. Calculate the resistance of the shunt required. Solution: fs 0x03, M 40 Ω, 500x03 By applying the equation of shunt, thus fs S M fs S.7 Ω Example : galvanometer has an internal resistance of 30 Ω deflects full scale for a 50 µ current. Describe how to use this galvanometer to make a. an ammeter to read currents up to 30. b. a voltmeter to give a full scale deflection of 000.(Giancoli,pg.68.no.50) Solution: fs 50x06, M 30 Ω a. We make an ammeter by putting a resistor in parallel ( S ) with the internal resistance, M of the galvanometer as shown in figure below. SF07 78
40 Given 30. fs S M G Since S connected in parallel with M hence M S S fs fs M ( fs ) S 6 S 50x0 Ω in parallel. b. We make a voltmeter by putting a resistor in series ( B ) with the internal resistance, M of the galvanometer as shown in figure below. fs Since B connected in series with M hence the current through them are the same, fs. B S SF07 79 M G Given 000 fs Therefore + B M fsb + fs fsm B fs 6 B 0x0 Ω M in series. Example 3: (exercise) milliammeter of negligible resistance produces a full scale deflection when the current is m. How would you convert the milliammeter to a voltmeter with full scale deflection of 0? ns. :.0 x 0 4 Ω in series Example 4: (exercise) movingcoil meter has a resistance of 5.0 Ω full scale deflection is produced by a current of.0 m. How can this meter be adapted for use as : a. a voltmeter reading up to 0, b. a ammeter reading up to? ns. :9995 Ω in series,.5 x03 Ω in parallel. SF07 80
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