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©, Current Electricity ten 12th Science 12th - Physics Syllabus 1. Introduction 2. Kirchhoff’s laws of electrical network 2.4 Kirchhoff’s first law (current law/junction 2.2 Kirchhoff’s voltage law 3. Wheatstone bridge 3.1 Metre Bridge 4. Potentiometer 4.1 Potentiometer principle 4.2 Use of potentiometer 4.3 Advantages of potentiometer over a voltmeter 5. Galvanometer 5.1 Galvanometer as an ammeter 5.2 Galvanometer as an voltmeter law) Theory Notes Q. What is current electricity? Q. Define i) Junction ii) Loop tii) Branch Q. State and explain Kirchhoff's laws of electrical circuit. Current Electricity 1) Electrostatics is the study of charges at rest. , 2) Current electricity is the study of charges in motion. | 3) The rate of flow of charge is called electric current. 4) The current which does not change with time is known as steady current. i) Junction A point in electric circuit where two o together is called a junction. x more conductors are joined ii) Loop A closed conducting path in an electric network i mesh, s called a loop or iii) Branch A part of electric network between two junctions is called a branch. Kirchhoff’s laws Gustav Kirchhoff has given two laws, one is about current and ot is about voltage. her 1) Kirchhoff’s first law (Current law or Junction law) Statement The algebraic sum of electric currents at any junction in a circuit is always equal to zero. “wDh=0 i Where /, is the current in the i” conductor at a junction having n conductor. Sign Conventions 1) The current entering the junction is taken as positive. | 2) The current leaving the junction is taken is negative. Explanation Junction is a point in the circuit where the current can split. Consider a junction Pin the circuit. Fig. Kirchhoff’s first law Q. How to solve a problem using Kirchhoff slaws? Q. Obtain balancing condi- tioning case of Wheatstone’s network. Q. Explain Wheatstone’s Bridge. Steps usually followed to solve a problem using Kirchhoff’s laws Wheatstone’s Bridge Select some direction of the currents. , Reduce the number of variables using Kirchhoff’s first law, Determine the number of independent loops. Apply Kirchhoff’s second law to all the loops. Solve the equations obtained simu taneously. ; If the answer of a current variable is negative, the direction of the conventional current is opposite to the selected direction. The Wheatstone’s bridge network is used to find unknown resistance, Explanation y 6) It consists of four resistances Rj,R2,R:and R; which forms a quadrilateral ABCD as shown in figure. The source of emf E is connected across diagonally opposite points A and C and the galvanometer is connected across the points B and D. + - { II 1} E * Fig. Wheatstone’s Network Let Vi,B:.Veand Vp are the respectability and V, > V.. Let V, = Vp(Potential at points B an D are equipotential points) Due to this no current flows from B to D no deflection i.e. null point. RR In this balanced condition, R. = R 2 4 This equation is called as balancing condition or Wheatstone’s principle. From this equation unknown resistance can be de mined, if other three values are known. potential at points A, B, C and D id D is equal i.e. points B and and galvanometer shows ter- Proof (By Kirchhoff’s Law) Let / be the current sent by the cell. at junction 'A’, it divides into two parts i.e. J, through R, and /;through k:. Applying Kirchhoff’ first law- T=h+h In balanced condition, Vo = Vo “L=0 Q. Describe the working of meter bridge. + Q. Explain with neat circuit diagram, how will you determine the unknown resistance by using a meter-bridge. Hence current through R; will be 4 and current through R:will be h. Applying Kirchhoff’s second law to the closed Ipp ‘ABDA’- —-1)R+0XG+hR=0 CAR = Reece) Applying Kirchhoff’s second law to the closed loop ‘BCDB’- -LR:+hbRi+0XG=0 AR: = BRa eee (2) Divide equation (1) by (2) RoR Ry. Rs If positions of galvanometer and cell are interchanged, same result is obtained. Hence opposite arms AC and BD are called conjugate arms. Proof (By Ohm’s Law) In balanced condition Vi = Bo * Va — Vo = Va Vow vee) And Ve — Ve = Vow Vows ses nl) Applying Ohm’s law Vi — Ve = 4A, and Va- Vo = bRs Similarly Ve- Ve = (R2, and Vo— Ve = hRs Therefore equation (1) and (2) becomes- TR = Das. ee (3) LRr = DaR4 ee oe (4) Divide equation (3) by (4) Ro_ Rs Ri Rs; Metre Bridge It is modified form of Wheatstone’s network which is used to measure the unknown resistance. It works on the Wheatstone’s principle Hence it is known as Wheatstone’s metre-bridge. The length of the wire used is one meter, so it is called as metre-bridge. mi agin meter scale /}—{0)}— VAN Rh E K Fig. Wheatstone’s metre-bridge Q. Explain Wheatstone post office box. 'B' when the bridge is balanced. 4) A suitable resistance 'X' is taken in resistance box and deflection in galvanometer is noted without touching jockey on wire Pe . 5) By touching the jockey at different points on wire AC, { nd a point D such that galvanometer shows the same deflection as before. Hence it is equal deflection method and point D is called balance point. | 6) Let /, and be the lengths of segments AD and DC respectively. 7) From the balancing condition of Wheatstone’s network- G _ resistance of wire of length /, “R ~ Fesistance of wire of length /, G_ ol “R pl, Where Gis the resistance per unit length of the wire. Gob "Rd , ne = 7 Ge R( 4) or o= Rpt) wf l +1, = 100 em} Using this relation, we can find out the unknown resistance of galvanometer. Post Office Box . A post office box (PO box) is a practical form of wheatstone bridge used to measure small unknown resistance. Fig. PO Box Construction 1) It consists of three arms P,Q and R with adjustable resistances. 2) Pand Q are called ratio arms with 10Q, 100Q and 1000 each. 3) Third arm R contains resistances from 1Q to 5000Q. 4) Unknown resistance X connected between points C and D forms the fourth arm of the bridge circuit. 5) There are two tap keys K, and K; battery E is connected. Working 1) The resistances in the arms P and Q are fixed to desired ratio. 2) Resistance in arm R is adjusted so that galvanometer shows no deflection. 3) For this balanced condition xX RP QR x= Thus unknown resistance X can be determined from the known values of Q,R and P. Q. Explain the use of Wheatstone bridge for strain measurement. Q. What are the possible errors while using meter bridge? How these errors can be minimized? Q. State any two possible sources of error in the meter bridge experiment. How can thy be minimized? (Oct. 15, Mark 2) Q. Explain: In Wheatstone’s meter-bridge experiment the null point is obtained in middle one third portion of wire. Q. Explain working of Potentiometer. Q. Explain the principle of a Potentiometer. (Oct. 13, Mark 2) : : t Wheatstone Bridge for Strain Measuremen : 1) For strain measurement, a resistor (Ry) is the wheatstone bridge is replaced with a strain gauge as shown in figure. strain gauge 2) Resistors R, and R: are equal while the resistor R; is variable. 3) With no force applied to the strain gauge, R; is adjusted so that the voltmeter shows zero deflection i.e. the bridge is balanced. In this condition strain is zero. 4) When force is applied to the strain gauge, its resistance changes and causes unbalancing of the bridge. S) Deflection in voltmeter corresponds to the strain change. Greater the strain, more is the deflection. Possible Errors while using Meter Bridge (or Kelvin’s method) 1) Wire may not be uniform, therefore resistance per unit length of wire may not be constant. There may be, contact resistance between copperplates and end terminals of wire. 3) The end of wire may not coincide with zero and 100 cm mark on meter scale. 4) |, and lr may not be accurately measured. 2) Minimization of Errors (Precautions) These error can be minimized by taking following precautions- 1) The wire used must be uniform i.e. having uniform cross section. 2) The value of R should be chosen such that the null point is obtained as near to the centre of wire as possible. If the number of readings to be taken is more, the null points should lie in the middle one third of the wire. 3) The experiment should be repeated by interchanging the positions of X and R. 4) Top the jockey on the wire instead of sliding. Potentiometer 1) Potentiometer consists of a long uniform conducting wire of length 2 to 10 meter. 2) This wire is fixed between two point A and B on wooden platform. The source of emf FE and negligible internal resistance is connected in series with the wire through key K. Q. Explain direct method of comparison of e.m.f's of two cells using potentiometer. Q. Describe how potentiom- eter is used to compare the emfs of two cells by direct method, Q. Explain combination method of comparison of em fis of two cells using potentiometer. Q. Describe how potentiom- eter is used to compare the emfs of two cells by combi- nation method. . e e.m.f.s of Direct Method (Individual Cell Method) to compar two cells using potentiometer Let E, and FE, be the e.m.f.s of two cells to a be compared. Le | ni — Rh Fig. Comparison of e.m.f.s of two cells by direct method. 1 Battery E, key K and rheostat Rh are connected in series with potentiometer wire AB such that point A is at higher potential than point B. Battery must have e.m.f. greater than both the cells to be compared. . Positive terminals of E, and E, are connected to point A of wire AB and negative terminals are connected to K,and K, of the two way keys respectively. Common terminal of K, and K) is connected to jockey through galvanometer G. . Key K is closed so that current / is passed through the circuit. Key K; is closed and key K; is kept open, so the current / passes through E,. By touching the jockey at different points on the Wire, null point P is obtained for which galvanometer shows zero deflection. Let AP = |, be the balancing length. 3 By = Toh...) Where @ is the resistance per unit length. 5) Now key Kiis opened any key K2 is closed, so the current / Passes through E). By touching the jockey at different points on the wire, null point P' is obtained. Let AP' = |; be the balancing 2) 8) length. “Ea = Ilr... (2) From equations (1) and (2) Be doh _h E: lol 1, Thus e.m.f.s of two cells can be com ‘ pared with the help of potentiometer. Combination Method (Sum and Difference Method) to compare e.m.f.s two cells using potentiometer 1) The circuit diagram for comparison of emfs of two cells using sum and difference method is shown below- Q. Enlist the precautions while using potentiometer. Q. State the precautions which must be taken while performing experiment with potentiometer. I: | Rh Fig. Comparison of e.m.f.s of two ells by combination method. 2) A battery of e.m.f. E,key K' and rheostat Rh are in series with potentiometer wire AB such that point A is at higher potential | than point B. 3) When the keys (a) and (b) are closed and keys (c) and (d) are | open, the cells E, and E,are said to assist each other. The total | e.m.f. is E, + E:. By touching the jockey at different points of wire | AB, null point P is obtained for which the galvanometer G shows | zero deflection. Let AP' = hb be the balancing length. Ey t Ey = U0b ce 1) Where o is the resistance per unit length. 4) Now the keys (a) and (b) are opened and keys (c) and (d) are closed, the cells , and E,are said to oppose each other. The total e.m.f. is E;— Es. By touching the jockey at different points of wire AB, null point P' is obtained for which the galvanometer G shows zero deflection. Let AP' = |;be the balancing length. «E,- E=Ioh.......(2)..4E, > Es assumed} From equations (1) and (2) At _ Joh _ bh “EE lok bh By componendo and dividendo method AL bth “EO ho-h Thus e.m.f.s of two cells can be compared by sum and difference method using potentiometer. Precautions while using potentiometer 1) The potentiometer wire must be uniform. 2) The resistance of potentiometer wire should be high. 3) Positive terminals of all the cells must be connected at the common point A. 4) The e.m.f. of cell E must be greater than the sum of e.m-fs of E, and E,. §) Applying Ohm’s law for closed circuit of R,r and E;, we get- Ey = MRAP) csc oe (4) P.D. across R is given as- V= IR. (5) From (4) and (5) & - Aer (6) From equations (3) and (6) Rtr oh R L W1tE= t bab r= a(t) =n(44) Using this relation, internal resistance of the cell is determined. Q. Give the application of Application of Potentiometer potentiometer. 1) Voltage Divider Potentiometer can be used as a voltage divider to continuously change the output voltage of a voltage supply. 2) Audio Control Sliding and rotary potentiometers are used for frequency attenua- tion, loudness control and for controlling different characteristics of audio signals. 3) As a Sensor Potentiometer can be used as a motion sensor by connecting its slider to the moving part of a machine. Advantages of potentiometer over voltmeter Potentiometer measures the e.m.f. of cell and small terminal P.D. While voltmeter measures the terminal P.D. 2) The accuracy of potentiometer can be increased by increasing length of wire while voltmeter is less accurate. 3) Potentiometer can measure very small value of e.m-f. while voltmeter cannot measure such a small value. 4) Potentiometer can be used to find out internal resistance of cell. Q. State the advantages of potentiometer over voltmeter. 1) Disadvantages of Potentiometer 1) Voltmeter gives direct reading while potentiometer does not. 2) Voltmeter is portable but potentiometer is not. Q. What is galvanometer? Q. Explain ammeter. Q. Explain how a moving coil galvanometer is converted into an ammeter. Derive the necessary formula. Galvanometer . 1) Galvanometer is a device used to detect weak electric currents in a circuit. oles of 2) It has a coil pivoted or suspended between concave P a strong horse shoe magnet. . . 3) A pointer is attached to the coil to indicate its deflection. 4) When an electric current passes through the coil, it deflects and hence pointer also deflects. ; 5) Deflection of the pointer is proportional to the electric current. 6) Galvanometer can be converted into ammeter or voltmeter with suitable modifications. Ammeter R 1) A shunted permanent magnet moving coil (PMMC) galvanom- eter is called as ammeter. It is a device used to measure electric current. 2) Galvanometer is a very sensitive device which gives large deflec- tion for a small current. It gives full scale deflection for a current of the order of 10mA. If large current flows through the coil, the coil may get damaged. 3) Also galvanometer coil is having large resistance. The resistance is introduced into the circuit when galvanometer is connected in series with circuit. 4) All these drawbacks are minimized by connecting small value of resistance called shunt in parallel with galvanometer. This galvanometer having small resistance in parallel is called as ammeter. Equation for shunt (S) 1) Suppose '/' is total current flowing through the circuit, '/,' is the current flowing though galvanometer and '/,' is the current flowing though shunt, then- 1=1L+1, I-k, s e i 2) Let Gand S be the resistance of galvanometer and shunt respec- tively. As galvanometer and shunt resistance are in parallel combination- Q. Explain voltmeter. Q. Explain how a moving coil galvanometer is converted into voltmeter, Derive the necessary formula. Q. Give the function high value series resistor in voltmeter. Voltmeter To | Ss . : eries 1) A Permanent Magnet Moving Coil (PMMC) Galva nO ee atagure with high resistance is called as voltmeter. bi is use potential difference (p.d.) between two points. A. Gi 2) MCG gives full scale deflection for current of the order palin when a small potential difference is applied across its Oe ol 3) To measure larger potential difference, effective res. earstor in galvanometer can be increased by connecting high r series with moving coil galvanometer. . 4) Voltmeter is always connected in parallel with two points across which potential difference is to be measured. | Equation for high resistor in series (R.) ; . . 1) Let 'G' is the resistance of galvanometer and 'R,' is the high resis- tance in series , 'V' is the potential difference to be measured. Let 'l,' is the current for full scale deflection. As galvanometer and high resistance are in series combination- ~. p.d. = p.d. across galvanometer + p.d, across high resistance VE1G+IR, OV=L(G+R) | Lv =GtR, From this relation we can find out the value of resistance of high resistor, which should be connected in series with MCG so as to convert it into voltmeter. Functions of high value series resistor (R.) in voltmeter 1) It increases the effective resistance of galvanometer. 2) It increases the range of voltmeter. 3) It protects the galvanometer from damage due to large current. _— Q. Distinguish between ammeter and voltmeter. Q. What is thermoelectric effect? Q. What is Seebeck effect? Ans. Ammeter Voltmeter 1) It measures current. 1) It measures potential difference. 2) It is connecied in series. | 2) It is connected in parallel. 8) It is an MCG with low] 3) It is an MCG with high resistance. (ideally zero) resistance. (Ideally infinite) 4) Smaller the shunt,| 4) Larger its resistance greater will be the current| greater will be the potential measured. difference measured. 5) Resistance of ammeter is | 5) Resistance of voltmeter is R, = SG = G Rv = G+X = Gi “"S4+G on Thermoelectric Effect ‘ Conversion of thermal energy directly into electric energy 1S = thermoelectric effect. It was discovered by Seebeck. led eS Seebeck Effect 1) If two different metals ar their junctions are kept at different temperature produced and a current flows through the meta called Seebeck effect. 2) The emf produced in Seebeck effect is known as thermo emf. 3) The pair of two different metals forming the junction in Seebeck effect is called a thermocouple. Ex. An antimony - bismuth ther- mocouple is shown below. ~O- e joined to form a closed circuit and s, a small emf is Is. This effect is Sb Bi Sb Hot junction Cold junction Fig. Seebeck Effect For Sb-Bi thermocouple the current flows from Sb to Bi at the cold junction. Similarly for Cu-Fe thermocouple the current flows from Cu to Fe at the hot junction. 4) If the hot and cold junctions are interchanged, the direction of the current can be reversed. 5) If the cold junction is at 0°C and the hot junction is at TC, the thermo emf developed in a thermo couple is given as - E=ar+her Where @ and B are called thermoelectric constants. It shows that the graph of E with Tis a parabola. _——_—_ a Numerical From Text Book (Solved) a) 2) 3 4) 5) 6) 7) 8) 9) 10) A Galvanometer has a re Figure shows currents i : s ts in a part of electrical circuit. Find the current X? (fig. from textbook) ohm respectively are qwo batteries of 7 volt and 13 volt and internal resistances 1 ohm and 2 branch of ted i ri Coo eireuit nee with a resistance of 12 ohm. Find the current through each e potential difference across 12-ohm resistance. For the given network, find the current through 4 ohm and 3 ohm. Assume that the cells have negligible internal resistance. At what value should the variable resistor be set such that the pridge is balanced? of the metre bridge idge is balanced and stance of the Two resistances 2 ohm and 3 ohm are connected across the two gaps as shown in figure. Calculate the current through the cell when the bri the specific resistance of the material of the metre bridge wire. Given the resi bridge wire is 1.49 ohm and its diameter is 0.12 cm. ° 1.5 V, the balance point n the external re, Determine In an experiment to determine the internal resistance of a cell of emf in the open cell condition at is 76.3 cm. When a resistor of 9.5 ohm is used i circuit of the cell the balance point shifts to 64.8 cm of the potentiometer wi the internal resistance of the cell. vurrent is 100/ A. A galvanometer has a resistance of 100 Q and its full scale deflection ¢ ge of 0 to 10mA? What shunt resistance should be added so that the ammeter can have a ran, ance that allows 20% of the main current through a What is the value of the shunt resist galvanometer of 99 QP Galvanometer has a resistance of 25 Q and its full scale deflection current is 25uA. What resistance should be added to it to have a range of 0 -10 V? sistance of 40 Q and a current of 4 mA is needed for a full scale deflection . What is the resistance and how is it to be connected to convert the galvanometer (a) into an ammeter of 0.4 A range and (b) into a voltmeter of 0.5 Vrange? From Text Book (Unsolved) 1) A battery of emf 4 volt and battery of emf 1 V and intern combination is used to send current through the external internal resistance 1 Q is connected in parallel with another al resistance 1 Q (with their like poles connected together). The current through an external resistance of 2 Q. Calculate the resistance. Ans: 1A 2) Two cells of emf 1.5 Volt and 2 Volt having respective internal resistances of 1 Q and 2 Q are connected in parallel so as to send current in same direction through an external resistance of 5 Q. Find the current through the external resistance. Ans: 5/17 A 3) A voltmeter has a resistance 30 Q. What will be i di hen it is connected across a cell of emf 2 V having intern: ea ne woe al resistance 10 Q? Ans: 1.5 V 4) A set of three coils having resistances 10 Q, 12 Q and 15 Q are connected in parallel. This combination is connected in series with series combination of three coils of the same resis- tances. Calculate the total resistance and current through the circuit, ifa battery of emf 4.1 Volt is used for drawing current. Ans: 0.1 A 5) A potentiometer wire has a length of 1.5 m and resistance of 10 Q. It is connected in series with the cell of emf 4 Volt and internal resistance 5 Q. Calculate the potential drop per centimeter of the wire. Ans: 0.0178 V/em 6) When two cells of emfs. ¢, and €: are connected in series so as to assist each other, their balancing length on a potentiometer is found to be 2.7 m. When the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3 m. Compare the emfs of the two cells. Ans: 1.25 7) The emf of a cell is balanced by a length of 120 cm of potentiometer wire. When the cell is shunted by a resistance of 10 Q, the balancing length is reduced by 20 cm. Find the internal resistance of the cell. Ans: r= 2 Ohm 8) A potential drop per unit length along a wire is5x10“V/m. If the emf of a cell balances against length 216 cm of this potentiometer wire, find the emf of the cell. Ans: 0.01080 V 9) The resistance of a potentiometer wire is 8 Q and its length is 8 m. A resistance box and a 2 V battery are connected in series with it. What should be the resistance in the box, if it is desired to have a potential drop ofluV/mm? Ans: 15992 chm 10) Find the equivalent resistance between the terminals of A and B in the network shown in the figure below given that the resistance of each resistor is 10 ohm. (figure from text book) Ans: 14 Ohm 11) 20. A voltmeter has a resistance of 100 Q. What will be its reading when it is connected across a cell of emf 2 V and internal resistance 20 Q? Ans: 1.66 V 20