13.Current Electricity Marks :03/04

Transcription

1 13.Current Electricity Marks :03/04 Q. State and explain kirchhoff s laws for an electric network. Ans:- Kirchhoff s law for an electric network :- In 1842 Kirchhoff s stated two laws to determine currents and potential differences in the complicated circuits 1)Kirchhoff s first law (Current law/ Junction law) The algebraic sum of the electric currents at any junction is always equals two zero. i.e. I = 0 2) Kirchhoff s second law (Voltage law/loop law/mesh law) :- The algebraic sum of potential differences and electromotive forces applied in a close loop in an electric network is zero. i.e. IR + E = 0 Sign conventions:- 1)The e.m.f. of a cell is taken as positive, if the cell sends the current in the direction in which the loop is traced, otherwise e.m.f. is a taken as negative. 2) If the closed loop is traced in the direction of the current then the potential difference across the resistance is taken as negative and in opposite direction of the current, the potential difference is taken as positive For better understanding, we consider following circuit. Sign Conventions :- 1)The currents flowing towards the junction (or incoming currents ) are considered as positive. 2) The currents flowing away from the junction (or outgoing currents) are consider are negative. In the above figure, we get, I 1 + I 2 I 3 I 4 I 5 = 0 I 1 + I 2 = I 3 + I 4 + I 5 Total incoming current = Total outgoing current. Thus, the sum of the currents flowing toward the junction is equal to the sum of the current flowing away from the junction. Kirchhoff s current law deals with the law conservation of charge since there is no accumulation (Gain or lose) of charge at the junction. For loop ABCDEFA, I 1 R 1 + I 2 R 4 E 2 + I 2 R 5 I 1 R 3 + E 1 = 0 I 1 R 1 I 2 R 4 I 2 R 5 + I 1 R 3 = E 1 E 2 E 1 E 2 = I 1 R 1 + R 3 I 2 R 4 + R 5 For loop ABEFA I 1 R 1 (I 1 + I 2 ) R 2 I 1 R 3 + E 1 = 0 E 1 = I 1 R 1 + I 1 + I 2 R 2 + I 1 R 3 = I 1 R 1 + R 3 + I 1 + I 2 R 2 1

2 For loop CBEDC I 2 R 4 I 1 + I 2 R 2 I 2 R 5 + E 1 = 0 E 2 = I 2 R 4 + I 1 + I 2 R 2 + I 2 R 5 E 2 = I 2 R 4 + R 5 + I 1 + I 2 R 2 Kirchhoff s voltage law deals with the law of conservation of energy. Q. Obtain the balancing condition in case of Wheat stone s network. OR With the help of neat diagram obtain the condition for balanced with stone network. Ans:- WheatStone s Network:- galvanometer (Ig).The networks said to be balanced when the potentials at points B and D are same i. e. V B = V D (Ig = 0) Resistances is the network are adjusted to make the galvanometer current zero (Ig = 0).Hence the network is balanced. Apply the kirchhoff s voltage law to the loop ABDA, we get, I 1 R 1 + I g G + I 2 R 3 = 0 I 1 R I 2 R 3 = 0 I 1 R 1 = I 2 R Similarly for loop CDBC, I 2 R 4 + I g G I 2 R 2 = 0 I 2 R I 2 R 2 = 0 I 1 R 2 = I 2 R 4 2 Divide equation (1) by equation (2), we get, I 1 R 1 I 1 R 2 = I 1R 3 I 1 R 4 R 1 R 2 = R 3 R 4 This is the balancing condition for wheatstone s network. The value of unknown resistance can be determined if the remaining three values are known.. Wheatstone s network is used two determine an unknown resistance. It consist of four resistance R 1, R 2, R 3, R 4 connected end to end to form four sides of a quadrilateral ABCD as shown in the above fig. Q. Explain with the neat circuit diagram, how will you determine the unknown resistance by using the Wheatstone s meter bridge what are the errors introduced. While performing this experiment and how are they eliminated. Ans:- Wheatstone s meterbridge experiment :- A source of e.m.f. E is connected between the points A and C. Galvanometer G is connected between the points B and D.. When circuit is closed, current flow in a different branches and some current flows through the 2

3 4) Now, X, R and the resistances of length l x and l R form the four resistances of wheatstone s network. In the balancing condition. X R = Resistance of the wire of the length l x Resistance of the wire of length l R X R = σ l x σ l R X R = l x l R Fig. Wheatstone s meterbridge. Construction :- This is the simple form of Wheatstone s network. It is used to find an unknown resistance X. It consist of a thin uniform wire of length one meter, fixed on a wooden board, connected in series with a battery (E), plug key (K) and a Rheostat (R h ). The ends of the wire are soldered to two thick L shaped copper strips as shown in fig. A third copper strip is placed such that two gaps are formed. The unknown resistance (X) is connected in one gap and the resistance box (R) is connected in other gap. One terminal of the galvanometer (G) is connected to the point C and the other terminal to the jockey which can be touched to different points of the wire AB. Working :- 1)A suitable resistance (R) is taken from the resistance box and the current is sent around the circuit by closing the key (K). X R = l x 100 l x By using this equation, we can determine the value of unknown resistance (X). Errors :- 1) If the wire is not uniform, an error is introduced in the determining the unknown resistance because resistance per unit length may not be constant for it. 2) Another error is introduced due to the contact resistances developed at the points where the ends of the wire are forced to the copper strip. These errors are minimized in the following manner :- 1)The position of X and R are interchanged and the experiment is repeated. 2) The value of R is chosen in such a way that the null point D is obtain at the centre of the wire. Q. With the help of neat circuit diagram, describe the Kelvin s method to determine the resistance of a galvanometer by using Wheatstone s method. 2) Now touch the jockey to the different points of the wire AB and obtain the point D on the wire AB where the galvanometer shows zero deflection (null deflection). This point is the null point. 3) The distance l x and l R from the two ends of the wire are measured as shown in fig. 3

4 Now, G, R and resistances of wire of length l G and l R form four resistances of the wheatstone s network. G R = Resistance of the wire of the length l G Resistance of the wire of length l R Fig. Kelvin s Method G R = σ l G σ l R G R = l G l R Construction :- The electric circuit for Kelvin s method is shown in the above figure. The galvanometer whose resistance is to be determined is connected in one gap and the resistance box R is connected in other gap of Wheatstone s meter bridge. A cell of e.m.f.(e), a plug key (K) and the Rheostat (R h ) are connected in series with the wire AB as shown in the above circuit diagram. A jockey (J) is connected to the point C on the third copper strip. Working :- 1)A suitable resistance (R) is taken from the resistance box. Close the key and take the galvanometer deflection without touching the jockey to any point on the wire AB. The rheostat (R h ) is adjusted to get a suitable deflection in the galvanometer. 2) Now, touch the jockey to the different point of the wire AB and obtain the point D where the galvanometer shows same deflection as before. It means that the galvanometer shows same deflection with or without contact of the jockey. It must be noted that null point is not obtained in this method. Thus, the Kelvin s method is an equal deflection method and not the null deflection method. G R = Potentiometer :- l G 100 l G A potentiometer is device used to measure potential difference by balancing it with a known potential difference. Potentiometer is used :- 1)To measure the e.m.f. of a cell. 2) To compare the e.m.f. of two cells by. i) individual method. ii) Sum and difference method. 3) To determine the internal resistance of cell. Q. State & explain the principle of potentiometer. Principle of potentiometer :- The potential difference between any two points on a potentiometer wire is directly proportional to the length of the wire between that two points. Explanation :- Measure the distances l G and l R from the two ends of the wire AB. 4

5 Consider a long uniform wire AB of length L and resistance R is stretched on a wooden board and a cell is connected across a wire as shown in fig. Let V AB is the potential difference across the wire AB. Explain the use of potentiometer to compare e.m.f. s of two cell using two cells separately. Derive the necessary formula. Ans:-Comparison of two cells by individual method. Let σ is the resistance per unit length of the wire AB. σ = R L R = σ L Let, I is the current passing through the wire V AB = IR.(Ohm s law) I = V AB R I = V AB σl..(1) Consider any point P on the wire such that AP = l. Let, r is the resistance of the wire AP. σ = r l r = σ l Let V AP is the potential difference across the two points A and P of the wire AB. V AP = I r V AP = I σl..(2) Now, Substitute the value of I from equation (1) in equation (2), we get, V AP = V AB σl V AP = V AB L σl l V AP l [ V AB & L are constant] Q. Explain the use of potentiometer to compare the e.m.f. s of two cells by individual method. (Separate cell method). OR Fig:-Comparison of e.m.f. s of two cells by individual method Consider a battery of e.m.f.(e), key (K) and rheostat (R h ) are connected in series with the wire AB, as shown in the above circuit diagram. Let, E 1 and E 2 be the e.m.f. s of two cells which are to be compared by using the potentiometer (or individual method). First take a battery of e.m.f. E 1 and connect it s positive terminal to the battery of e.m.f. E and negative terminal to the jockey through the galvanometer. Touch the jockey on different of the wire AB and find the null point P. The null points means the point at which galvanometer shows zero deflection. If no current flows through the galvanometer then, the potential at point P is same as the potential at negative terminal of battery of e.m.f. E 1. E 1 = fall of potential from A to D = fall of potential along l 1 E 1 = I σ l 1..(1) 5

6 Where, σ = resistance per unit length of wire AB, I = current flowing through wire. Now, replace the cell E 1 by the cell E 2 and find the null point as before. If l 2 is the length of the wire between the point P, in this case then we get E 2 = I σ l 1 (2) Dividing equation (1) by equation (2), we get, E 1 E 2 = I σ l 1 I σ l 2 E 1 E 2 = l 1 l 2 In this way the e.m.f. s of two cells can be compared with the help of potentiometer by individual method. Explain the use of potentiometer to compare the e.m.f.s of two cells by sum & difference method. OR Explain the use of potentiometer to compare the e.m.f. s of two cells by sum & difference method. Derive the necessary formula. OR Describe sum & difference method for comparison of e.m.f. s of two cells. State two precautions in it. Ans:- Fig (a) Sum method (assist each other) (b) Difference method (oppose each other) Consider a battery of e.m.f.(e), Key (K) and Rheostat(Rh) are connected in series with the potentiometer wire as shown in the above fig. Let E 1 and E 2 be the e.m.f. s of the two cells which are to be compared but it should be noted that E 1 is greater than E 2 (E 1 > E 2 ) and E > (E 1 + E 2 ) By using the commutater key, the two cells. E 1 and E 2 are connected in series i. e. in sum method as shown in the above fig. (a).this combination is connected to the jockey through the galvanometer. Touch the jockey on different points of the wire AB and find the null point as P. Measure the length of the wire AP and say it as l 1. Therefore, in this method, E 1 + E 2 = I σ l 1 1 Now, by using the commutater key, connect the two cells E 1 and E 2 in opposite manner i. e. in difference method as shown in fig. Now, their combined e.m.f. is (E 1 -E 2 ). Touch the jockey on different points of the wire AB & find the null point as P.. Measure the length of the wire A P and say it as l 2. Therefore, in this case E 1 E 2 = I. σ l (2) Dividing equation (1) by (2), we get, Fig:-Comparison of e.m.f. s of two cells by sum and difference method 6 E 1 + E 2 E 1 E 2 = I σ l 1 I σ l 2

7 E 1 +E 2 E 1 E 2 = l 1 l 2 By using componendo and dividend, we get, E 1 + E 2 + E 1 E 2 E 1 + E 2 E 1 E 2 = l 1 + l 2 l 1 l 2 2E 1 2E 2 = l 1 + l 2 l 1 l 2 E 1 E 2 = l 1+l 2 l 1 l 2 In this way, the e.m.f/s of two cells can be compared by using sum and difference method. Precautions to use potentiometer:- 1)The e.m.f. of battery E must be greater than the sum of the e.m.f. s of two cells which are to be compared ( i. e. E > (E 1 + E 2 ) and also E 1 >E 2. 2) The positive terminal of E 1 or E 2 of the combination must be connected to that end the of the potentiometer wire where the positive terminal of the battery( driving cell) is connected. 3) The potentiometer wire must be uniform. 4) The resistance of potentiometer wire should be high. Describe with the help of a neat circuit diagram, how will you determine the internal resistance of a cell by using potentiometer. Ans:-Determination of internal resistance of cell Fig.Internal resistance of a cell Consider a battery of e.m.f.(e), key (K 1 ) and rheostat (Rh) are connected in series across the potentiometer wire AB as shown in the above fig. Let E 1 be the e.m.f. of the cell whose internal resistance is to be determined ( E > E 1 ). Connect the positive terminal of E 1 to the positive terminal of E and it s negative terminal to jockey through a galvanometer. A resistance box R and key K 2 are connected across the cell E 1. Initially the key K 2 is open. By closing the key K 1, the current is passed through the wire AB. Now touch the jockey at different point of the wire AB and find out the null point P on it. Measure the length of the wire AP as l 1 Therefore in this case, E 1 l 1 E 1 = I. σ l (1) Take a suitable resistance from the resistance box (R) and key K 2 is closed. Again find out the null point P and measure the length of wire AP as l 2 7

8 In this case, the terminal potential difference of the cell is equal to fall of potential along the length l 2.. V = I. σ l (2) Divide equation (1) by (2), we get E 1 v = l 1 l 2 3 But, E 1 = IR + Ir and E 1 = I R + r V = IR equation (3) becomes r = R I R + r IR R R = l 1 l 2 + r R = l 1 l r R = l 1 l 2 r R = l 1 l 2 l 2 l 1 l 2 l 2 Q. What are the advantages of potentiometer over a voltmeter. 1)Potentiometer can be used for the measurement of very small potential difference accurately. A voltmeter can not be used to measure very small potential difference. ii) Potentiometer is used to measure the e.m..f of a cell while the voltmeter measures the terminal potential difference. iii) The accuracy of voltmeter can be not be increased beyond certain limit while the accuracy of potentiometer can be increase to any extent by increasing the length of the potentiometer wire. 8 iv) Potentiometer is used to find out the internal resistance of a cell but the voltmeter cannot be used for such purpose. Disadvantages of potentiometer:- 1) Voltmeter is a direct reading instrument while a potentiometer is not so. 2) Voltmeter is portable whereas a potentiometer is non-portable. 1)We Know that, V= IR Important Formulae: Type I I = q t =ne t ( Current is rate of flow of charge) 2) If the two resistances R 1 and R 2 are connected a) In series combination R s = R 1 + R 2 b) In parallel combination 1 R p = 1 R R 2 Rp = R 1R 2 R 1 +R 1 The resistance R is given by R = V I SI unit : ohm (Ω) The reciprocal of resistance of the wire (conductor) is called aas conductance (k) k = 1 R SI unit : mho (Ω 1 ) 3) The specific resistance or resistivity (ρ) of the material of the conductor is given by ρ = RA l Where, R is the resistance of the conductor. A its area of cross section L is the length of the conductor

9 SI unit :- ohm-metre (Ωm) The reciprocal of resistivity of the conductor is called conductivity of the conductor σ = 1 ρ = l RA SI unit :- mho/metre 4)If a cell of e.m.f.e and internal resistance r is across an external resistance, the current passing through the circuit is E I = (R+r) Kirchhoff s current law:- I = 0 Kirchhoff s voltage law:- E + IR = 0 Type II 1)Wheatstone s Network:- If R1, R2, R3 and R4 are the resistances forming the wheatstone s network, then the network is said to be balanced if, R 1 R 2 = R 3 R 4 2)Wheatstone s Meterbridge:- In Wheatstone s meterbridge the value of unknown resistance is calculated by the following equation X R = l x l R X = l x l R R X = l x (100 l x ) R 2)Kelvin s method:- The resistance of the galvanometer is calculated by equation G R = l G l R G = l G R l R G = 1)Potentiometer:- l G (100 l G ) R Type III i) If a battery of emf E is connected in series with a resistance R and a potentiometer wire of length L and resistance R p and suppose r is internal resistance of the cell. The current along the potentiometer is given by E I = R+R p + r Therefore, the potential difference across the wire is given by V=I R p = ER p R+R p + r Therefore, the potential drop per unit length of potentiometer wire is given by Potential drop = V L = 1 L ER p R+R p + r 2) If l 1 is the balancing length of the potentiometer wire for the cell of emf R then, E 1 = potential drop per unit length / Balancing length E 1 = ER p R+R p + r l 1 L 2)Comparision of e.m.f. s of two cells by individual method:- If l 1 and l 2 are the balancing lengths for the two cells of e.m.f. s E 1 and E 2 then by the individual method E 1 E 2 = l 1 l 2 2)Comparision of e.m.f. s of two cells by sum and difference method:- If l 1 is the balancing length when the cells assist each other (sum method ) and l 2 are the balancing length when the cells oppose each other (difference method ) then 9

10 E 1 E 2 = l 1 + l 2 l 1 l 2 3)Internal resistance of a cell :- r= R l 1 l 2 l 2 Where, l 1 = the balancing length when the cell is not delivering any current (i.e. it is in an open circuit) l 2 = the balancing length when the cell is sending a current through a resistance R connected in parallel with the cell Problems Type I 1. What length of a wire of specific resistance 2 x l0-6 Ωm and cross-sectional area 5mm 2 should be taken to form a coil of resistance 10Ω? (Ans.25 m) 2. A wire of resistance l6ω is bent into a circle and a cell of e.m.f. 2V and internal resistance 1Ω is connected between two points of the wire, a quarter circumference apart. Determine the current in each segment of the wire. (Ans. l/8 A in longer segment, 3/8 A in shorter segment) 3. A nichrome wire of length 250 cm and cross-section 0.5mm 2 has a resistance of 5.5Ω Detemine the resistivity of nichrorne. (Ans. l.l x l0-6 Ωm) 4. A wire of resistance l2ω is stretched uniformly till its length becomes three times the original length. What is the resistance of the new length of the wire? (Ans. 108Ω) 5. A cell of e.m.f. l.5 V and internal resistance 2Ω has its terminals joined by resistances of 5Ω and 12Ω connected in parallel. Find the current delivered by the cell and current in each resistance. (Ans A, A, A) 6. What would be the reading on a voltmeter of resistance 50Ω when it is connected across a cell of e.m.f. 2V and internal resistance 2.5Ω? (Ans V) 7. What would be the reading on a voltmeter of resistance 40Ω when it is connected in series with a resistance of 7.5Ω and a cell of e.m.f. 1.5 V and internal resistance 0.5 Ω? (Ans Ω) 8. A battery of e.m.f. 4 V and internal resistance 1Ω is connected in parallel with another battery of e.m.f. 1V and internal resistance 1Ω. The combination is used to send current through external resistance of 2 Ω. Calculate the current through the external resistance. (Ans. 1 A) 9. Determine the current in each branch of the circuit shown in Fig. (Ans. l/34 A from B to A; 10/34 A from C tod:9/34a from F to E) 10. Two cells of e.m.f.s 2V and l.5v respectively with internal resistance 1Ω each, are connected in parallel with similar poles joined together. The combination is connected to an external resistance of l0ω. Find the current through the external resistance. (Ans.l/6 A) 11. Determine the current flowing through the galvanometer G, of resistance 8 Ω, shown in Fig. (Ans A) 10

11 12. Twelve equal wires, each of resistance 6 ohms, are connected to form a cube. Find the resistance between two diagonally opposite corners of the cube. (Ans.5Ω) 13. The current flowing through an external resistance of 5Ω is 1A when. it is connected to the terminals of a cell-the current reduces to 0.6 A when the external resistance is 10Ω. Find the internal resistance of the cell using kirchhoff s law. (2.5Ω) 14. The resistance of a wire of length 40m and radius 0.25 mm is 10Ω. Find the specific resistance and conductivity of the material of the wire. (Ans x 10-8 ohm metre, k=2.037 x 10 7 siemens / metre) Type II 15. Four coils of resistances 3Ω,6Ω,9Ω and 30Ω respectively are arranged to form a Wheatstone's bridge. Determine value oft the resistance with which the coil of 30Ω, resistance should shunted,so as to balance the bridge.(ans. 45 Ω) 16. The resistances in the four arms of wheatstone's bridge are 10Ω, 20Ω, 30Ω and 40Ω respectively. What modificatibn in the fourth arm is necessary to balance the bridge? (Ans. Connect a resistance of 20Ω in series with 4OΩ resistance) 17. The resistances in the arms of a wheatstone s meterbridge are 4Ω,6Ω,16Ω and 24Ω respectively.if the e.m.f. of the cell is 2V,find the current passing through the cell.(ans.0.25 A) 18. In a Wheatstone bridge, the four resistances in the arms of the bridge are AB = 2Ω, BC = 4Ω, AD = 1Ω and DC = 3Ω. The terminals of a cell of e.m.f. 2 volts and negligible resistance connected by wire of negligible resistance to A and C. If a galvanometer of resistance l0ω is connected between B and D, find the current in the galvanometer. 19. In a balanced metre bridge, the segment of wire opposite to a resistance of 30 ohms is 30 cm. Calculate the unknown resistance. (Ans.70Ω) 20. An unknown resistance X is placed in the left gap and a known resistance of 60 ohms is placed in the right gap of a metrebridge. The null point is obtained at 40 cm from the left end of the bridge. Find the unknown resistance. (Ans. 40Ω) 21. When a resistance of 60 Ω is joined in the left gap and standard resistance in the right gap of a metre bridge, the null point is obtained at 60 cm from the left end. If the known resistance is replaced by 40 Ω resistance, where would be null point? (Ans. 50 cm from the left end) 22. A resistance of 8Ω is connected in the left gap of a bridge. In the right gap, two resistances of 18Ω and 36Ω, are connected in parallel. Find the position of the null point. (Ans..40 cm from left end) 23. In a metre bridge experiment, with resistance R 1 in left and resistance X is right gap, null point is obtained at 40 cm from the left end. With a resistance R 2 in left gap and the same resistance X in right gap, null point is obtained at 50 cm from the left end. what will be the null point, if R 1 and R 2 are put in series in the left gap, the right gap containing X? (Ans. 62.5cm from left end) 24. Two coils are connected in series in one gap of a bridge and the null point is obtained in the middle of the wire by putting a resistance of 75 ohms in the other gap. The two coils are connected in parallel and the null point is obtained in the middle of the wire by connecting a resistance of l8 ohms in the other gap. Find the resistance of each coil. (Ans. 45Ω and 30Ω) 25. An unknown resistance X is connected in the left gap and known resistance R in the right gap of the metre bridge. The balancing point is obtained at 60 cm from the left end of the (Ans. 2/145 A) 11

12 wire. when R increased by 2 ohm, the balance point shifts by l0 cm. Find X and R. (Ans.X=6Ω,R=4Ω) 26. Two coils are connected in series in one gap of Wheatstone s meterbridge and null point is obtained at the centre of the wire, with a resistance of 100 ohms in the other gap. when the two coils a connected in parallel in the same gap, the known resistance has to be changed by 84 ohm to obtain the null point at the centre once again. calculate the resistance of the two coils. (Ans. 80Ω and 20Ω) 27. With two unknown resistances in series in the left gap of metre bridge and a resistance of 27Ω in the right gap, the null point is obtained at 40 cm from the left end. If the unknown resistances are connected in parallel, in the left gap, a resistance of 6Ω in the right gap gives the same null point as before. Determine the unknown resistances. (Ans. l2ω & 6Ω) 28. With an unknown resistance in one gap and a known resistance of 80 Ω in the other gap of metre bridge, the null point is found. The unknown resistance is then shunted by a resistance l/l9th of its value. Calculate the known resistance which, when connected in other gap, would give the same null point as before. (Ans. 4Ω) 29. Two resistances R 1 and R 2 connected in the left and right gap of a metre bridge respectively, give a null point dividing the wire in the ratio 4:3.If each resistance is increased by 5Ω, the null point divides the wire in the ratio 5 : 4. Determine each resistance. (Ans. 20Ω and 15Ω) 30. A meter bridge is balanced by putting 20 Ω resistance in the left gap and 40 ohm in the right gap. If 40 ohm resistance is not shunted with 40 Ω resistance, find the shift in the null point. (Ans cm towards right) 31. An unknown resistance is placed in the left gap of a meterbridge and a resistance 'R' is placed in the right gap. The null point is obtained at a distance of 40 cm from the left end. When the resistance of l0 ohm is connected in series with the unknown resistance and the same resistance 'R' is kept in right gap, the null point is obtained at the centre of the wire. Calculate the unknown resistance. (Ans.20Ω) 32. A wire of uniform cross-section is bent in the shape of complete ring. Two diametrically opposite points on the wire an connected to the two terminals of left gap of a meter bridge. The resistance of l5 ohm is connected in the right gap. If the null point is obtained at 70 cm from the left end of metre bridge wire, find the resistance of the wire before bending it to a ring. (Ans. 140 Ω ) 33. A resistor of 10Ω is connected in the left gap of a metre bridge Two resistors of 20Ω and 16Ω are connected in parallel in the right gap. Find the position of the null-point on the bridge wire. (52.5 cm from left end) 34. Two resistances of values 20Ω and 30Ω are connected in left an right gap of a metre bridge. Determine the shift in null point when the resistance of 20Ω is shunted by another resistance of 2OΩ. (15 cm to left) 35. With resistances X and Y ohm in the left and right gap respectively of a metre bridge, the null point is obtained at 30cm from the left end of the wire. When Y is shunted by a l5ω resistor, the shift in the null point is l0 cm. Find X and Y. (Ans. 25/7Ω; 25/3Ω) 36. Equal lengths of wires A and B are connected in the left and the right gap respectively of a metre bridge. The nullpoint is obtained at 0.4 m from the left end of the bridge wire. If the diameters of the wires A and B are in the ratio 2 : 3, compare the specific resistances of the materials of the wires. (Ans. p A : P B :8 :27) 37. In Kelvin's method of determination of the resistance of a galvanometer, the galvanometer is connected in the left gap and a resistance of 60Ω in the right gap of the Wheatstone metre bridge. If the balance point is obtained at 60 cm from the left end of the wire, find the resistance of the galvanometer. (Ans. 90 Ω) Type -III 12

13 38. A potentiometer wire has a length of 1.5 m and resistance of l0 ohms. It is connected in series with a cell of e.m.f.4 volts and internal resistance 5 ohms. Calculate the potential drop per unit length cf the wire. (Ans V/m or V/cm) 39. A potentiometer wire of length 4m has a resistance of 4ohms. What resistance must be connected in series with the wire and an accumulator of e.m.f. 2 volts, so as to get a potential drop of l0-3 V/cm along the wire? (Ans. 16Ω) 40. An accumulator of e.m.f. 2V and internal resistance 2Ω is connected between the terminals of a potentiometer wire of length 5m md resistance 8 Ω. Find the balancing length for a cell of e.m.f. 1.5 V. (Ans cm) 41. A potentiometer wire is l0 metres long and a potential difference of 6 volts is maintained between its ends. Find the e.m.f. of a cell which balances against a length of 180 cm of the potentiometer wire. (Ans V) 42. A potentiometer wire of length l0 m has a resistance of l0ω. A battery of e.m.f. 4V and negligible internal resistance is connected across the wire. Determine the potential drop per cm of the wire. If the balancing length for a cell of unknown e.m.f. is 3.75 m, find the unknown e.m.f. (Ans V/cm, 1.5 V) 43. Resistance of a potentiometer wire is 1 ohm per metre.a Daniel cell of e.m.f volts balances at 2l6cm on this potentiometer.calculate the current through the wire. Also calculate the balancing length of another cell of e.m.f. 1.5 volts on the same potentiometer. (Ans. 0.5 A, 300 cm) 44. A potentiometer wire of length 8 m and resistance 16Ω is connected in series with a resistance box and an accumulator of e.m.f.2v and neglibible resistance. What resistance should there be in the resistance box, so that the potential drop per e.m.f. 4 volts and internal resistance 2 ohms. Find the potential drop per cm of the wire. What length of the wire will balance a cell of e.m.f. 1.5 V? (Ans x l0-3 V/cm, 180 cm) 46. In a potentiometer experiment, the balancing length, when a cell of unknown e.m.f. and a cell of e.m.f. 1.5 V were connected so as to assist each other, was found to be 7 m; while the balancing length when the cells were connected so as to oppose each other, was found to be lm. Calculate the unknown e.m.f. (Ans.2V) 47. In a potentiometer experiment, the balancing length for a cell is 250 cm. On shunting the cell by a resistance of 10Ω, the balancing length was found to be 200cm. Calculate the internal resistance of the cell. (Ans. 2.5 Ω ) 48. A potentiometer wire of length 4m has a resistance of 8ohms. It is connected in series with a battery of e.m.f. 2 volts and negligible internal resistance. If the e.m.f. of a cell balances against a length 217 cm of the wire, find the e.m.f. of the cell. When the cell is shunted by a resistance of l5ohms, the balancing length is reduced to 200 cm. Find the internal resistance of the cell (Ans V, 1.275Ω) 49. When a resistor of 5 ohms resistance is connected across a cell, its terminal potential difference is balanced by 150 cm. of potentiometer wire and when a resistor of l0 ohms resistance is connected across the cell, the terminal potential difference is balanced by 175 cm of the potentiometer wire. Calculate the internal resistance of the cell. (Ans. 2Ω) 50. The e.m.f. of a cell is balanced by the length of m of a potentiometer wire, in the potentiometer experiment. When a resistance of 5 Ω is connected across the cell, the balancing length is found to be l.5l m. Find the internal resistance of the cell. (Ans. 1 Ω) 51. A wire of resistance 20Ω is uniformly stretched until its length becomes double of its original length. What will be its cm of the wire is 20 micro-volts? (Ans. 1984Ω) new resistance? (Ans.80Ω) 45. A potentiometer wire has a length of 4 metres and a 52. The resistance of a potentiometer wire is l0ω and its resistance of l0 ohms. The wire is connected to a battery of length l0 metre. A resistance box and 2V accumulator are 13

14 placed in series with it. What should be the value of the resistance in the box, if it is desired to have potential drop of 1 microvolt/mm? (Ans Ω) 53. In a potentiometer experiment, e.m.f. of a cell is balanced by a length of 150 cm on the potentiometer wire. When the cell its shunted by a 10 Ω resistance, the balancing length reduces to 90 cm. What is the internal resistance of the cell? (Ans Ω) 54. When a resistor of 5 Ω is connected across a cell, is terminal potential difference is balanced by 150 cm of a potentiometer wire and resistor of l0 Ω resistance is connected across the cell, the terminal potential difference is balanced by 175 cm of the same potentiometer wire. Find the balancing length when the cell is on open circuit and the internal resistance of the cell. (Ans. 210 cm, 2ohm) 55. A potentiometer wire has length 10m and resistance 20 ohm. Its terminals are connected to a cell of e.m.f.5v and internal resistance 5 Ω. What are the distances at which null points areobtained when two cells of e.m.f. 1.5 V and 1.3 V are connected so as to (a) assist, (b) oppose each other. 59. In the given figure, find each branch and the P.D. acris the 5Ω resistance. (16/A; 9/17A; 7/17A; 35/17V) 60. Find the length of metal wire of diameter 0.5 mm needed to prepare a coil of resistance of 10Ω. Specific resistance of metal = 4.4 x 10-7 Ωm (Ans m) 61. Determine the current flowing through the galvanometer G as shown in figure (Ans. (a) 7m (b) 0.5 m) Problems for Home work 56. Find the length of the wire of diameter 1mm needed to prepare a coil of resistance 25 ohm. (Specific resistance of the metal = 4.4 x l0-7 Ωm) (Ans m) 57. A wire of length 90 cm and area of cross-section 0.5 mm 2 has a resistance of 2Ω. Find the specific resistance of wire. (Ans. l.ll x l0-6 ohm- metre) 58. A cell of E.M.F. of 3 volt and internal resistance 4 ohm is connected to two resistances of 10 ohm and 24 ohm joined in parallel. Find the current through each resistance using (Ans. l/22 A) 62. Two cells of E.M.F. 3 volt and 1 volt and each of internal resistance 2 ohm are connected in parallel to send a current through an external resistance of 5 ohm. Find the current in the 5 ohm resistance. (Assume that the positive terminals of cells are connected together) (1/3 A) 63. The specific resistance of aluminium is 2.65 x 10-8 ohm metre. Find the resistance of an aluminum wire of length 100m and diameter 0.25 mm. (Ans Ω) Kirchhoffs Laws. (0.19A;0.0794; 0.272A) 14

15 64. A metal wire has length 3.14m and diameter 2mm. If the resistance of the wire is 0.01Ω, find the resistivity of the metal. (Ans Ωm) 65. A metal wire of length 100cm and area of cross section 0.5 mm 2 has resistance 2.2 x10-2 Ω. Find the specific resistance of the metal. (Ans. 1.1 x 10-8 Ωm) 66. What length of a metallic wire of area of cross section 5 x 10-2 mm 2 should be taken to prepare a coil of resistance 10Ω? The specific resistance of the metal is 2x 10-8 Ωm. (Ans. 25m) 67. Find the length of a metal wire of radius 0.15 mm needed to prepare a coil of resistance 10Ω. The resistivity of the metal is 2.63 x 10-8 Ωm. (Ans m) 68. Find the (i) resistivity (ii) conductivity of a metal wire of resistance 0.2Ω having area of cross section 1 x 10-4 cm 2 and length 10cm. (Ans. (i) 2 x 10-8 Ωm, (ii) 5 x 10 7 siemens / metre) 69. A wire of resistance 25Ω is uniformly stretched until its length becomes three times its original length. What will be its new resistance. (Ans. 225Ω) 70. The resistance of a wire is 100 Ω. Find the resistance of the wire when it is stretched so that its length increases by 1%. Assume that the volume of the wire remains constant. (Ans. 102 Ω) 71. Two wires p and Q have resistance in the ratio 1:4, lengths in the ratio 1:2 and radii in the ratio 3:1. Compare the specific resistance of their materials. (Ans. p : Q =9:2) 72. A uniform wire 1m long and having diameter 0.3 mm, has resistance 4.2Ω. Calculate the resistance of another wire of the same material having length 1.5m and diameter 0.15 mm. (Ans. 25.2Ω) 73. A voltmeter has a resistance 3990Ω. What will be the potential difference between its terminals when it is connected across a cell of e.m.f. 2V and having internal resistance 10Ω? (Ans V) Three coils having resistances 10Ω, 12Ω and 15Ω are connected in parallel. This combination is connected in series with a series combination of three coils having the same resistances. Calculate (i) the total resistance (ii) the current through the circuit if a battery of e.m.f. 4.1V is used for drawing current. (Ans. (i) 41Ω (ii) 0.1A) 75. A current of 3A flows through a certain resistance when a cell is connected across it. The potential difference across the resistance is found to be 4.8V. The e.m.f. of the cell is 5V. Calculate the internal resistance of the cell. (Ans. 0.2/3Ω = x 10-2 Ω) 76. Two cells of e.m.f.s 1.5V and 2V having internal resistance 1Ω and 2Ω respectively are connected in parallel, with their similar poles joined together, to send a current through an external resistance of 5Ω. Find the current thourgh the external resistance. (Ans. 5/17 A = A) 77. The resistance in the four arms of a Wheatstone s bridge are 15Ω, 10Ω, 10Ω and 10Ω respectively. What modification is necessary to balance the bridge?(ans. Connect a resistance of 30Ω in parallel with 15Ω resistance) 78. Four resistances 4, 4,4, and l2 Ω. Find the resistance which when connected across the l2 ohm resistance to balance the network. (Ans.6Ω) 79. Resistances P = 10Ω, Q = l5ω, S = 50Ω and R = 25Ω are connected in order in the arms AB, BC, CD and DA respectively of a Wheatstone' s network ABCD. The cell is connected between A and C. What resistance has connected in parallel with S to balance the network? (150Ω) to be 80. Two coils are connected in series in one gap of a metre bridge and null point is obtained at the middle of the wire by ptrtting 50Ω resistance in the other gap. The two coils are then connected in parallel atrd it is found that the resistance in the other gap has to be decreased by 38Ω to get the null point

16 at the same place as before. Find the resistance of the coils. (30Ω and 20Ω) 81. In a balanced metre bridge, the segment of wire opposite to a resistance of 40Ω is 40cm in length. Determine the unknown resistance. (Ans. 60Ω) 82. An unknown resistance is connected in the left gap and a resistance of 50Ω in the right gap of a metre bridge. The null point is obtained at 40cm from the left end of the wire. Determine the unknown resistance. (Ans. 100/3 Ω = Ω) 83. Four resistance 5Ω, 10Ω, 15Ω and X are connected in series so as to form Wheatstone s network. Determine X if the network is balanced with these resistance. (Ans. 7.5Ω) 84. In a metre bridge experiment, with a resistance R 1 in the left gap and a resistance X in the right gap, the null point is obtained at 60cm from the left end of the wire. With a resistance X in the right gap, the null point is obtained at 50cm from the left end of the wire. Find the null point if R 1 and R 2 are connected in series in the left gap and the right gap contains X. (Ans. At 71.4 cm from the left end of the wire.) 85. In a metre bridge experiment, with a resistance R 1 in the left gap and a resistance X in the right gap, the null point is obtained at 60 cm from the left end of the wire. With a resistance R 2 in the left gap and the resistance X in the right gap, the null point is obtained at 50cm from the left end of the wire. Find the null point if R 1 and R 2 are connected in parallel in the left gap and the right gap contains X. (Ans. At 37.5cm from the left end of the wire) 86. Two equal resistances are introduced in two gaps of a metre bridge. Find the shift in the null point if an equal resistance is connected in series with the resistance in the left gap. (Ans cm to the right ) 87. A resistance of 60Ω is connected in the left gap and a resistance X in the right gap of a metre bridge. The null point is obtained at 60 cm from the left end of the wire. Find the 16 shift in the null point when a resistance of 120Ω is connected in parallel with the 60Ω resistance and the right gap contains the same resistance X. (Ans. Shift in the null point: 10 cm towards the left end of the wire) 88. Two resistances X and Y in the two gaps of a metre bridge give a null point dividing the wire in the ratio 2: 3. If each resistance is increased by 30 ohms, the null point divides the wire in the ratio 5 : 6. Find X and Y. (3 marks) (Ans. X =20Ω,Y = 30Ω) 89. Two resistances X and Y are connected in the left and the right gap respectively of a metre bridge. A null point was found on the bridge wire such that the ratio of the lengths of two segments of the wire is 2 : 3. The distance of the null point was measured from the left end of the wire. When X is changed by 20 Ω the position of the null point divides the wire into two segments of lengths in the ratio 1 :4. Determine X and Y. (Ans. X : 32 Ω, Y :48 Ω ) 90. An unknown resistance X is connected in the left gap and a known resistance R in the right gap of a metre bridge. The balance point is obtained at 60 cm from the left end of the wire. When R isincreased by 20Ω, the balance point shifts by 10 cm. Find X and R. (Ans. X : 60 Ω, R : 40 Ω) 91. Four resistances 10 Ω, 10 Ω, 10 Ω and 20 Ω form a Wheatstone network. Calculate the shunt required across the 20Ω resistor to balance the network. (Ans. 20 Ω) 92. Resistances P = 10 Ω, Q= 15 Ω, R=25 Ω and S=50 Ω are connected in the arms AB, BC, CD and DA respectively of awheatstone network ABCD. A cell is connected between the points A and C.What resistance has to be connected in parallel with S to balance the network? (Ans. 25 Ω ) 93. Resistances 20 Ω and 30 Ω are connected in the left gap and right gap respectively of a metre bridge. Determine the nuli point when the resistance of 20 Ω is shunted by another resistance of 20 Ω. (Ans. At 25 cm from the left end of the wire) 94. Two resistances Prepared from the wires of the same material having diameters in the ratio 3 : 1 and

17 lengths in the ratio 3 : 1 are connected in the left and the right gaps respectively of Wheatstone's metre bridge. Determine the distance of the null point from the left end of the bridge wire. (Ans.25 cm) 95. A potentiometer wire has a length of 2 m and a resistance of 10 ohm. It is connected in series with a cell of e.m.f. 4 volt and internal resistance 6 ohm. Find the potential gradient on the wire. Find also where a cell of e.m.f. 1 volt will balance on the wire. (1.25V/m; 0.8m) 96. In a potentiometer experiment, the balancing length for a cell of e.m.f. E volt is 165cm. For another cell whose e.m.f. differs from the first cell by 0.1V, the balancing length is 160cm. Find the e.m.f. of each cell. (3.3V, 3.2V) 97. A potentiometer wire has length 2m and resistance 10Ω. It is connected in series with a resistance of 990Ω and a cell of e.m.f. 2V. Calculate the potential gradient along the wire. (Ans. 1 x 10-2 V/m) 98. A cell of e.rn.f. 2 V and negligible internal resistance is connected to a potentiometer wire of length 4 m and resistance 25Ω to form a closed circuit. Find the potential gradient along the wire. (Ans.0.5V/ m) 99. A cell of e.m.f. 2 V and internal resistance 2 Ω is connected to a potentiometer wire of length 2 m and resistance 8Ω to form a closed circuit. Find the (i) potential gradient along the wire(ii) balance length for a cell of e.m.f.1.2v. [Ans. (i) 0.8 V/m (ii) 1.5 m] 100. In a potentiometer circuit, the length of the wire is 4 m. When two ceils of e.m.f.s E 1 and E 2 are connected in series so as to assist each other, the balance length is found to be 2.5 m. When the cells are connected in series so as to oppose each other, the balance length is found to be 0.5 m. Compare the e.m.f.s of the two cells. (Ans. E 1 : E 2 = 3 : 2) 101. A potentiometer wire is 5 m long and a potential difference of 4 volts is maintained between its ends. Find the e.m.f.of a cell which is balanced against a length of 240 cm of the potentiometer wire. (Ans V) 102. Two cells having e.m.f.s E 1 and E 2 (E 1 > E 2 ) are connected in a potentiometer circuit so as to assist each other. The null point is obtained at 8.125m from fie higher potential 17 end. When the cell of the e.m.f. E 1 is connected so as to oppose the cell with e.m.f. E 2, the null point is obtained at 1.25 m from the same end. Compare the e.m.f.s of two cells. (Ans. E 1 :E 2 = 15: 11 OR E 1 /E 2 = ) 103. In a potentiometer experiment to compare the e.m.f.s of two cells by the sum and difference method, the balancing lengths of wire are found to be 240 cm and 60 cm respectively. Determine the ratio of the two e.m.f.s. (Ans. E 1 : E 2 = 5 : 3) 104. A potentiometer wire of length 4 m has a resistance of 4 ohms. What resistance must be connected in series with the wire and an accumulator of e.m.f. 2 volts and internal resistance 2 ohms so as to obtain a potentiometer drop of 10-3 volt/cm of the wire? (Ans. 14 Ω) 105. The resistance of a potentiometer wire is 10 Ω per metre. A cell of e.m.f. 1.5 volt balances at 300 cm on this potentiometer. (i) Find the current through the wire. (ii) Find the balance length for another cell of e.rn.f. 1.4 volt on the same potentiometer.[ans. (i) 0.05 A (ii) 280 cm] 106. In a potentiometer experiment, the balancing length is found to be 1.80 m for a cell of e.m.f. 1.5 V. Find the balancing length for a cell of e.m.f. 1 V. (Ans m ) 107. The potential drop per unit length of a potentiometer wire is 5 x 10-3 V / cm. If the e.m.f. of a cell balances against a length of 276 cm of the wire, find the e.m.f. of the cell. (Ans V ) 108. A potentiometer wire of length 10 m and resistance 9 Ω is connected to a battery of e.m.f. 2.1 V and internal resis tance 1.5 Ω. Find (i) the potential gradient along the wire (ii) the balancing length for a cell of e.rn.f V. [Ans. (i) 0.18 V/m (ii) 6 m] 109. The resistance of a potentiometer wire is 0.1 Ω/cm. A cell of e.m.f. 1.5 V balances at 300 cm on this wire. Find the balancing length for another cell of e.m.f. 1.2 V on the same wire. (Ans. 240 cm ) 110. A potentiometer wire of length 4 m has some resistance. The resistance connected in series with the wire and an accumulator of e.m.f. 2V is 16Ω. If the potential gradient

18 along the wire is 10-3 V/cm, find the resistance of the wire. The internal resistance of the accumulator is negligible.(ans. 4 Ω) 111. A potentiometer wire of length 10 m and resistance 20 Ω is connected in series with a battery of e.m.f. 4V and obtained internal resistance 5Ω. What are the distances at which the null points are obtained when two cells of e.m.f.s 1.5V and 1.3V are connected so as to (i) assist and (ii) oppose each other? [Ans. (i) I 1 = 8.75m (ii) I 2 = 0.625m] 112. The resistance of a potentiometer wire is1 Ω / m. A cell of e.m.f. 1.4 V balances against a length of 280 cm on the wire. (i) Find the current through the wire. (ii) Also find the balancing length for another cell of e.m.f V on the same wire. [Ans. (i) 0.5A (ii) 216 cm] 113. A potentiometer wire of length 2 m and resistance 5 Ω is connected in series with a resistance of 998 Ω and a cell of e.m.f. 2 V and internal resistance 2 Ω. Find (i) the potential drop along the wire (ii) the length required to balance a potential difference of 4 mv (4 marks) [Ans. (i) x 10-3 V (ii) m ] 114. A potentiometer circuit is made by connecting a wire 4 m long in series with an accumulator of e.m.f. 2 volts and an external resistance of 3994Ω. The resistance of the potentiometer wire is 4 e and the internal resistance of the cell is 2 Ω. It is found that at a certain temperature, the e.m.f. of thermocouple is balanced on a 3 m length of the potentiometer wire. Find the thermo e.m.f(ans. 1.5 x 10-3 V) 115. The length of a potentiometer wire is 10 m. It is connected in series with an accumulator. The e.m.f. of a cell balances against 250 cm of the wire. If the length of the potentiometer wire is increased by 1 m, calculate the new balancing length of the wire. (Ans. 275 cm) 116. An accumulator of e.m.f. 2 V and internal resistance 1 Ω is connected to a potentiometer wire of length 4 m and resistance 24 Ω. Find (i) the resistance to be connected in series with the wire so that the potential gradient along the wire is A.24 V lm (ii) the balancing length for a cell of e.m.f V. [Ans. (i) 25 Ω (ii) 3.5 m] A cell of e.m.f V is balanced by a length of 150 cm of a potentiometer wire. When the cell is shunted by a resistance of 4 Ω, the balancing length reduces to 120 cm. Find the internal resistance of the cell. (Ans. 1 Ω) 118. The e.m.f. of a cell is balanced by a length of 120 cm of a potentiometer wire. When the cell is shunted by a, resistance of 10 Ω the balancing length i is reduced by 20 cm. Find the internal resistance of the cell.(ans. 2 Ω) 119. A cell balances against a length of 250 cm of a potentiometer wire when it is shunted by a resistance of 10Ω. The balancing length reduces to 200 cm, when it is shunted by a resistance of 5 Ω. Calculate the internal resistance of the cell. (Ans. 10/3Ω = 3.333Ω ) Board Book Problems 120. Three cells are connected in parallel with their like poles connected together with wires of negligible resistance. The e.m.f. s of the cells are 2V, 3V and 4V respectively and their internal resistance are 1Ω, 2Ω and 3Ω respectively. Find the current through each branch (or each cell) The resistances P = 10Ω, Q= 15Ω, R = 25Ω and S = 50Ω are connected in the arms AB, BC, CD and DA respectively of a Wheatstone s network ABCD. The cell is connected between points A and C. What resistance has to be connected in parallel with S to balance the network? 122. Find the current through galvanometer in the following fig., if resistance of galvanometer is 200Ω. (Ans :- Ig = A) 123.Two diametrically opposite points of a metal ring are connected to two terminals of the left gap of meter

19 bridge. In the right gap, resistance of 15Ω is introduced. If the null point is obtained at a distance of 40 cm from the left end, find resistance of the wire bent in the shape of the ring.(ans :- X=40 Ω) 124. Two resistance of values 20Ω and 30Ω are connected in left and right gap of meterbridge. Find the shift in null point when resistance of 20 Ω is shunted by another resistance of 20 Ω. (Ans :- Shift in the null point is 15 cm towards left) 125. Two resistance X and Y are connected in the left and right gap of meter-bridge. A null point was found on the bridge wire such that the ratio of lengths of two segments of wire is 2:3. The distance of the null point was measured from the left end of the wire. When the value of X is changed by 20Ω, the position of null point divides the wire into segments of lengths in the ratio 1 : 4. Determine X and Y. (Ans :- X=32 Ω and Y=40 Ω) 126. Two coils are connected in series in one gap of the Wheatstone s meterbridge and null point is obtained at the center with the resistance of 100Ω in the other gap. When two coils are connected in parallel in same gap the known resistance is to be changed by 84 Ω to obtain the null point at the centre again. Calculate the resistance of the coils.(ans :- R 1 =20 Ω and R 2 =80 Ω) 127.A potentiometer wire has a length 4 m and resistance 4Ω. What resistance must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient of 10-3 V/cm on the wire? (Ans :- R e =16 Ω) 128, The resistance of a potentiometer wire is 0.1Ω per cm. A cell of e.m.f. 1.5 V balances at 300 cm on this potentiometer wire. Find the current and balancing length for another cell of e.m.f. 1.4 V on the same potentiometer wire. (Ans :- l 2 =2.8 m) 129. The length of potentiometer wire is 10 m and is connected in series with an accumulator. The e.m.f. of a cell balance against 250 cm length of wire. If the length of potentiometer wire is increased by 1m, calculate the new balancing length of wire. (Ans :- l 2 =2.75 m) 130. A cell balances against a length of 250 cm on potentiometer wire when it is shunted by a resistance of 10Ω.The balancing length reduces by 50 cm, when it is shunted by a resistance of 5Ω. Calculate the balancing length when the cell is in open circuit and the internal resistance of a cell. (Ans :- r=3.333 Ω) 131. Two resistance prepared from the wire of the same material having diameters in the ratio 3:1 and lengths in the ratio 3:1 are connected in the left and right gap of 19 Wheatstone s meter-bridge. Determine the distance of null point from the left end of wire. (Ans :- l =25 cm) 132.With a resistance R 1 in the left gap and a resistance R 2 in the right gap of a meter-bridge, the null point is obtained at a distance of 70 cm from the left end. When R 1 is reduced by 2Ω and R 2 is increased by 2Ω, the neutral point is obtained at 30 cm from the left end. Find the values of resistance R 1 and R 2 (Ans :- R 1 =3.5 Ω and R 2 =1.5 Ω) 133. A Skelton cube is made of 12 conductors, each of resistance 6 Ω and connected to cell. (Ans :- 5 Ω) 134. Find the currents through different branches and P.D. across 10 Ω resistor in the network / circuit shown in the following figure. What is the P.D. across BE? PROBLEMS FOR PRACTICE 135. A cell of e.m.f. 3V and internal resistance 4Ω is connected to two resistances 10Ω and 24Ω joined in parallel. Find the current through each resistance using Kirchhoff s laws. ( Ans : I = A, I I = A, I 24 = A) 136. The current flowing through an external resistance of 2Ω is 0.5 A when it is connected to the terminals of a cell. This current reduces to 0.25 A when the external resistance is 5Ω. Use Kirchhoff s laws to find e.m.f. of cell. ( Ans : E = 1.5 V, r = 1Ω) 137. Four resistances 5Ω, 10Ω, 15Ω and an unknown XΩare connected in series so as to form Wheatstone s network. Determine the unknown resistance X,if the network is balanced with these numerical values of resistances. ( Ans : X = 7.5 Ω) 138. In a meter-bridge experiment with resistance R 1 in left gap and resistance X in right gap, null point is obtained at 40 cm from left end. With a resistance R 2 in left gap and same resistance X in right gap, null point is obtained at 50 cm from left end. Where will be null point if R 1 and R 2 are put first in series and then in parallel, in the left gap and right gap still containing X? ( Ans : For series 62.5 cm from left end, For parallel cm from left end) 139. A potentiometer wire has a length of 2m and resistance 10Ω. It is connected in series with resistance 990Ω and a cell of e.m.f. 2V. Calculate the potential gradient along the wire. ( Ans : V L = 10 2 V/m)