CURRENT ELECTRICITY. Q1. Plot a graph showing variation of current versus voltage for a material.

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1 CURRENT ELECTRICITY QUESTION OF ONE MARK (VERY SHORT ANSWER) Q. Plot a graph showing variation of current versus voltage for a material. Ans. Q. The graph shown in the figure represents a plot of current versus voltage for a given semiconductor. Identify the region, if any, over which the semiconductor has a negative resistance. Ans. Region BC shows negative resistance. Q3. Plot a graph showing the variation of resistance of a conducting wire as a function of its radius, keeping the length of the wire and its temperature as constant. l Ans3. R = ρ πr, R α r

2 Q4. Define the term Mobility of charge carriers in a conductor. Write its S.I. unit. Ans4.Mobility is defined as the magnitude of the drift velocity acquired by it in a unit electric field. μ = Iv d I E, S.I. unit m V - s -. Q5. The V I graph for two ohmic conductors are as shown in figure. What is the ratio of resistances of conductor and conductor in terms of θ and θ. Ans5. cot θ / tan θ Q6. A copper wire of length m and radius mm is joined in series with an iron wire of length m and radius 3mm and a current is passed through the wires. What will be the ratio of the current density in the copper and iron wires? Ans6: J J = I π r I πr = 9 : Q7. A wire 50cm long and mm in cross-section carries a current of 4A when connected to a V battery. What is the resistivity of the wire? Ans7: ρ=r A l =V I A l = x0-6 Ω m Q8. Show variation of resistivity of Si with temperature in a graph. Ans8. Resistivity of Si decreases rapidly with increasing temperature. Q9. Two wires one of maganin and the other of copper have equal length and equal resistance. Which one of these wires will be thicker? Ans9. R = ρ l πr

3 Resistivity ρ of maganin is much greater than that of copper, therefore to keep same resistance R for same length of wire, the maganin wire must be thicker. Q0. The variation of potential difference V with length L in case of two potentiometer P & Q is as shown in fig. Which of these two will you prefer for comparing emfs of two primary cells? Ans0. The potential gradient (V/ l) must be small. The slope V/l is smaller for a potentiometer Q, hence we shall prefer potentiometer Q for comparing the emfs of two cells. Q. Why do we prefer a potentiometer to measure e.m.f of a cell rather than a voltmeter? Ans. A potentiometer does not draw any current from the cell whose emf is to be determined, whereas a voltmeter always draws some current. Therefore, emf measured by voltmeter is slightly less than actual value of emf of the cell. Q.A potential difference V is applied to a copper wire of length l and thickness d. What will be the drift velocity when V is doubled? Ans: v d = ee m τ = e m τ(v l ) or v d V The drift velocity will be doubled Q3.A capacitor of 0µF has a potential difference of 40 volts across it. If it is discharged in 0. seconds, what is the average current during discharge? Ans3. Current I= q t = CV t =x0-3 A= ma Q4.A charge of x0 - C moves at 30 revolutions per second in a circle of diameter 80cm.What is the current linked with the circuit? Ans4: I= q/t = q ν = x0 - x30=0.60a Q5.A steady current is set up in a metallic wire of non-uniform cross-section.how is the rate of flow of electrons(r) related to the area of cross-section (A)? Ans5: Rate of flow of electrons will depend upon drift velocity and drift velocity is inversely proportional to area of cross-section, i.e. v d. Therefore R / A A

4 Q6.In the circuit shown potential difference between X and Y is Ans6: As the circuit is open, therefore no current flows through the circuit. Hence potential difference across X and Y= emf of battery =0V Q7.The V-I graph for a good conductor makes an angle 40 0 with V-axis. Here V denotes voltage and I denote current. What will be the resistance of the conductor? Ans7: Resistance R= V I = cot 400 Q8.A uniform wire of resistance 36Ω is bent in the form of a circle. What is the effective resistance across the points A and B Ans8: R= =.75Ω Q9.The amount of charge passed in time t through a cross-section of a wire is Q (t) = At + Bt + C. If the numerical values A,B and C are 5,3 and respectively in SI units, find the value of the current at t=5s. Ans9. I = dq = At + B, substituting the values we get- dt I = 53 A Q0.An electron gun emits.0x0 6 electrons per second.what electric current does this correspond to? Ans0: I = ne /t = =3.x0-3 A Q.The given figure shows a network of currents. The magnitude of currents is shown in figure. What is the value of I in the figure?

5 Ans. Applying Kirchhoff s first law I=+3+5= 0A QUESTION CARRIES TWO MARKS Q. A battery of emf V and internal resistance 0.5Ω is connected across a resistance of 9.5Ω. How many electrons cross through a cross-section of the resistance in second? V Ans: The current in the circuit is- i = = 0.A 9.5Ω+0.5Ω Thus, a net transfer of 0. C per second takes place across any cross-section in the circuit. The 0.C number of electrons crossing the section in second is.6x0 9 C =.5x08 Q3.Masses of the three wires of same material are in the ratios of ::3 and their lengths in the ratio of 3::. In what ratio will the electrical resistance of these wires be? Ans3: Mass =volume density = Al d A=M/l d / Resistance R= ρl / A= ρl d / M R l /M R : R : R 3 = 7 : 6 : Q4. Why are alloys used for making standard resistance coils? Ans4. Alloys have (i) low value of temperature coefficient and the resistance of the coil do not vary much with rise in temperature. (ii)high resistivity, so that even a smaller length of the material is sufficient to design high standard resistance. Q5.A current through a wire depends on time as I=0+4t. What will be the charge that crosses through the section of the wire in 0 seconds? Ans5: I = dq =0+4t dt dq =(0+4t)dt, Integrating Q =0 t + t

6 Putting t=0 sec Q=300C Q6. Two wires of the same material but of different diameters carry the same current I. If the ratio of their diameters is :, then what will be the ratio of their mean drift velocities? Ans6: v d = = 4 i.e. v nae nπ ed d D v d = D D = 4 v d Q7. A copper wire of cross-sectional area.0mm, resistivity =.7x0-8 Ω m, carries a current of A.What is the electric field in the copper wire? Ans7: E= V l = I R l = I ρ A Substituting the values E = 8.5x0-3 V/m Q8. The specific resistance of a wire is ρ, its volume is 3m 3 and its resistance is 3 ohm, then what will be the length of such a wire? ρ l Ans8: R= = A R V l = ( ρ l V/l = ρl V ρ )/ = ( 3 3 ρ ) = 3 / ρ Q9., 4 and 6 S are the conductances of three conductors. When they are joined in parallel, what will be the equivalent conductance of the combination? Ans 9: R = Ω, R = Ω, R 3 = Ω 4 6 In series R= R + R +R 3 = / Ω Equivalent conductance G=/R = S Q30.Two resistances are joined in parallel whose resistance is (3/5) Ω.One of the resistance wire is broken and the effective resistance becomes 3Ω. What is the resistance of the wire in ohm that got broken? Ans30. R p = R R R +R =3/5 and R=3Ω, substituting the values, so we get R = (3/4) Ω Q3. Nine resistors each of resistance R are connected in the circuit as shown in the figure.what is the effective resistance between points A and B

7 Ans3. The equivalent circuit is shown in figure (R 3 )(R/3) Effective resistance is R eff = R = 3 +R 9 R 3 Q3. Two cells of emfs E = 6V and E =5V are joined in parallel, without any external load.if their internal resistances are r = Ω, r = 3 Ω respectively. What is the terminal potential difference across the combination? Ans3. I = 6 5 (3+) = 5 A Potential drop across 3Ω = 5 3=0.6 V Terminal potential difference across 5V= = 5.6V Q33.A wire of resistance R is divided in 0 equal parts. These parts are connected in parallel, what will be the equivalent resistance of such a connection? Ans33. Resistance of each part of wire r = R/0

8 Total resistance of 0 parts of wire connected in parallel = r/0= (R/0)/0 =R/00 = 0.0R Q34. Two wires of same metal have the same length but their cross sections are in the ratio 3:. They are joined in series. The resistance of the thicker wire is 0Ω. What is the total resistance of the combination? Ans34. For the same length and material, R = A = 3 R A R = 3R and R =0Ω The resistance of thin wire R =30Ω Total resistance = 0+30 = 40Ω Q35. What is terminal potential difference of a cell? Can its value be greater than the e.m.f of a cell? Explain. Ans35. Terminal potential difference of a cell is defined as the potential difference between the two electrodes of a cell in a closed circuit. The value of terminal potential difference of a cell is less than emf of a cell, when current is drawn from the cell. During charging of a cell the value of terminal potential difference is greater than the emf of a cell. Q36. A voltage of 30V is applied across a carbon resistor with first, second and third rings of blue, black and yellow colour respectively. Find the value of current through the resistor. Ans36. Resistance = Ω. I = V R = A QUESTIONS CARRY THREE MARKS Q37. To get a maximum current through a resistance of.5 Ω, one can use m rows of cells each row having n cells. The internal resistance of each cell is 0.5Ω. What are the values of m and n if the total number of cells is 0? Ans37. mn =0, R =.5Ω, r = 0.5Ω R = nr/m, Substituting the values n=5m m 5m=0 or m = 4 So m =, n = 0 Q38.Two identical cells connected in series send.0a current through a 5Ω resistor. When they are connected in parallel, they send 0.8A current through the same resistor. What is the internal resistance of the cell? Ans38: case (i) E + E = ( r + r + 5 ).0

9 E = r (i) case (ii) E = ( r r r+r + 5 ) 0.8 E = 0.4 r (ii) Multiplying (ii) by and equating with (i) and then on solving we get, r =.5 Ω Q39. Two identical cells send the same current in 3Ω resistance, whether connected in series or in parallel.what is the internal resistance of the cell Ans39. Let E,r be the e.m.f and internal resistance of each cell. In series combination of two cells the current through external resistance R will be E I = r+r In parallel combination of two cells the current through external resistance r will be, I' = E E = )+R r+r ( r If I = I' then r + R = r + R R = r = 3 Ω Q40.In figure shown, AB is metre long uniform wire of 0Ω resistance. Other data are shown in the diagram. Calculate (i) potential gradient along AB, (ii) length AO, when galvanometer shows no deflection. 3 Ans40. (i) Potential gradient along AB = ( ) 0 = V cm-.5 (ii) Current through 0.3 Ω = = A.+0.3 Potential difference across 0.3 Ω = 0.3 = 0.3 V Across length AO, V = 0.3 V is to be balanced. Length AO = 0,3 = 37.5 cm Q4. kg piece of copper is drawn into a wire mm thick and another piece into a wire of mm thick. What will the ratio of their resistances?

10 Ans4. Let d be the density of the material of copper wire. Let l, l be the lengths of copper wires of diameter mm and mm respectively. As Mass= volume density = (π D /4) l d / From this we get l =4 l ρl Now R= / π D /4 R = l D R l D =6 So ratio is 6: Q4.Determine the current drawn from a V supply with internal resistance 0.5Ω by the following infinite network; each resistor has Ω resistance. 3 Ans4. Let R be the equivalent resistance between A & B. Since the network is infinite, therefore resistance between C & D is same as between A & B, then equivalent resistance of R & Ω in parallel is R = R R+ R AB = R + + R + + = R R R+ + = R R - R = 0, on solving R = ( + 3) Ω =.73 Ω Current drawn I = = 3.7 A ½ Q43. Establish a relation between drift velocity and time of relaxation. Use this relation to deduce the expression for the electrical resistivity of the material. Ans43. See NCERT Book Article No. 3.5 /

11 e E Prove that v d = - τ m & use the above relation to prove ρ = m ne τ Q44. A potentiometer wire of length m has a resistance of 5Ω. It is connected to a 8 V battery in series with a resistance of 5Ω.Determine the emf of the primary cell which gives a balance point at 60 cm. Ans44. Given l = m =00cm, R= 5Ω, emf of battery = E = 8 V, R = 5 Ω, x = 60 cm. E 8 I = = R+R 5+5 = 0.4 A Potential drop across potentiometer wire V = I R = =.0 V EMF of primary cell = V x = 60 =. V l 00 Q45. A battery of emf E volt & internal resistance r ohm is joined in series with two resistances X & Y ohm in a closed circuit. A standard cell of emf.06 V and a galvanometer are joined in series & the combination is connected across X. The galvanometer shows no deflection when X = 60Ω & Y = 4 Ω or when X = 40 Ω & Y = 40 Ω. Calculate the values of E & r. E Ans =.06 ( 60+4+r ) E 40 =.06 (40+40+r ) On solving, we get E = 5.5 V & r = 8 Ω Q46. In the potentiometer circuit shown in figure, the balance point with R = 0Ω when switch S is closed & S is open is 50 cm, while that when S is closed & S is open is 60 cm. What is the value of x? What will you do if you fail to find a balance point with the given cell E? Ans46. If k is the potential gradient & I is the current through resistances 0Ω & x Ω due to cell E'. Potential difference across 0 Ω = I 0 volt Potential difference across x Ω = I x volt When switch S is closed & S is open

12 I 0 = k (i) / When switch S is closed & S is open I (0 + x) = k (ii) / Solving equations (i) & (ii) 0 + x = = x = Ω If we fail to get balance point on the potentiometer wire, it shows that potential drop across (0 +x) is greater than potential difference across potentiometer wire due to cell E. To find the balance point (i) (ii) We can increase the voltage of cell E by adding one or more cells in series with E or We can reduce the current in resistance (0+ x) & hence potential difference across it by using a suitable resistance in series with cell E. Q47. A potential difference V is applied to a conductor of length L, diameter D. How are the electric field E, the drift velocity v d & the resistance R affected when (i) V is doubled (ii) L is doubled? Ans47. E = V e E e V, drift velocity vd = τ = τ, R= ρ L / A L m m L (i) When V is doubled, E & v d become double, R remains unchanged. (ii) When L is doubled, E & v d become half, R becomes double. Q48. The heating element of an electric toaster is of nichrome. When a very small current passes through it, at room temperature 7 o C, its resistance is 75.3Ω. When toaster is connected to a 30V supply, the current settles after a few seconds to a steady value of.68 A. What is the steady temperature of nichrome element? The temperature coefficient of resistance of nichrome over averaged temperature range is o C -. Ans48. R = 75.3 Ω,α = o C -,V= 30 V, I =.68 A, t = 7 o C R = = 85.8Ω (t t ) = R R R α Substituting the values, t t = 849 o C Q49.(i) Calculate the value of R in the balance condition of the wheatstone bridge, if the carbon resistor connected across the arm CD has the colour sequence red, red & orange as shown in the figure.

13 (ii)if now the resistance of arms BC & CD are interchanged, to obtain the balance condition, another carbon resistor is connected in place of R. What would now be sequence of colour bands of the carbon resistor? Ans49. (i) Let a carbon resistor S is given to the bridge, Then R R S R = S R = S = 0 3 Ω (ii)after interchanging the resistance R X = X = 4R = = 88 kω The sequence of colour bands is Grey, Grey, and Orange. Q50. A few storage cells in series are to be charged from a 30 V DC supply. The emf of each cell is.35 V & internal resistance is 0. Ω. The charging current is 3.0 A. In this arrangement how many cells can be charged and what extra resistance is required to be connected in the circuit? Ans50. Let n be the maximum number of cells in series, which can be charged & R is the extra resistance required. If I is the current in the circuit, then (n r + R ) I = 30 n E E =.35 V, r = 0. Ω, I =3.0 A Substituting the values, we get 3 n +.65 R = 30.3 = n = 8, R = 0. Ω QUESTION CARRIES FIVE MARKS

14 Q5.State the principle of potentiometer.draw a circuit diagram used to compare the emfs of two primary cells. Derive the formula used. How can the sensitivity of a potentiometer be increased? Ans5. See NCERT Book Article No. 3.6 Principle V = k L, k = V L Prove E E = l l, using circuit diagram. + The sensitivity of a potentiometer can be increased by decreasing its potential gradient. The same can be achieved (a) by increasing the length of potentiometer wire. (b) by reducing the current in the potentiometer wire. Q5. (i) Using Kirchhoff s rules obtain the balance condition in terms of the resistances of four arms of Wheat stone bridge. (ii) First a set of n equal resistors of R each are connected in series to a battery of e.m.f ԑ and internal resistance R. A current I is observed to flow. Then the n resistors are connected in parallel to the same battery. It is observed that the current is increased 0 times. What is n? Ans5. (i) See NCERT Book Article No. 3.4 Prove R = R 4, using Fig. + R R 3 (ii). When n resistors are in series E I = / R+nR When n resistors are parallel R eff = R / n E ne I' = = / R( n+) R+ R n According to question, I' = 0 I Substituting the values of I & I n = 0 Q53. (a) A resistance of R Ω draws current from a potentiometer as shown in figure. The potentiometer has a total resistance R o Ω. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer. (b) Why are the connections between the resistors in a meter bridge made of thick copper strips? (c)why is it generally preferred to obtain the balance point in the middle of the meter bridge wire? Ans53. (a) The equivalent resistance of the circuit

15 R eq = R o R = R. R o R + R o + R.R o R + R o Current in the circuit I = V R eq = V = ( 4 R + R o ) R o ( R + R o ) ( R + R o ) ( 4 R + R o ) R o (i) V = I R Substituting the values of I & R V R We get V = (4 R+ R o ) (b) Resistivity of copper wire is very less. The area of thick copper strip is more. So it offers negligible resistance in the circuit. (c) The error in the measured value of unknown resistance using meter-bridge will be minimum, when the null point is obtained at the middle of meter-bridge wire.in this situation, the end error of the bridge will become ineffective. Q54. State Kirchhoff s laws. Calculate the value of current I, I & I 3 in the following circuit. Ans54. Statements of Kirchhoff s laws Applying Kirchhoff s rule to the loop PRSP 4 I I = (i) / Loop PRQP 5 I I = (ii) / From first rule I 3 = I + I (iii) Solving the above equations I = I = 4 5 I 3 = A = A = 5 A = ma

16 Q55.(a) Six lead-acid type of secondary cells each of e.m.f.0v and internal resistance 0.05Ω are joined in series to provide a supply to a resistance of 8.5Ω. What are the current drawn from the supply and its terminal voltage? (b) A secondary cell after long use has an e.m.f of.9v and a large internal resistance of 380Ω.What maximum current can be drawn from the cell? Could the cell derive the starting motor of a car? (c) The emf of the driver cell in the potentiometer experiment should be greater than the emf of the cell to be determined. Why? (d) Why do we prefer potentiometer of longer length for accurate measurements? ne Ans55. (a) I = R+nr = ( 6 ) =.4 A Terminal voltage, V = I R = =.9 V / (b) I max = E =.9 = A, r 380 This amount of current cannot start a car because to start the motor, the current required is 00A for few seconds. / (c) If it is not so, there will be smaller fall of potential across the potentiometer wire than the emf of the cell to be determined and hence the balance point will not be obtained on the potentiometer wire. (d) The value of potential gradient decreases. Therefore the measurement is more accurate. Q56. (a)draw the circuit diagram of a potentiometer which can be used to determine the internal resistance r of a given cell of emf (E). Describe a method to find the internal resistance of a primary cell. (b) Two students X & Y perform an experiment on potentiometer separately using the circuit given. Keeping other parameters unchanged, how will the position of the null point be affected if (i) X increases the value of resistance R in the set-up by keeping the key K closed and the key K open? (ii) Y decreases the value of resistance S in the set-up, while the key K remains open and the key K closed? Justify your answer in each case. Ans56. (a). See NCERT Book Article No. 3.6 Circuit diagram Prove r = R ( l l ) (b) (i) By increasing resistance R the current through AB decreases, so potential gradient decreases. Hence a greater length of wire would be needed for balancing the same potential difference. So the null point would shift towards B.

17 (ii) By decreasing resistance S, the current through AB remains the same, potential gradient does not change. As K is open hence there is no effect of S on null point. Q57. (a) Find the potential drops across the two resistors shown in figure. (b)a voltmeter of resistance 600Ω is used to measure the potential drop across the 300Ω resistor. What will be the measured potential drop? (c) As the temperature of a conductor increases, its resistivity & conductivity change. What will be the effect on the ratio of resistivity to conductivity? Ans57. (a) The current in the circuit = 00V 300Ω+00Ω =0.A The potential drop across the 300Ω resistor is =60V The drop across the 00Ω resistor is 40V. / (b)the equivalent resistance, when the voltmeter is connected across 300Ω 600Ω x 300Ω R=00Ω + 600Ω+300Ω Thus, the main current from the battery is I = 00V 400Ω = 0.5A / The potential drop across the 00Ω resistor is 00Ω x 0.5A=50V and that across 300Ω is also 50V.This is also the potential drop across the voltmeter and hence the reading of the voltmeter is 50V. (c) Increases. Q58. (a) Derive the formula for the equivalent EMF & internal resistance for the parallel combination of two cells with EMFs E & E and internal resistances r & r respectively. (b).each of the resistors shown in figure has a resistance of 0Ω and each of the batteries has an emf of 0V. Find the currents through the resistors a and b in the two circuits. Ans58. (a). See NCERT Book Article No. 3.

18 Cells in parallel combination Prove E eq = E r + E r r +r r r r eq = r + r (b). Fig. (a) Current across a = 0 V 0 Ω = A / Current across b = 0, First cell, resistance a & (series combination of b and second cell) are parallel. So potential difference is same. / Fig. (b) Current across a = 0 V 0 Ω = A / Current across b is zero. / Q59.(a). The circuit in figure shows two cells connected in opposition to each other. Cell E is of e.m.f 6V and internal resistance Ω; the cell E is of e.m.f 4V and internal resistance 8Ω. Find the potential difference between the points A and B. E A E B (b) State, with the help of a circuit diagram, the working principle of a meter-bridge. Write the expression used for determining the unknown resistance. What happens if the galvanometer & cell are interchanged at the balance point of the bridge. Ans59. (a). I = = 5 A Potential difference across A & B is the terminal voltage of cell E = E I r = 6 - = 5.6 V. 5 (b)see NCERT Book Article No.3.5 Circuit diagram Working principle Expression No change in the position of the balance point of the bridge. / ½ Q60.Find the equivalent resistances of the networks shown in figure between the points a and b.

19 Ans60. Fig. (a) R = r + r =r, R = R + r R = r 3 R 3 = R + r = 5r / 3 = + R net R 3 r R net = (5r / 8) Ω Fig. (b) R = r + r + r R = r / 3 R eq = R + r = r + r = (4r / 3) / 3 Fig. (c) It is balanced condition of wheat stone bridge. Therefore resistance in diagonal are neglected. / Equivalent resistance between a & b R eq = r + r, R eq = r Ω