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NUMERICALS-ANSWERS QUESTIONS How many electrons pass through a conductor in one minute if the current through it is 0.32 mA? A current of 1.6 ampere is flowing through a copper voltameter Find the charge passing through it per minute Calculate the electric current through a conductor if 6.25 × 1018 electrons are crossing a point per second In hydrogen atom the electron moves in an orbit of radius 5.0 × 10– 11 m with a speed of 2.2 × 106 m/s Find the equivalent current In copper the number of free electrons per meter3 is 8.4 × 1028 Calculate the drift velocity of electrons in a copper wire of cross-sectional area × 10– m2, if a current of 0.63 A is flowing through it The specific resistance of a material is × 10– Ω m Find the resistance of a piece of that material of length 10 cm and area of cross-section × 10– m2 A copper wire is stretched so that its length is increased by 0.2% What is the percentage change in its resistance ? A current of A exist in a 10 Ω resistor for minutes Find the charge and the number of electrons that pass through any cross-section of the resistor in this time A 10 Ω thick wire is stretched so that its length becomes three times Assuming that there is no change in its density on stretching, calculate the resistance of the new wire (AISSCE 1994) 10 Calculate the conductivity of a conductor of resistance 0.05 ohm, area of cross section 10– m2 and length 50 cm 11 Calculate the potential difference required to maintain a current of A through a resistance of 45 ohm S Chand & Company Limited 12 A parallel combination of resistances takes a current of 7.5 amperes from a 30 volt battery If two of the resistances are 10 Ω and 12 Ω , find the third resistance 13 Calculate the current shown by ammeter A in the circuit shown below: 5Ω 10 Ω Ω 10 Ω 10 10 Ω 5Ω A 10 V Fig A.12 (AISSCE 2000) 14 Two cells of emfs 6V and 12V and internal resistances Ω and Ω, respectively, are connected in parallel so as to send current in the same direction through an external resistance of 15 Ω (a) Draw the circuit diagram (b) Using Kirchhoff’s laws calculate (i) the current through each branch of the circuit, (ii) the potential difference across the 15 Ω resistance (AISSCE 1995) 15 Using Kirchhoff’s laws in the given electrical network calculate the values of I1, I2 and I3 (AISSCE 1994) S Chand & Company Limited C I2 B A I1 5Ω 3Ω I1 2Ω I2 I3 12 V F 6V E Fig A.13 D 16 A silver wire has a resistance of 2.1 Ω at 27.5° C and a resistance of 2.7 Ω at 100°C Determine the temperature coefficient of resistance of silver 17 Two wires of the same material, having lengths in the ratio 1:2 and diameters in the ratio 2:3, are connected in series with an accumulator Compute the ratio of the potential differences across the two wires 18 The size of a carbon block is 1.0 cm × 1.0 cm × 50 cm Find the resistance (a) between the square ends of the block, (b) between the two opposite rectangular faces Resistivity of carbon is 3.5 × 10–3 Ω cm 19 Two cells of emfs E1 and E2 (E1>E2) produce a current of A through a resistance when they are in opposition The current through same resistor is 5A when they are in series Find the ratio of the emfs of the two cells S Chand & Company Limited 20 Calculate the resistance between A & B of the given network: A 10 B Fig A.14 (AISSCE Delhi 1999) 21 Two cells E1 & E2 of emfs 4V and 8V respectively, having internal resistances 0.5 Ω and 1Ω respectively, are connected in opposition to each other This combination is connected in series with resistances of 4.5 Ω and Ω Another resistance of 6Ω is connected in parallel across the 3Ω resistor (i) Draw the circuit diagram (ii) Calculate the total current flowing through the circuit (AISSCE 1995) 22 A current of 3A is flowing in an aluminium wire of cross-section × 10–6 m2 Find the current density in the wire 23 When a current of A is drawn from a battery, the potential difference between its terminals is 20 V The potential difference becomes 16 V when a current of A is drawn from it Find the emf and the internal resistance of the battery 24 The emf of a battery is V and its internal resistance is 0.5 Ω The external resistance in the circuit is Ω Calculate terminal voltage of the battery 25 Two resistance of 25 Ω and 75 Ω are connected in series across a source of 120 V Calculate the current through them and the potential drop across each resistance 26 In the above problem, if the resistances are connected in parallel then calculate (a) the total resistance of the combination, and (b) the current through each resistance S Chand & Company Limited 27 A cubical network consists of 12 resistances, each of 1Ω Determine the equivalent resistance between diagonally opposite corners 28 Find the resistance between the points X and Y: 6Ω 4Ω X 12Ω Y Fig A.15 29 Find the potential difference between the points A and B in the network shown in Fig A.16 30 Calculate the equivalent resistance between the point P and T in the network shown in Fig A.17 A 2Ω 3Ω D C 2A 3Ω 2Ω B Fig A.16 Fig A.17 S Chand & Company Limited 31 A battery of emf 3.5 volt and internal resistance r is connected in series with a resistance of 55Ω through an ammeter of resistance 10Ω The ammeter reads 50 mA Draw the circuit diagram and calculate the value of r (AISSCE Delhi 1995) 32 Calculate the equivalent resistance between points A and B in Fig A.18 2Ω 2Ω 2Ω A 2V 30 Ω 30 Ω B 2Ω 2Ω Fig A.18 30 Ω Fig A.19 33 Calculate the current through the battery in the circuit of Fig A.19 34 Find the potential difference between the points A and B in the following circuit: S Chand & Company Limited 5V 10 Ω B A 20 Ω 2V Fig A.20 35 A battery of emf V and internal resistance 0.1 ohm is being charged by a current of A What is the direction of current inside the battery? What is the potential difference between the terminals of the battery? 36 Calculate the value of R if the equivalent resistance of the network is 2Ω in the figure given below 4Ω R A 12 Ω B Fig A.21 S Chand & Company Limited 37 The current in a potentiometer wire is adjusted to give a null point at 56 cm with a standard cell of emf 1.02 V Calculate the emf of another cell for which the null point is at 70 cm 38 Two resistances of Ω and Ω are connected in parallel in the right gap of a meter bridge with a known resistance of 1.5 Ω in the left gap Find the position of the null point when the bridge is balanced 39 Two cells of emfs E1 and E2 are connected as shown in figure A.22 When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300 cm On connecting the same potentiometer between A and C, the balancing length is 100 cm Calculate the ratio of E1 and E2 (AISSCE Delhi 1994) + A – B – + C E1 E2 Fig A.22 40 A potentiometer wire of length 100 cm has a resistance of 10 Ω It is connected is series with a resistance and an accumulator of emf V and negligible resistance A source of emf 10 m V is balanced against a length of 40 cm of potentiometer wire What is the value of the external resistance? 41 Calculate the electrical conductivity of the material of a conductor of length m and area of cross-section 0.02 mm2, having a resistance of Ω (AISSCE 1996) S Chand & Company Limited 42 A wire has a resistance of 32 Ω It is melted and drawn into a wire of half its original length Calculate the resistance of the new wire What is the percentage change in resistance ? (AISSCE Delhi 1997) 43 Why is a potentiometer preferred over a voltmeter to measure the emf of a cell? The potentiometer wire AB shown in the figure is 400 cm long Where should the free end of the galvanometer be connected on AB so that the galvanometer shows zero deflection? (AISSCE 1997) 12 Ω 8Ω G B A Fig A.23 44 In the given circuit diagram, what should be the value of R so that there is no current in the branch of V battery 6Ω 6V 12 Ω 12 V R Fig A.24 S Chand & Company Limited 45 A voltmeter V of resistance 300 Ω is used to measure the potential difference across the resistor R1 in the circuit shown (a) What will be the reading of the voltmeter? (b) Calculate the p.d across R1 before the voltmeter is connected V = 45 V R1 = 100 R2 = 125 V Fig A.25 ANSWERS Let n be the number of electrons We have Q = ne = it or Here n= it e i = 0.32 mA = 0.32 × 10– A t = = 60 s S Chand & Company Limited (b) For rectangular faces : l = cm 19 A = 1× 50 cm2 –3 3.5 × 10 × ∴ R = = × 10 Ω × 50 When the cells are in opposition, the net emf of the circuit is (E1 – E2) So, E – E2 I= R or = E1 – E2 R When the cells are in series, we have 5= E1 + E2 (2) R Dividing Eq (1) by Eq (2), E1 (1) = E1 – E E1 + E = E2 20 The given network is balanced Wheatstone bridge So we can ignore the 10 Ω resistance and redraw the network as Solving, S Chand & Company Limited Ω B A Fig A.42 If R is the equivalent resistance between A and B, then 1 = + R 1+ = 6 or R = = 2Ω S Chand & Company Limited 21 (i) 3.0 4.5 E1 (4 V, 0.5 E2 ) (8 V,1.0 ) Fig A.43 (ii) The equivalent resistance of 3Ω Ω and Ω in parallel is 3× R′ = 3+6 Total resistance of the circuit, R = 4.5 + + 0.5 + 1.0 =8Ω Net emf of the circuit = – = V V i= = R = 0.5 A i 22 Current density j = A = –6 ×10 = 1.5 × 106 Am–2 S Chand & Company Limited 23 Terminal p.d V = E – Ir When A current is drawn we have 20 = E – r When A current is drawn 16 = E – r Solving (1) and (2), E = 24 V, r=4Ω  E  24 Terminal voltage of battery V =  R  R+r  Here E=3V r = 0.5 Ω R = 1Ω    ×1  + 0.5  ∴ V = =2V 25 Equivalent resistance in series R = R1 + R2 = 25 + 75 = 100 Ω I= V = 120 = 1.2A R 100 Potential drop across the 25 Ω resistance = 25 × 1.2 = 30 V Potential drop across the 75 Ω resistance = 75 × 1.2 = 90 V S Chand & Company Limited (1) (2) 26 (a) Total resistance R = = R1 + R2 R1 + R2 25 × 75 25 + 75 = (75/4) Ω 25Ω I1 I2 75Ω Fig A.44 (b) Current through the 25 Ω resistance = Current through the 75 Ω resistance = 120 25 120 = 4.8 A = 1.6 A 75 27 Let the current entering at A be I From symmetry, the current distribution is as shown in Fig A.45 It is obvious that the points B, D and E are at the same potential Similarly, the points H, C and F are also at the same potential S Chand & Company Limited H I G I /2 D I /2 C I I /2 F E I /2 I I 3I I I /2 A I B I /2 Fig A.45 Therefore, the equivalent circuit becomes: 1Ω 1Ω 1Ω A 1Ω 1Ω 1Ω 1Ω 1Ω 1Ω 1Ω B,D,E 1Ω C,F,H 1Ω Fig A.46 S Chand & Company Limited G which reduces to Ω Ω Ω Fig A.47 Equivalent resistance R = + + = 28 All the three resistance are in parallel 1 1 So, = + + R R1 R R3 or R = + + +1+ = Ω 12 + +1 = = 12 12 or R = 2Ω 29 Current through arm CA = current through arm CB = A VC – VA = × = V VC – VB = × = V ∴ VA – VB = (VC – VB) – (VC – VA) = – = V 30 The given net work may be simplified as in Fig A.48 The effective resistance between points Q and S is Ω Thus there are three Ω resistances in series between the points P and T So the equivalent resistance of the network is S Chand & Company Limited Req = + + = 15 Ω 5Ω P 5Ω Q R 5Ω 10 Ω S Fig A.48 31 Total resistance of the circuit = 55 + 10 + r = (65 + r) Current I = 50 mA = 50 × 10–3 A 55 Ω A 10 Ω E r E = 3.5 V r =? Fig A.49 S Chand & Company Limited 5Ω T E Now I= or 50 ×10 or 65 + r = R –3 = 3.5 (65 + r ) 3.5 50 ×10 –3 = 70 r =5Ω or 32 Simplified circuit of the net work is shown in Fig A.50 It is a balanced wheatstone Bridge So the resistance between C and D can be ignored We now have two 4Ω resistances in parallel Ω The equivalent resistance of the net work is therefore 4/2 = 2Ω C 2Ω 2Ω 2Ω A B 2Ω 2Ω D Fig A.50 S Chand & Company Limited 33 Let R be the equivalent resistance of the circuit Then R or = R = 30 60 + 60 = +1 60 = 60 = 20Ω Current through the battery, I = V R = 20 = 10 = 0.1A 34 Net emf of the circuit = – = V Total resistance of the circuit = 20 + 10 = 30 Ω Therefore, current in the circuit, I = V R = 30 = 0.1 A Potential drop across the 10 Ω resistance = 0.1× 10 = V ∴ Potential difference between A and B, VA– VB = – = – 4V 35 Inside the battery the direction of current is from the positive to the negative terminal The potential difference between the terminals of the battery during charging is given by S Chand & Company Limited V = E + Ir Here E = 2V , I = 5A r = 0.1Ω So, V = + 5(0.1) = + 0.5 = 2.5V 36 All the resistance are connected in parallel Therefore, 1 1 = + + R 12 or R = – – = 12 12 R = 6Ω E1 l = 37 We know that E2 l2 or Here E1 = 1.02 V, E2 = ? l1= 56 cm, l2 = 70 cm So, E2 = E1l l1 = 1.02 × 70 56 = 1.275 V 38 Equivalent resistance in the right gap = 3× 3+6 = 2Ω S Chand & Company Limited If the null point is obtained at a distance x from the left end, then 1.5 x = 2.0 (100 – x ) or x = 42.9 cm 39 Potential difference between points A and B = E1 Potential difference between points A and C = E1– E2 Now E1 E1 – E2 = l1 l2 = 300 =3 100 E1 = E2 40 Resistance of potentiometer wire = 10Ω or R 2V D 40 cm A 10 mV Fig A.51 Current through the battery, I = Resistance of AD = Ω Potential drop across AD, 10 + R S Chand & Company Limited C   4  10 + R  V = ⇒ or   –3 10 + R  = 10 ×10 10 + R = or 10 ×10 R = 790 Ω 41 Electrical conductivity σ = Here = 800 –3 ρ = l RA l = 3m A = 0.02 mm2 = 0.02 × 10–6 m2 R = 2Ω 3×10 σ= = ∴ –6 × 0.02 ×10 × 0.02 = −1 × 10 = 7.5 × 10 Sm 42 Let the initial length of the wire be l and the area of cross section be A When the length is reduced to l/2, area of cross section becomes 2A Original resistance R = ρl/A = 32 Ω If R′ is new resistance then R′ = ρ l/2 2A = ρl 4A = 32 = 8Ω S Chand & Company Limited Percentage change in resistance R′ – R 24 × 100 32 = 75% 43 For I part see Short-Answer Q 87 Let the required point be at a distance x from A Using the balancing condition, we have, = 12 R = × 100 = x (400 – x ) or x = 160 cm 44 Applying Kirchhoff’s loop rule in loop (2) (12 + R) i2 = 12 For loop (1), i1 – 12i2 = – 12 or i1 – i2 = –1 Putting i1 = (given), i2 = 1/2 A ⇒ (12 + R ) × or = 12 R = 12Ω S Chand & Company Limited i1 6V i1 i2 i2 12 12V i1 + i2 R Fig A.52 45 (a) Equivalent resistance of R1 and voltmeter is 100 × 300 R= = 75Ω 100 + 300 VR 45 × 75 = Reading of voltmeter = = 16.875 R + R2 75 + 125 =16.9V (b) P.D across R1 when the voltmeter is not connected VR1 45 ×100 = = = 20V R1 + R2 100 + 125 S Chand & Company Limited