CURRENT ELECTRICITY-III ¾ ELECTRIC CURRENT ¾ OHM’S LAW AND ¾ KIRCHHOFF’S LAW 1 1. Assume that each atom of copper contributes one electron electron. If the current flowing through a copper wire of 1 mm diameter is 1.1 A, the drift velocity of electrons will be (density of Cu =9 gcm-3 , atomic wt. of Cu =63) a)) 0.3 mm/s c) 0.1 mm/s b)) 0.5 mm/s d) 0.2 mm/s 2 1 Solution: 1. No. = of atoms in 63g of Volume of 63 g of copper = = 7 cm3 = n= copper = m3 = A=πr2= 3.142X( X m-3 )2 m2 D ift velocity, Drift l it = = = 0.1X 10-3 ms-1 =0.1mms-1 Ans : (c) 3 02.The potential difference applied to an X-ray tube is 5 kV and the current through g it is 3.2 m A. Then the number of electrons striking the target per second is a) 2 x 1016 b) 5 x 106 c) 1 x 1017 d) 4x 1015 4 2. Solution: As = = = = = Ans: (a) 5 3.The current in a conductor varies with time t as I = 2t + 3t2, where I is in ampere and t in seconds. Electric charge flowing through a section of the conductor during t=2 t 2 sec to t = 3 sec is a) 10 C b) 24 C c) 33 C d) 44 C 6 3. Solution: As I= I dq = I dt = dt Charge passed from t=2s to t= 3s is q= = 2 3 = 3 q= t )2 + t ) 2 + )23 3 =(9-4) ( ) + ((27-8)) = 5+ 19 = 24C Ans : b) 7 4.The current I and voltage V curves for a given metallic wire at two different temperatures T1 and T2 are shown in the T1 fi figure. Then, Th a) T1 > T2 c) T1= T2 b) T1 < T2 d) T1 = 2T2 I T2 V 8 4. Solution: for the same potential v0 , T1 I At temperature T1, R1 = , At temperature T2, R2= , As , T2 I1 I2 R2 R1. V0 V Since resistance Si i off metall increases i with temperature, So, T2 T1 or Ans : (b) 9 5. The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in diameter diameter, the change in the resistance of the wire will be a) 200% b) 100% c) 50% d) 300% 10 5. Solution: When the length is increased by 100%, it becomes 2l and area of cross section decreases to N New resistance, i t = = = = 4R Ch Change iin resistance i t 100%= ( Ans: (d) )100 = 300% 11 6. To verify Ohm’s law, a student is provided with a test resistor RT, a high resistance R1, a small resistance R identical ,two galvanometers G1 and G2 and a variable voltage source V. The y out the correct circuit to carry experiment is 12 13 6.Solution:G1 behaves as voltmeter byy connecting g it in series with high resistance R1 and applying it across RT in parallel G2 behaves as ammeter by shunting it with parallel. small resistance R2 and connecting in series with RT. Ans : (c) 14 7. By increasing the temperature, the specific resistance of a conductor and a semiconductor (a) Increases for both (b) decreases for both (c) Increases, decreases respectively (d) d decreases, iincreases respectively ti l 15 Solution(7):On increasing the temperature , the specific resistance of a conductor increases and that of a semiconductor decreases Ans . (c) 16 8. The resistance of a wire is 5Ω at 50 ˚C and 6 at 100 ˚C 6Ω ˚C. Th The resistance i off the h wire i at 0 ˚C will be (a) 3 Ω (b) 2 Ω (c) 1Ω (d) 4Ω 17 8. Solution: Rt-Ro = R50-R Ro = t - ------- (1) And R100 - Ro = -----(2) = Or = Or 10-2Ro = 6-Ro ; R0 = 4 Ans: (d) 9. I-V characteristics of a copper wire of length L and area of 9 cross-section A is shown in figure. The slope of the curve becomes (a) More if the experiment is performed at higher temperature (b) More if a wire of steel of same dimension is used (c) More if the length of the wire is increased (d) Less if the length of the wire is increased I V 19 9.Solution: 9 Solution: Slope of II-V V graph = = I If the experiment is performed at a higher temperature, the resistance R of copper increases and hence slope d decreases, so th the option ti ((a)) iis wrong. IIn option (b) and (c), the resistance more and d so slope l b becomes lless. IIn option ti (d), the resistance R increase and so slope l d decrease. H Hence only l option ti (d) iis correct Ans: (d) V I V 20 10.The ratio of masses of three wires is 1 : 2 : 3 and that of their lengths is 3 : 2 : 1. If the wires are made of same material, the ratio of their resistances will be ( )1:1:1 (a) (b) 1 : 2 : 3 (c) 9 : 4 : 1 (d) 27 : 6 : 1 21 10 Solution: 10. Mass= m= volume X density = AlXd Or A = R = = = i.e; R given m1: m2: m3: = 1:2 :3 and l1 : l2 : l3 = 3 : 2: 1 R1 : R2 : R3 = : : =9 : 2 : = 27: 6 : 1 Ans: (d) 22 11. The colour sequence in a carbon resistor is red brown, red, brown orange and silver. silver The resistance of the resistor is (a) 21X103 Ω ± 10% (b) 23 X 101Ω ± 10% (c) 21 X 103Ω ± 5% (d) 12 X 103 Ω ± 5% 23 11 Solution: 11. S l ti Red 2 Brown Orange 1 3 R =21 X 10 3 Silver 0% 0% Ans: (a) 24 12. The three resistances of equal value are arranged in 12 the different combinations shown below. Arrange them in increasing order of power dissipation. (a) III < II < IV < I ( ) I < IV < III < II (c) (b) II < III < IV < I (d) I < III < II < IV 25 12 Solution: 12. I Req = R + R + R = 3R ; PI = I2 (3R) II Req = = ; PII = I2 PIII = I2 III Req = ; IV Req = Clearly PIII +R = PII PIV = I2 ; PIV PI Ans: (a) 26 13. A 100 W bulb 13 b lb B1, and d ttwo 60 W bulbs B2 and B3, are connected to a 250 V source as shown in the figure. Now W1, W2 and W3 are the output powers of the bulbs B1, B2 and d B3 respectively. i l Then (a) ( ) W1 > W2 = W3 (b) W1 > W2 > W3 (c)W1 < W2 = W3 (d) W1 < W2 < W3 27 13. S 13 Solution: l i V Voltage l across B3 is i maximum, i so B3 gglows with maximum brightness. g In the series combination of B1 and B2 , the bulb B2 of lesser wattage glows brighter than B1, i.e. i e w2 w1. Hence w1 w2 w 3. Ans: (d) 28 14. Figure shows three resistor configurations R1, R2 and R3 connected to 3 V battery. If the power dissipated by the configuration R1, R2 and R3 is P1, P2 and P3, respectively, then (a)P1 > P2 > P3 (c)P2 > P1 > P3 (b) P1 > P3 > P2 (d) P3 > P2 > P1 29 14. S 14 Solution: l ti E Equivalent i l t resistance i t off the th three th combinations are R1 = 1 , R2 = 1/2 Power p= As R2 P2 R1 P1 , R3 = 2 ; for same V, p 1/R R3 P3 Ans: (c) 30 15. Six identical resistors are connected as shown in the figure. The equivalent resistance will be (a)Maximum between P and R (b) Maximum between Q and R (c) Maximum between P and Q (d) all are equal 31 15. Solution: = R; = R ; RPR = R is maximum Ans: (c) 32 16 Th 16. The resistance i t off the th series i combination bi ti of two resistances is S. When they are joined in parallel, the total resistance is P. If S = nP,, the minimum possible value of n is (a) 4 (b) 3 (c) 8/9 (d) 2 33 16 Solution: In series , 16. R1 + R2 = S In parallel, =P But S = nP R1 + R2 = n For n to be minimum, R1 = R2 = R (say) R+R=n ; Or 4 R2 = n R2 or n = 4 Ans: ((a)) 34 17.In 17 I the th circuit i it shown h iin th the fi figure, the current through the (a) 3 Ω resistor is 0.50 A. (b) 3 Ω resistor is 0.25 A. (c) 4 Ω resistor is 0 0.50 50 A A. (d) 4 Ω resistor is 0.25 A. 35 17.Solution: Resistance between B and C in the RHS of the circuit = =4 Resistance between A and B in the RHS of the circuit = =4 Equivalent q resistance of the circuit,, R = 3+4+2 = 9 C Current td drawn ffrom th the b battery tt I= = 1A 36 At the junction A, the current of 1A is divided equally q y between the 8 resistance and the remaining circuit of resistance 8 . At the junction B B, the current of 0.5A 0 5A is divided equally between the 8 resistance and the remaining i i circuit i it off resistance i t 8 . current through the 4 resistor = 0.25A Ans: (d) 37 18.The total current supplied to the circuit b th by the b battery tt wire i iis (a) 1A ((b)) 2A (c) 4A (d) 6A 38 18.Solution: 18 Solution: The equivalent circuit is as shown in the fig. Net resistance in the upper arm = 1.5 + 2 6 = 1.5 + 1.5 = 3 Ω 2 6 Now we have 3 Ω and 3 Ω resistance in parallel; the net resistance is 2 R = 3 3 3 3 2 = 1.5 Ω 1.5 1 .5 6 6 3 3 6 V Current I = Ans: (c) 6 1.5 = 4A 19 For a cell 19. cell, the terminal difference is 2 2.2 2V V, when circuit is open and reduces to 1.8 V, when cell is connected to a resistance R=5Ω The internal resistance of cell r is (a) ((c)) 10 9 Ω 11 Ω 9 (b) (d) 9 10 Ω 5 Ω 9 40 19. Solution: the internal resistance of the cell is r=( )R=( )5=( ) Ans: (a) 41 20. If two identical cells, when p , connected in series or in parallel, supply same amount of current through an external resistance of 2Ω. The internal resistance of each cellll iis ((a)) 8 Ω ((b)) 2Ω (c) 4Ω (d) 1Ω 42 20 Solution: 20. Current for series combination = current in parallel combination = 2( ) =2r + 2 or r = 2 Ans: (b) 43 21. An ideal gas is filled in a closed rigid and thermallyy insulated container. A coil of 100 Ω resister carrying current 1 A for 5 minutes supplies heat to the gas. The change in internal energy of th gas iis the ((a)) 10 kJ ((b)) 30 kJ (c) 20 kJ (d) 0 kJ 44 21 .Solution: S l ti The change in internal energy of the gas = Heat produced in the coil 2 2 = I Rt = 1 X 100 X 5 X 60 J = 30,000J = 30 kJ. Ans: (b) 45 22. When 22 Wh two t identical id ti l b batteries tt i off internal resistance 1Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2,. If J1 = 2.25 2 25 J2 , then what is the value of R in Ω? (a) 2Ω (b) 1Ω (c) 4Ω (d) 0.5Ω 46 22 .Solution S l ti I case; J1= = II case; J2 = = Or r= 1 == = 2.25 = = R=4 Ans: ((c)) 47 23.A steady current flows in a metallic conductor of non-uniform non uniform cross cross-section section. The quantity/quantities constant along the length of conductor is/are a) Current, electric field and drift speed b) Drift speed only c) Current and drift speed d) Current only. only 48 23. Solution:- When a steady current flows through a metallic conductor of non-uniform cross section, = = or and or E (Note: j= (Note:- ;j= = ) That is both and E change g with A . Only current I remains constant. Ans: (d) 49 24 A material B has twice the specific 24. resistance of the material A. A circular wire made of B has twice the diameter of the wire made of A. Then for the two wires to have the Then, same resistance, the ratio l A / l B of th i respective their ti llengths th mustt b be 1 ((a)) 4 ((b)) 2 (c) 1 (d) 21 50 24. Solution:- given RA = RB or or = = = 2 = Ans: (d) 51 25. An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3 then the ratio of the currents passing through the wire will be (a) 3 (b) 1/3 (d) 2 (c) 8/9 52 25 S 25. Solution:l i = X = = = X =3 For wires connected in parallel parallel, V1 =V2 or I1R1 = I2R2 or = = Ans: (b) 53 26. A 25 W-220 V bulb and a 100 W- 220 V bulb are joined in series and connected to the mains. mains Which bulb will glow brighter? (a) 25 W bulb (b) 100 W bulb (c) First 25 W bulb and then 100 W bulb. (d) Both will glow with same brightness. 54 26. Solution:- In the series circuit, same current g each bulb. But the 25W bulb has flows through higher resistance (R = V2 / P) . It produces more heat per second (P = I2 R) and hence glows brighter than 100 W bulb. Ans: (a) 55 27. A 30 V, 90W lamp is to be operated on a 120 V DC line line. For proper glow glow, a resistor of ....Ω Ω should be connected in series with the lamp. (a) 40 (b) 10 (c) 20 (d) 30 56 2 Solution:27. S l i Resistance of the lamp, R = =10 Safe current through the lamp, I = = = 3A When the lamp is operated on 120V line, the current should not exceed 3A, Then , = = =40 Required resistance to be put in series= R1-R = 40-10 = 30 Ans: (d) 57 28.The 28 Th equivalent i l t resistance i t b between t points i t A and B of an infinite network of resistances, h off 1Ω connected t d as shown, h iis each (a) infinite (c) 1 √5 Ω 2 (b) 2 Ω (d) zero 58 28.Solution:- Let X be the equivalent resistance between A and B. the network consists of infinite units of two resistors of 1 1 . The addition of one more such unit across AB will not affect total resistance. The network will then appear a shown below; 59 P 1 1 1 1 = Or Or X 1 2 = X Q =X 1 √1 4 1 √5 – X – 1 = 0 or X = = 2 2 (since X cannot be ‐ ve) Ans: (C) 60 29.In the circuit shown the value of I in ampere is (a) 1 (c) 0.4 (b) 0.60 (d) 1.5 61 29.Solution-:-the equivalent circuit is as shown in the fig. Current I = I = 1A i 4 2 .4 6 = I i i 4V 1 .6 1 .6 Now I1 X 4 = 6 X I or I1 = But I = I + I1 or 1 = I + =+ Or I = + Ans: (c) 62 30. Two cells, having the same emf are connected in series through g an external resistance R. Cells have internal resistances r1 and r2 (r1 > r2) respectively. When the circuit is closed, the potential difference across the th fifirstt cellll iis zero. Th The value l off R iis 63 30 .Solution:- given; V1 = E – Ir1 =0 I R E 1 ,r1 I E 2 ,r2 Or I = Also I = Or = Or R + r1 + r2 = 2r or R = r1 - r2 Ans: (b) 64 All the Best Th k you Thank 65