Current Electricity MCQ Class XII. Physics pptx

MCQs in Current Electricity' Based on Class XII (TN State Board Syllabus) Physics. Answers with explanation.

1. Current Electricity MCQ Class XII By Dr M. Arunachalam Head, Department of Physics ( Rtd .) Sri SRNM College, Sattur (TN State Board MCQs with explanation)

2. 1 . The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor? (a) 2 ohm (b) 4 ohm (c) 8 ohm (d)1 ohm R = V/I = 4/2 = 2ohm Ans a

3. 2. A wire of resistance 2 ohms per meter is bent to form a circle of radius 1m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is π Ω (c) 2 Ω (c) 2π Ω (d) 4Ω Length of the wire between A and B is πr = πx1m A B Resistance of the semicircular wire = π x2 = 2π Ω We can assume that the circular wire is the combination of two resistances (each of 2π Ω ) connected in parallel. The equivalent resistance is R P 1/ R P = 1/ 2π + 1/ 2π = 2 / 2π = 1/ π ie. R P = π Ω Ans: a

4. 3. A toaster operating at 240 V has a resistance of 120 Ω. Its power is a) 400 W b) 2 W c) 480 W d) 240 W Power P = VI But I = V/R ie . I = 240/120 = 2A ie . P = 240x2 = 480W Ans: C

5. 4. A carbon resistor of (47 ± 4.7 ) k Ω to be marked with rings of different colours for its identification. The colour code sequence will be a) Yellow – Green – Violet – Gold b) Yellow – Violet – Orange – Silver c) Violet – Yellow – Orange – Silver d) Green – Orange – Violet – Gold B B R O Y G B VG W Y-Yellow-4 V-Violet-7 O-Orange-3zeros 0 1 2 3 4 5 6 7 8 9 47 k Ω Silver-10% tolerance ( + 4.7 k Ω) Hence we have, Yellow – Violet – Orange – Silver Ans: b

6. 5. What is the value of resistance of the following resistor? ( a)100 k Ω (b)10k Ω (c) 1k Ω (d)1000 k Ω B-Brown=1 Black -0 Yellow- 4 zeros 10,0000 Ω = 100 k Ω Ans: a B B R O Y G B VG W 0 1 2 3 4 5 6 7 8 9

7. 6. Two wires of A and B with circular cross section are made up of the same material with equal lengths. Suppose R A = 3 R B , then what is the ratio of radius of wire A to that of B? 3 (b) √ 3 (c) 1/ √ 3 (d) 1/3 R = ρ L/A = ρ L/ πr 2 Both the wires have equal length and of same materials. So L and ρ are the same for A and B R A = ρ L/ πr A 2 and R B = ρ L/ πr B 2 But R A = 3 R B ie . ρ L/ πr A 2 = 3 ρ L/ πr B 2 ie. r A 2 / r B 2 = 1/3 ie . r A / r B = 1/ √ 3 Ans. C

8. 7. A wire connected to a power supply of 230 V has power dissipation P 1 . Suppose the wire is cut into two equal pieces and connected parallel to the same power supply. In this case power dissipation is P 2 . The ratio P 2 / P 1 is ( a)1 (b) 2 (c) 3 (d) 4 Let us assume that, before cutting, the wire has a resistance 2R ( ie . two resistors each of value R ohm connected in series). Now the power dissipation is P 1 = V 2 /2R ----- 1 After cutting, we have two resistors ( each of value R ohm) connected in parallel.

9. Contd … Now we have, 1/R P = 1/R + 1/R = 2/R ie R P = R/2 Power dissipation P 2 = V 2 / R/2 ----- 2 Dividing equation 2 by equation 1 we get P 2 / P 1 = [V 2 / (R/2)] x (2R/ V 2 ) = (2 V 2 / R) x (2R/ V 2 ) = 4 ie . P 2 / P 1 = 4 Ans: d

10. 8. In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60W bulb for use in India is R , the resistance of a 60W bulb for use in USA will be R (b) 2 R (c) R/4 (d) R/2 P = V 2 /R ie . R = V 2 / P For India R = R I = (220) 2 / 60 ie . R I = (2x110) 2 / 60 ie . R I = 4x(110) 2 / 60 ----- 1 For USA R USA = (110) 2 / 60 ----- 2 Dividing equation 2 by equation 1, we get R USA = (110) 2 / 60 ie . R USA = R I /4 = R /4 ie . R USA = R /4 Ans: c R I 4x(110) 2 / 60

11. 9. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1k W are connected. The voltage of electric mains is 220 V. The maximum capacity of the main fuse of the building will be 14 A (b) 8 A (c) 10 A (d) 12 A Power P = VI ie. I = P/V I = 2500/220 = 11.1A Max.capacity of main fuse is 12 A Ans: d 15 bulbs x 40W = 600W 5 bulbs x 100W = 500W 5 fans x 80W = 400W 1Heater x 1000W= 1000W Total - 2500W

12. 10. There is a current of 1.0 A in the circuit shown below. What is the resistance of P ? 1.5 Ω b) 2.5 Ω c) 3.5 Ω d) 4.5 Ω By Ohm’s law, R = V/I ie . R = 9/1 = 9 Ω Resistors are connected in series ie . R = 9 = 3 + 2.5 + P ie . 9 = 5.5 + P ie . P = 9 – 5.5 = 3.5 Ω Ans: C

13. 11. What is the current drawn out from the battery? 1A b) 2A c) 3A d) 4A By Ohms’s law, I = V/R Resistors are connected in parallel 1/R P = 1/15 + 1/15 + 1/15 = 3/15 ie . R P = 15/3 = 5 ohm ie . I = V/R = 5/5 = 1A Ans: a

14. 12. The temperature coefficient of resistance of a wire is 0.00125 per °C. At 20°C, its resistance is 1 Ω. The resistance of the wire will be 2 Ω at 800 °C b) 700 °C c) 850 °C d) 820 °C R 2 = R 1 [1+ α (T 2 - T 1 )] R 1 = Resistance at temperature T 1 (20 °C) = 1 Ω R 2 = Resistance at temperature T 2 = 2 Ω α – Temperature coefficient of resistance = 0.00125 per °C 2 = 1[1+ 0.00125 (T 2 - T 1 )] ie . 2 = 1+ 0.00125 (T 2 - T 1 )

15. Contd … 2 -1 = 0.00125 (T 2 - T 1 ) 1 = 0.00125 (T 2 - T 1 ) (T 2 - T 1 ) = 1/0.00125 = 800 ie . T 2 = 800 + T 1 = 800 + 20 = 820°C Ans: d

16. 13. The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is 0.2 Ω b) 0.5 Ω c) 0.8 Ω d) 1.0 Ω Current I = V/ R T R T = Total Resistance = R+r R – Resistance = 10 Ω r – Internal resistance of the battery V – Potential of the battery = 2.1V I – Current = 0.2A R T = R+r = V/I ie . r = V/I – R = (2.1/0.2) – 10 ie . r = 10.5 – 10 = 0.5 Ω Ans: b

17. 14. A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of a) each of them increases b) each of them decreases c) copper increases and germanium decreases d) copper decreases and germanium increases Ans: d) copper decreases and germanium increases Copper is a conductor. Its resistance is directly proportional to temp. It has positive temp. coefficient Germanium is a semi-conductor. It has negative temperature coefficient

18. 15. In Joule’s heating law, when R and t are constant, if the H is taken along the y axis and I 2 along the x axis, the graph is straight line b) parabola c) circle d) ellipse By Joule’s law, H = I 2 Rt When R and t are constants, H α I 2 I 2 R ie . The equation represents a straight line Ans: a

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