E M induction& AC current MCQ. Class XII TN State Board pptx

Electromagnetic Induction and Alternating current. (Chapter - IV) Physics Multiple choice questions and answers with explanation. (Class XII Physics TN State board)

1. Electromagnetic induction and Alternating current MCQ Class-XII (TN State Board) (With explanation)

2. By Dr M. Arunachalam Head, Department of Physics ( Rtd .) Sri SRNM College, Sattur

3. 1. An electron moves on a straight line path XY as shown in the figure. The coil abcd is adjacent to the path of the electron. What will be the direction of current, if any, induced in the coil? ( a) The current will reverse its direction as the electron goes past the coil (b) No current will be induced (c) abcd a (d) adcb b d c x electron y

4. Solution We know that, the induced current will be in a direction so that it opposes the cause ( ie . Motion of electron) that produces it. Hence the Answer is a

5. 2. A thin semi-circular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure. The potential difference developed across the ring when it moves with a speed v , is Zero (b) Brv 2 /2 and P is at higher potential (c) π rBv and R is at higher potential (d) 2 rBv and R is at higher potential

6. Solution We know that, Induced emf e = Blv B- Magnetic field v- Speed of the ring l-length = 2r ie . e = 2rBv P will be at lower potential and R will be at higher potential Ans: d

7. 3. The flux linked with a coil at any instant t is given by φ B = 10t 2 − 50t + 250. The induced emf at t = 3 s is −190 V (b) −10 V (c) 10 V (d) 190 V e = -d φ /dt = d/dt(10t 2 − 50t + 250) Differentiating the above equation, we get, e = -[2x10t -50] at t = 3 seconds, we get e = -[2x10x3 -50] = - [60 – 50] = -10 e = -10V Ans: b

8. 4. When the current changes from +2A to − 2A in 0.05 s, an emf of 8 V is induced in a coil. The co-efficient of self-induction of the coil is 0.2 H (b) 0.4 H (c) 0.8 H (d) 0.1 H e = L dI /dt ie . L = e/( dI /dt) L – Coefficient of self induction Induced emf = 8V dI – Change in current = 2A- (-2A) = 4A dt – Change in time = 0.05 seconds L = 8/(4/0.05) = 8x0.05/4 = 0.1 L = 0.1H Ans: d

9. 5. The current i flowing in a coil varies with time as shown in the figure. The variation of induced emf with time would be

10. Solution We know that, e = -L di /dt When i increases with time, e will decrease and negative. When I is constant e = 0. Thus Answer a is correct

11. 6. A circular coil with a cross-sectional area of 4 cm 2 has 10 turns. It is placed at the centre of a long solenoid that has 15 turns/cm and a cross-sectional area of 10 cm 2 . The axis of the coil coincides with the axis of the solenoid. What is their mutual inductance? 7.54 μ H (b) 8.54 μ H (c) 9.54 μH (d) 10.54 μH M = µ n 1 n 2 A 2 l 2 M – Mutual inductance µ – Permeability of free space = 4 π x10 -7 H/m n 1 - No. of turns /m in the solenoid = 15/cm = 1500/m

12. Contd … n 2 l 2 = N 2 – No. of turns in the coil = 10 l 2 – Lenth of the coil A 2 – Area of cross-section of the coil = 4 cm 2 = 4x 10 - 4 m 2 Substituting the values in the equation for M, we get, M = µ n 1 n 2 l 2 A 2 = 4 π x10 -7 x1500x10x 4x 10 - 4 ie , M = 12,56x60x10 +3 x10 -11 = 753.6x10 -8 H = 7.536x10 -6 H ie . M = = 7.54µH Answer a

13. 7. In a transformer, the number of turns in the primary and the secondary are 410 and 1230 respectively. If the current in primary is 6A, then that in the secondary coil is 2 A (b) 18 A (c) 12 A (d) 1 A N P – No. of turns in the primary = 410 N S – No. of turns in the secondary = 1230 i P – Current in the primary = 6A i S – current in the secondary

14. Solution We know that, N S / N P = i P / i S i S = i P N P / N S = 410 x6/ 1230 3 = 6/3 = 2 Current in the secondary coil i S = 2A Ans: a

15. 8. A step-down transformer reduces the supply voltage from 220 V to 11 V and increase the current from 6 A to 100 A. Then its efficiency is 1.2 (b) 0.83 (c) 0.12 (d) 0.9 Efficiency = V S i S / V P i P = 11x100/220x6 = 1100/1320 = 0.83 Efficiency = 0.83 Ans: b

16. 9. In an electrical circuit, R, L, C and AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and current in the circuit is π /3. Instead, if C is removed from the circuit, the phase difference is again π /3. The power factor of the circuit is 12 (b) 12 (c) 1 (d) 32 Power factor = cos φ φ – Phase difference = π /3 – π /3 = 0 Therefore, Power factor = cos φ = cos(0) = 1 Power factor = 1 Ans: c

17. 10. In a series RL circuit, the resistance and inductive reactance are the same. Then the phase difference between the voltage and current in the circuit is 4 (b) 2 (c) 6 (d) zero For an LR circuit, φ = tan -1 (X L /R) ie . tanφ = (X L /R) Φ – Phase difference R – Resistance = R X L - Inductive reactance = R Therefore, tan φ = R/R = 1, ie . Φ = tan -1 (1) = 45 = π /4 Φ = π /4 Ans: a

18. 11. In a series resonant RLC circuit, the voltage across 100 Ω resistor is 40 V. The resonant frequency ω is 250 rad/s. If the value of C is 4 μF , then the voltage across L is 600 V (b) 4000 V (c) 400V (d) 1 V Voltage across L = V L = IX L = I ώ L ----- 1 At resonance, X L =X C ie . ώ L = 1/ ώ c ώ – Resonant angular frequency = 250 rad/s C – Capacitance = 4 μF = 4x 10 - 6 F Hence , X L =X C = 1/250x 4x 10 - 6 = 10 +6 /1000 = 10 +3 = 1000ohm = X L

19. Contd … Voltage across resistance = V = 40Volt Resistance= R = 100 ohm Current through the circuit is given by I = V/Z = V/R (At resonance) = 40/100 = 0.4A Substituting the values in equation 1 , we get V L = 0.4x 1000 = 400volt Ans: c

20. 12. An inductor 20 mH , a capacitor 50 μF and a resistor 40 Ω are connected in series across a source of emf V = 10 sin 340 t. The power loss in AC circuit is 0.76 W (b) 0.89 W (c) 0.46 W (d) 0.67 W capacitance = 50 μF = 50 x 10 - 6 F X L = ώ L and X C = 1/ ώ c Angular frequency ώ = 340 Self inductance = 20 mH = 20 x 10 - 3 H X C = 1/ ώ c = 1/340x 50 x 10 - 6 = 1/17000 x 10 - 6 = 10 + 6 / 17000 = 10 -+3 /17 X C = 1000/17 = 58.8 Ω

21. Contd … X L = ώ L = 340x 20 x 10 - 3 = 6800x 10 - 3 Ω = 6.8 Ω Z = R 2 + ( X C -X L ) 2 = (40) 2 + (58.8 – 6.8) 2 = (40) 2 + (52) 2 Z – Impedance in the circuit R – Resistance = 40 Ω  

22. Contd … Z = 1600 + 2704 = 4304 = ie . Z = 4304 Power = I 2 rms R = (V rms / Z) xR = [(V P / V P – Peak value of the AC voltage = 10V ie . Power = (V P / Z = (100/2)x40/4304 = 2000/4304 = 0.46 ie . Power = 0.46W Ans: c  

23. 13. The instantaneous values of alternating current and voltage in a circuit are i = 1 / and v = (1/ The average power in watts consumed in the circuit is   1/4 W (b) 3/4 W (c) 1/2W (d) 1/8 W P = V rms I rms cos φ φ – Phase difference = = 60 cos φ = cos 60 = 1/2 V rms = V P / x = 1/2 I rms = I P / x = 1/2 ie . P = (1/2) (1/2) (1/2) = 1/8 Ans: d  

24. 14. In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic fields is Q 2 (b) Q 3 (c) Q 2 (d) Q At any instant, the total energy is the sum of magnetic energy q 2 /2C and electrostatic energy Li 2 /2. q – Charge on the capacitor at any instant Q – Maximum charge stored in the capacitor L – Self inductance of the coil

25. Contd … i - Current Total energy = Q 2 /2C = q 2 /2C + Li 2 /2. When both magnetic energy and electrostatic energy are equal q 2 /2C = Li 2 /2 So, we can write, Q 2 /2C = q 2 /2C + q 2 /2C = q 2 /C ie q 2 = Q 2 /2 Therefore, q = Q/ 2 Ans: c  

26. 15. 20/ π 2 H inductor is connected to a capacitor of capacitance C. The value of C in order to impart maximum power at 50 Hz is 50 μ F (b) 0.5 μ F (c) 500 μF (d) 5 μF The maximum power will be imparted at resonance At resonance X L =X C ie . ώ L = 1/ ώ c ie . 2 π fL = 1/2 π fC Therefore, C = 1/( 2 π f) 2 L L – Self inductance = 20/ π 2 H f- Frequency of AC = 50Hz ie . C = 1/4 π 2 x(50) 2 x20/ π 2 = 1/200000

27. Contd … ie . C = 1/4 π 2 x(50) 2 x20/ π 2 = 1/200000 = 1/2x10 +5 C = 0.5 x10 -5 = 5x10 -6 ie . C = 5 μF Ans: d

28. Thank You