Alternating Current JEE (2024) Practice Sheet PDF

1 Tejas Lakshya JEE (2024) Practice Sheet (Physics) Alt...

1 Tejas Lakshya JEE (2024) Practice Sheet (Physics) Alternating Current Single Correct Type Questions: (1 to 15) 6. A generator produces a voltage that is given by 1. The ratio of peak value and r.m.s value of an v = 240 sin120 t, where t is in seconds. The alternating current is frequency and r.m.s. voltage are 1 (1) 60 Hz and 240 V (1) 1 (2) (2) 19 Hz and 120 V 2 (3) 19 Hz and 170 V (3) 2 (4) 1/ 2 (4) 754 Hz and 70 V 2. If an ac main supply is given to be 220 V. What 7. If E0 represents the peak value of the voltage in an would be the average e.m.f. during a positive half ac circuit, the r.m.s. value of the voltage will be cycle E0 E0 (1) (2) (1) 198 V (2) 386 V  2 (3) 256 V (4) None of these E0 E0 (3) (4)  2 3. R.M.S. value of current i = 3 + 4 sin (t + /3) is: 8. The peak value of an alternating e.m.f. E is given by (1) 5A (2) 17 A E = E0 cos t is 10 volts and its frequency is 50 Hz. 1 (3) 5 A (4) 7 A At time t = sec , the instantaneous e.m.f. is 2 2 600 (1) 10 V (2) 5 3 V 2t (3) 5 V (4) 1 V 4. An AC voltage is given by : E = E0 sin Then the T 9. The frequency of ac mains in India is mean value of voltage calculated over time interval (1) 30 c/s or Hz of T/2 seconds (2) 50 c/s or Hz (1) is always zero (3) 60 c/s or Hz (2) is never zero (4) 120 c/s or Hz (3) is (2E0/) always (4) may be zero 10. The root mean square value of the alternating current is equal to 5. Find the Average and rms value for the saw-tooth (1) Twice the peak value (2) Half the peak value voltage of peak value V0 from t = 0 to t = 2T as 1 shown in figure. (3) times the peak value +V 0 2 (4) Equal to the peak value + + 2T V t 0– – T T 3 11. Voltage and current in an ac circuit are given by T 2 2     –V 0 V = 5sin 100t −  and I = 4sin 100t +   6  6 V0 V0 (1) 0, (2) 0, (1) Voltage leads the current by 30° 1 5 (2) Current leads the voltage by 30° V V (3) Current leads the voltage by 60° (3) 0, 0 (4) 0, 0 3 6 (4) Voltage leads the current by 60°  2 12. In an ac circuit, the reactance of the coil is 3 times L (4) At t = , voltage drop across the resistor is its resistance, the phase difference between the R voltage across the coil to the current through the coil  1 will be E 1 −   e (1) /3 (2) /2 17. Consider the circuit shown. The switch is closed at (3) /4 (4) /6 t = 0. Currents in various branches are marked. Which of the following is correct? 13. The reactance of a 25μF capacitor at the ac frequency of 4000 Hz is 5 5 (1) ohm (2) ohm   (3) 10 ohm (4) 10 ohm 14. In an ac circuit, the current lags behind the voltage by π/3. The components in the circuit are (1) R and L (2) R and C    (1) At t = 0, I1 = , I2 = , I3 = (3) L and C (4) Only R 3R 6R 6R   15. The circuit given in figure has a resistanceless choke (2) At t = 0, I1 = , I3 = 3R 3R coil L and a resistance R. The voltages across R and    L are also given in the figure. The virtual value of (3) At t = , I1 = , I2 = , I3 = the applied voltage is 2R 4R 4R   (4) At t = , I1 = , I2 = 2R 2R Passage Type Questions (18 to 20) Passage - I In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch (1) 100 V (2) 200 V S1. Also when S1 is opened and S2 is closed the (3) 300 V (4) 400 V capacitor is connected in series with inductor (L). One or More than One Correct Type Questions: (16 to 17) 16. An inductor L and a resistance R are connected in series with a battery of emf E and a switch. Initially the switch is open. The switch is closed at an instant t = 0. Select the correct alternatives (1) At t = 0, induced emf in inductor is zero L (2) At t = , energy stored in the inductor is 18. At the start, the capacitor was uncharged. When R switch S1 is closed and S2 is kept open, the time  1   LE  2 2 − constant of this circuit is . Which of the following  e   2R2  1     is correct? L E (1) after time interval , charge on the capacitor is (3) At t = , induced emf in the inductor is CV R e 2 (2) after time interval 2, charge on the capacitor is CV(1 – e–2)  3 (3) the work done by the voltage source will be half of the heat dissipated within the capacitor is fully charged LV 2 (C) (r) joule (4) after time interval 2, charge on the capacitor 2R2 is CV(1 – e–1) 19. When the capacitor gets charged completely, S1 is opened and S2 is closed. Then, (1) at t = 0, energy stored in the circuit is purely in the form of magnetic energy LV 2 (2) at any time t > 0, current in the circuit is in (D) (s) joule the same direction 36 R 2 (3) at t > 0, there is no exchange of energy between the inductor and capacitor (4) at any time t > 0, instantaneous current in the LV 2 (t) joule C 18R2 circuit may be V L (1) A→p, r; B→p, r; C→q, s; D→p, r, t (2) A→p, r, s; B→p, q; C→p; D→r 20. Given that the total charge stored in the LC circuit is (3) A→p; B→q; C→p, r, s; D→r, t Q0, for t  0, the charge on the capacitor is (4) None of these  t  (1) Q = Q0 cos  +   2 LC  22. Match the following:  t  (2) Q = Q0 cos  −  Column-I Column-II 2 LC  Resonant (A) (p) VI cos  d 2Q frequency (3) Q = −LC 2 dt 0 L (B) Quality factor (q) 1 d 2Q R (4) Q=− LC dt 2 1 (C) Average power (r) LC Match the Column Type Questions: (21 to 22) 21. Column I shows different L-R circuits connected (1) A→p; B→q; C→r with cell. Column II shows amount of current (2) A→q, r; B→p, q; C→q, r passing through the inductor & magnetic energy (3) A→r; B→q; C→p stored in inductor at steady state. Match the column (4) None of these I with appropriate value in column II. Column-I Column-II Integer Type Questions (23 to 25) 23. In the given circuit the potential difference across (A) (p) V/R ampere resistance is 54 V and power consumed by it is 16 W. If AC frequency is 60 Hz find the value of L (Use approximation) in Henry. (B) (q) V/3R ampere  4 24. A 10 ohm resistance coil has 100 turns. It is placed in magnetic field of magnetic induction 5 × 10–4 tesla in 0.1 sec. If the area of the coil is one square metre, then the induced emf is 5 × 10–k volt. Find the value of k. 25. In the circuit given below, if the reading of the voltmeter is 100k. Find the value of k.  5 Hints and Solutions 1. (3) 14. (1) 2. (1) 15. (2) 3. (2) 16. (2, 3, 4) 4. (4) 17. (2, 4) 5. (3) 18. (2) 6. (3) 19. (4) 7. (4) 20. (3) 8. (2) 21. (1) 9. (2) 22. (3) 10. (3) 23. (1) 11. (3) 24. (1) 12. (1) 25. (2) 13. (1) PW Web/App - https://smart.link/7wwosivoicgd4 Library- https://smart.link/sdfez8ejd80if