Title Alternating Current (ALL DPPs).pdf ✅

vha Alternating Current Part-1   Digital Pvt. Ltd. [1] 1. The peak value of an alternating e.m.f. which is given by E = E0 cos t  is 10 volts and its frequency is 50 Hz. At time t = 1 600 s, the instantaneous e.m.f. is (1) 10 V (2) 5 3 V (3) 5 V (4) 1 V 2. A current in circuit is given by i = 3 + 4 sin t . Then the effective value of current is : (1) 5 (2) 7 (3) 17 (4) 10 3. The relation between an A.C. voltage source and time in SI units is : V = 120 sin (100 t ) cos (100 t ) volt value of peak voltage and frequency will be respectively :– (1) 120 volt and 100 Hz (2) volt and 100 Hz (3) 60 volt and 200 Hz (4) 60 volt and 100 Hz 4. In an ac circuit, peak value of voltage is 423 volts. Its effective voltage is (1) 400 volts (2) 323 volts (3) 300 volts (4) 340 volts 5. In an ac circuit I = 100 sin 200 t . The time required for the current to achieve its peak value will be - (1) 1 sec 100 (2) 1 sec 200 (3) 1 sec 300 (4) 1 sec 400 Instantaneous, Peak and Average values DPP-01 TG: @Chalnaayaaar
vha Alternating Current Part-1   Digital Pvt. Ltd. [2] 6. The peak value of 220 volts of ac mains is (1) 155.6 volts (2) 220.0 volts (3) 311.0 volts (4) 440 volts 7. A 40  electric heater is connected to a 200 V, 50 Hz mains supply. The peak value of electric current flowing in the circuit is approximately (1) 2.5 A (2) 5.0 A (3) 7.07 A (4) 10 A 8. An ac generator produced an output voltage E = 170sin377t volts, where t is in seconds. The frequency of ac voltage is approximately : (1) 50 Hz (2) 110 Hz (3) 60 Hz (4) 230 Hz 9. In a circuit, the value of the alternating current is measured by hot wire ammeter as 10 ampere. Its peak value will be (1) 10A (2) 20A (3) 14.14A (4) 7.07A 10. The voltage of domestic ac is 220 volt. What does this represent (1) Mean voltage (2) Peak voltage (3) Root mean voltage (4) Root mean square voltage 11. A 280 ohm electric bulb is connected to 200V electric line. The peak value of current in the bulb will be (1) About one ampere (2) Zero (3) About two ampere (4) About four ampere TG: @Chalnaayaaar
vha Alternating Current Part-1   Digital Pvt. Ltd. [3] SOLUTIONS 1. (2) E E cos t =  0 Given, E 10V, 0 = t = 1 600 sec., f = 50Hz E = 10 cos 1 2 50 600         E = 10 cos E 5 3V 6   = 2. (3) I = 3 + 4 sin t  (i)  2 = 9 + 16 2 sin t  + 24 sin t  (ii) 2 I = 9 + 16 × 1 2 + 0 = 17 (iii) 2 I = 17 So, effective value of current (r.m.s.) rms I 17A = 3. (4) V = 120 sin (100 t) cos (100 t)   2 sin cos = sin 2    sin(2 100 t) sin (100 t) cos (100 t) = 2      V = 60 sin (200 t)  compare from V = V sin t O  V = 60 volt, O   = 200 2 f = 200   f = 100 Hz 4. (3) Effective voltage o r.m.s. V 423 V 300 V 2 2 = =  5. (4) Answer key Question 1 2 3 4 5 6 7 8 9 10 11 Answer 2 3 4 3 4 3 3 3 3 4 1 TG: @Chalnaayaaar
vha Alternating Current Part-1   Digital Pvt. Ltd. [4] The current takes T 4 sec to reach the peak value. So 2 2 1 T sec 200 100   = = =   So T 1 sec 4 400 = 6. (3) Peak value = 220 2 311 V = ( V V 2 P rms = ) 7. (3) r.m.s. r.m.s. V 200 i 5 A R 40 = = = 0 r.m.s.  = = i i 2 7.07 A 8. (3) 2 f 377 f 60.03 Hz  =  = 9. (3) Hot wire ammeter reads rms value of current. Hence its peak value rms =  = i 2 14.14 amp 10. (4) 11. (1) rms 200 5 i A. 280 7 = = So 0 rms 5 i i 2 2 1A. 7 =  =   TG: @Chalnaayaaar