12. In the given circuit the current I¡ is19. Two batteries each of emf E and internalresistance r are connected turn by turn in seriesand in parallel, and are used to find current in anexternal resistance R. If the current in series isequal to that in parallel, the internal resistance ofeach battery is:1) R 2) R/220. In the network shown the potential differencebetween A and B is (R=r,=r2=r3=12, E;=3V,Ez=2V, E3=1V)1) 1V2) 2V3) 3V4) 4V21. Consider the circuit shown in the figure. Thecurrent I3 is equal to1) 5 amp2) 3 amp3) -3 amp4) -5/6 amp22. The potential difference between points A and Badjoining figure is1) 0.4 A2) - 0.4 A303) 0.8 A4) - 0.8 A40an.3) R/313. The emf of a battery is 2 V and its internalresistance is 0.5 Q. The maximum power which it candeliver to any external circuit will be4) R/41) 8 Watt2) 4 Watt4) None of the aboveE13) 2 Watt14. A battery is charged at a potential of 15 V for 8 hourswhen the current flowing is 10 A. The battery ondischarge supplies a current of 5 A for 15 hours. Themean terminal voltage during discharge is 14 V. The"Watt-hour" efficiency of the battery is1) 82.5%3) 90%54 22) 80 %4) 87.5%15. When the resistance of 9 Q is connected at the ends ofa battery, its potential difference decreases from 40volt to 30 volt. The internal resistance of the battery is1) 6Ω2)3) 9 04)16. Two batteries, one of emf 18 volts and internalresistance 22 and the other of emf 12 volt andinternal resistance 12, are connected as shown. Thevoltmeter V will record a reading of2.1)3508.2) v9.3215 2sa43)V4) 2V323. The current in adjoining circuit will be1) 15 volt1)ampere452) 30 volt1812022)ampere30 n4,30 a3) 14 volt4) 18 volt151ampere101211230 n17. The n rows each containing m cells in series are joinedin parallel. Maximum current is taken from thiscombination across an external resistance of 3214)ampere24. The reading of the ammeter as per figure shownresistance. If the total number of cells used are 24 andisinternal resistance of each cell is 0.5 2 then31)2) A41) m = 8, n = 32) m = 6,n = 42V3) m = 12,n = 24) m = 2,n = 123)4) 220ww18. The magnitude and direction of the circuit in thecircuit shown will beww25. A battery of e.m.f 10V is connected to resistanceas shown in figure. The potential difference VA-VBbetween the points A and B is1) -2V2) 2 V3) 5 V201)7A from a to b through e37A from b to a through e10"ww iHwb202)310 A 30ww w10 V30ww3) 1 A from b to a through e4) 1 A from a to b through e4)V10 V111/81/2

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12. In the given circuit the current I, is 1) 0.4 A 2) - 0.4 A 30 3) 0.8 A 4) - 0.8 A 40 an 40 80

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter2: Fundamentals

Section: Chapter Questions

Problem 2.17MCQ: Consider the load convention that is used for the RLC elements shown in Figure 2.2 of the text. A....

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