A number of parallel connected 110V lamps are fed a 130V battery having an internal resistance of 2.6 ohm. If the resistance of each lamp is 200 ohm and the resistance of conducting wires is 0.40 ohm, the number of lamps which the battery can supply is (a) 5 (b) 8 (d) 15 (C) 12
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Answer:
d
Explanation:
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Explanation:
Emf of n no. of lamps=E=110V
resistance of n no. of lamps=r/n=200/n
current needed by n no. of lamps= 110/(200/n)= 11n/20
Current given by battery= 130/2.6+0.4+(200/n) =130n/3n+200
11n/20=130n/3n+200
11(3n+200)=2600
33n+2200=2600
n=400/33=12
that means approx 12 lamps
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