Question

50 bulbs are joined in parallel, each being a 100 V bulb taking a current of 0.6A. A battery of 54 accumulators , each of emf 2V and internal resistance 0.005Ω, is used to light up the bulbs.Will the bulbs be properly lighted,overlit or under-lit?

Answer 1

Answer:

the bulbs be properly lighted

Explanation:

50 bulbs are joined in parallel, each being a 100 V bulb taking a current of 0.6A. A battery of 54 accumulators , each of emf 2V and internal resistance 0.005Ω, is used to light up the bulbs

Bulb is 100 V bulb

Current = 0.6 A

Resistance of bulb = 100/0.6 = 500/3 Ω

Wattage of a bulb = VI  = 100 * 0.6 = 60 W

50 bulbs are in parallel

so Resistance of all bulb would be = (500/3)/50 = 10/3 Ω

54 accumulators of emf 2 V  = 108V

Internal Resistance = 54 * 0.005 = 0.27 Ω

108 - I(0.27) = I * 10/3    ( I is the total current generated )

=> 324 - 0.81I = 10I

=> 324 = 10.81 I

=> I = 324/10.81

=> I = 29.97 A

Current in each bulb = 29.97/50  = 0.599 A  ≈ 0.6A

As current is Same so the bulbs be properly lighted