The rating of bulb B1 is 60 V, 12 W & bulb B2 is 100 V, 100 W. Find the maximum emf of the battery so that all bulbs remain safe.

ShresthaGuru

# The rating of bulb B1 is 60 V, 12 W & bulb B2 is 100 V, 100 W. Find the maximum emf of the battery so that all bulbs remain safe.

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## Kashyap abhishek

For B_{1},V_{1}= 60 V, P_{1}= 120 WR_{1}=^{V1²}/_{P}_{1}=^{60²}/_{120}= 300 ΩFor B_{2},V_{2}= 100 V, P_{2}= 100 WR_{2}=^{V2²}/_{P}_{2}=^{100²}/_{100}= 100 ΩSince, B_{1}and B_{2}are connected in parallel.So,^{1}/_{R}_{p}=^{1}/_{R}_{1}+^{1}/_{R}_{2}=^{1}/_{300}+^{1}/_{100}=^{4}/_{100}⇒ R_{p}=^{100}/_{4}= 75 ΩSo, R_{eq }across circuit = 75 + 300 + 25 = 400 ΩCurrent that can flow through B_{1}=^{V1}/_{R}_{1}=^{60}/_{300}=^{1}/_{5}ACurrent that can flow through B_{2}=^{V2}/_{R}_{2}=^{100}/_{100}= 1 ASo, maximum current that can flow through the circuit so that all bulbs remain safe =^{1}/_{5}ATherefore, required e.m.f. = IR_{eq}=^{1}/_{5}x 400 = 80 V