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Three bulb each having resistance of 90 Ω are connected in parallel to an ideal battery of emf 60V. Then the current delivered by battery when all the bulb are ON is I = 0.5 AI = 1.0 AI = 1.5 AI = 2.0 A |
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Answer» a) R EFF = 3180 =60Ωi=60/60=1Ab) R eff = 2180 =90Ωi=60/90=0.67Ac) R eff =180Ω⇒i=60/180=0.33A |
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