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Two resistors of resistances 2 Omega and 6 Omega are connected in parallel. This combination is then connected to a battery of emf 2 V and internal resistance 0.5 Omega . What is thecurrent flowing through the battery ? |
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Answer» Solution :`R_(P) = (R_(1)R_(2))/(R_(1) + R_(2))` `R_(p) = (2 XX 6)/((2+6)) = 1.5 Omega` `I = (E)/(R_(p) + r) = (2)/(1.5 + 0.5) = 1A` |
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