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Maximum power in a 0.5Omega resistance connected with two batteries of 2V emf and 1Omega internal resistance in parallel, is ........ . |
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Answer» `(8)/(9) ` W Equivalent resistance of internal resistance `r. = (1 xx 1)/(1 + 1) = (1)/(2) = 0.5 OMEGA` TOTAL resistance of circuit = 0.5 + 0.5 = ` 1.0 Omega`  `therefore` Current flowing through the circuit `I = (E)/(R) = (2)/(1) = 2 `A `therefore ` As 2A current is flowing through the 0.5`Omega` resistor, so P `= I^(2) R = 4 x 0.5 = 2.0 ` W |
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