two resistors are connected in series with 5 volt battery of negligible internal resistance a current of 2 ampere flows through each resistor,if they are connected in parallel with the sane battery a current of 25/3 flows through combination.calculate value of each resistance
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herw is your answer
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For inspection method use:
(R1-R2)^2=(R1+R2)^2-4(R1×R2)
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