Question

When An external resistance R is connected with the battery of emf E and internal resistane r, what is the maximum power?

Read 1 Solution.

E2/ 4r     (maX )      by   current =  E/(r+R)    , by this total  power =  I^2  ( r)   +  i^2 (R) 

Best SolutionSARIKA 10 year ago is this solution helpfull: 2 0

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