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In the given circuit if the internal resistance of the batteries are negligible, then for what value of resistance R("in" Omega) will the thermal power generated in it be maximum. |
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Answer» <P> Solution :GIVEN circuit can be simplified as dotted part can be replaced as`epsilon=_(eq)=(6/3+0/6)/(1/3+1/6)=4V``1/r_(eq)=1/3+1/6 rArr r_(eq)=2Omega` then current `I=(10-4)/(2+R)=6/(2+R)` POWER in R, `P=(6/(2+R))^2R=(36R)/((2+R)^(2))` for P to be maximum `(dP)/(dR)=0` on solving `R=2Omega` 
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