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In the circuit shown, the resistances are given in ohms and the battery is assumed ideal withh emf equal to 3.0 volts. Q. The resistor that dissipates maximum power |
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Answer» `R_(1)` `i_(1)=i_(2)+i_(x)` ..(i) `i_(3)=i_(4)-i_(x)` ..(ii) `V=i_(1)R_(1)+i_(1)R_(2)+i_(4)R_(4)` ..(iii) and `i_(1)R_(1)+X i_(x)-i_(3)R_(3)=0` ..(iv) Solving these equation for `i_(x)` we get from (1) & (3) `V-i_(1)(R_(1)+R_(2))-i_(x)R_(2)` ..(v) and from (2) and (3) `V-i_(3)(R_(3)+R_(4))+i_(x)R_(4)` ...(vi) Sustituting for `i_(1)` from (5) and `i_(3)` from (vi) and (iv) `R_(1)[(V+i_(x)R_(2))/(R_(1)+R_(2))]+X i_(2)-R_(3)[(V-i_(x)R_(4))/(R_(3)+R_(4))]=0` Collecting thew terms in `i_(x),i_(x)[x+(R_(1)R_(2))/(R_(1)+R_(2))+(R_(3)R_(4))/(R_(3)+R_(4))]` `=V[(R_(3))/(R_(3)+R_(4))-(R_(1))/(R_(1)R_(2))]` `impliesI_(x)=(V[(R_(3))/(R_(3)+R_(4))-(R_(1))/(R_(1)+R_(2))])/(x+(R_(1)R_(2))/(R_(1)+R_(2))+(R_(3)R_(4))/(R_(3)+R_(4)))` Proceeding to limit `xto0` (`x=` m negligibly small) `i_(x)=(25[(R_(3))/(R_(3)+r_(4))-(R_(1))/(R_(1)+R_(2))])/([(R_(1)R_(2))/(R_(1)+R_(2))+(R_(3)R_(4))/(R_(3)+R_(4))])` substituting values, `i_(x)=(25[(3)/(7)-(1)/(3)])/([(2)/(3)+(12)/(7)])=(25xx2)/(50)=`lamp |
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