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Prove that total produced in different resistors of the circuit is minimum when the current is divided into a number of braches . |
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Answer» SOLUTION :When the current is divided into a number if parallel braches , then`I=I_(1)+I_(2)` Heat produced in the circuit in time t , `H=(I_(1)^(2)r_(1)t)/(J)+((I-I_(1))^(2)r_(2)t)/(J)`  DIFFERENTIATING both SIDE with respect to `I_(1)`, `(dH)/(dI_(1))=(2I_(1)r_(1)t)/(J)-(2(I-I_(1))r_(2)t)/(J)` When heat generated in the circuit is minimum , `(dH)/(dI_(1))=0" or, "(2I_(1)r_(1)t)/(J)-(2(I-I_(1))r_(2)t)/(J)=0` or , `(2I_(1)r_(1)t)/(J)=(2(I-I_(1))r_(2)t)/(J)=" or, "I_(1)r_(1)=(I-I_(1))r_(2)=I_(2)r_(2)` `therefore(I_(1))/(I_(2))=(r_(2))/(r_(1))` i.e., the heat produed in the circuit will be minimum if the current is divided into the braches such that the current in each branch is INVERSELY proportional to the resistance . |
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