InterviewSolution
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InterviewSolution
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An element with `emf epsilon ` and interval resistance `r` is connected across an external resistance `R`. The maximum power in external circuit is `9 W`. The current flowing through the circuit in these conditions is `3 A`. Then which of the following is // are correct ?A. ` epsilon = 6 V`B. ` r = R `C. `r = 1 Omega`D. `r = 3 Omega` |
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Answer» Correct Answer - A::B::C Current through the circuit is `I = epsilon //( r + R )`. Power dissipated in external resistance is `P = I^(2) R = (epsilon^(2))/(( r + R )^(2)) R` We know that power dissipated is maximum when `r = R`. `P_(max) = (epsilon^(2) r) /( 2r)^(2) = (epsilon^(2))/( 4 r ) = 9 W , i = ( epsilon ) /( 2r) = 3 A` Solve to get , ` epsilon = 6 V and r = 1 Omega`. |
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