1.

What maximum power can be obtained from a battery of emf epsilonand internal resistance r connected with an external resistance R ?

Answer»

`(epsilon^(2))/(4r)`
`(epsilon^(2))/(3r)`
`(epsilon^(2))/(2R)`
`(epsilon^(2))/(r)`

Solution :`(epsilon^(2))/(4r)`
CURRENT in CIRCUIT I = `(epsilon)/(R + r) `

If we get maximwn power through battery then EXTERNAL resistance becomes R = r.
`therefore " power P " = I^(2) R = (epsilon R)/((R + r)^(2)) `
`= (epsilon^(2) r)/((2r)^(2))[ because " PUTTING R " =r]` ,
` = (epsilon^(2))/(4r)`


Discussion

No Comment Found

Related InterviewSolutions