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 ?

Options

A. ` epsilon = 6 V`
B. ` r = R `
C. `r = 1 Omega`
D. `r = 3 Omega`

Answer

A, B, C

Explanation

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`.