(a) A cell of emf `E` and internal resistance `r` is connected to a resistance `R`.
(i) Relate `r` and `R`, so that power in `R` is maximum.
(ii) Maximum power consumed by `R`.
(iii) Efficiency.
(b) Find value `r` in terms of `R` so that power in external circuit is maximum.

(c) A battery of internal resistance `4Omega` is connected to the network of resistance as shown in the figure. what must be the value of `R` so that maximum power is delivered to the network? what is tha maximum power?

Explanation

(a)
Current in the circuit
`i=E/(R+r)`
(i) Power consumed in `R`,
`P=i^(2)R = ((E)/(r+R))^(2) R = (E^2)/((r+R)^(2))`
For `P ` to be maximum
`(dP)/(dR) = E^2((r+R)^(2)*1-R*2(r+R))/((r+R)^(4))`
`(r+R)-2R = 0`
`R=r`
(ii) When `r = R, i_(max) = E/(2R) or E/(2r)`
Power consumed in `R, P_("max") = i_("max")^2 R=(E^2)/(4R)`
(ii) Input power `P_("in") = E i_("max") = (E^2)/(2R)`
Output power `P_("out") = (E^2)/(4R)`
Effeciency `= ("Output power")/("Input power") xx100%`
`eta = (P_("out"))/((P_("in"))xx100-(E^2//4R)/(E^2//2R)xx100`
`=50%`
(b)
Power in external resistance is maximum if
`r = R/3`
(c)
Power in `2R` is maximum ify
`2R - 4 implies R = 2 Omega`
`i=E//(4+2R) = E//8`
`P_(max) = i^2xx2R=(E/8)^(2)xx4= (E^2)/(16)`.