In the given electric circuit find

(a) current
(b) power output
(c) relation between r and R so that the electric power output (that means power to R) is maximum.
(d) value of maximum power output
(e) plot graph between power and resistance of load
(f) from graph we see that for a given power output there exixts two values of external resistance, prove that the product of these resistances equal to `r^2`.
(g) what is the efficiency of the cell when it is used to supply maximum power.

Explanation

(a) In the circuit shown if we assume that potential at A is zero then potential at A is zero then potential at B is `epsi-ir`. Now since the connecting wires are of zero resistance
`thereforeV_(D)=V_(A)=0rArrV_(C)=B_(B)=epsi-ri`
Now current through CD is also i (because it's in series with the cell).
`therefore i=(V_(C)-V_(D))/R=((epsi-ir)-0)/R" Current i"=epsi/(r+R)`
Note : After learning the concept of series combination we will be able to calculate the current directly
(b) Power output `P=i^(2)R=epsi^(2)/((r+R)^(2))R`
( c) `(dp)/(dR)=epsi^(2)/((r+R)^(2))-(2epsi^(2)R)/((r+R)^(3))[R+r-2R]`

for maximum power supply `(dp)/(dR)=0`
`rArr r+R-2R=0rArrr=R`
Here for maximum power output outer resistance should be equal to internal resistance
(d) `P_("max")=epsi^(2)/(4r)`
( e) Graph between 'P' and R maximum power output at R=r
`P_("max")=epsi^(2)/(4r)rArri=epsi/(r+R)`
(f) Power output `P=(epsi^(3)R)/((r+R)^(2))`
`P(r^(2)+2rR+R^(2))=epsi^(2)R`
`R^(2)+(2r-epsi^(2)/P)R+r^(2)=0`
Above quadratic equation in R has two roots `R_(1)" and "R_(2)` for given values of `epsi,` P and r such that
`thereforeR_(1)R_(2)=r_(2)("product of roots")`
`r^(2)=R_(1)R_(2)`
(g) Power of battery spent `=epsi^(2)/((r+)^(2)),2r=epsi^(3)/(2r)`
Power (output)`=(epsi/(r+r))^(2)xxr=epsi^(2)/(4r)`
`"Efficiency ="("Power output")/("total power by cell")=(epsi^(2)/(4r)xx100)/(epsi^(2)/(2r))=1/2xx100=50%`