A battery has an open circuit potential difference of `6V` between its terminals. When a load resistance of `60 Omega` is connected across the battery,the total power supplied by the bttery is `0.4 W`. What should be the load resistance `R`, so that maximum power will be dissipated in `R`. Calculated this power. What is the total power supplied by the battery when such a load is connected?

Explanation

Whe the circuit is open `V=E`
`:. E=6V`
Let r be the internal resistance of the battery
Power supplied by the battery in this case is
`P=E^2/(R+r)`
Subsituting the value, we have `0.4=((6)^2)/(60+r)`
Solving this we get `r=30 Omega`
Maximum power is dissipated in the circuit when net external resistance is equal to net internal resistance or
`R=r`
`:. R=30Omega`
further, total power supplied by the battery under this condition is
`P_("total")=E^2/(R+r)=((6)^2)/(30+30)`
`=0.6W`
of this 0.6 W half of the power is dissipated in R and half in r. Therefore, maximum power dissipated in R would be
`0.6/2=0.3W`