A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is :

Given, ε=3 VTerminal potential difference = 2.5 V  The same potential difference is across external load  "R"=VR=2.5 VSo  i=PRVR=0.52.5=0.2 ASo power supplied  =ε i=30.2=0.6 WHence power dissipated in  r=0.6−0.5=0.1 W