Question

If a resistor rated at 5 watts and 6 volts are connected across a battery with an open circuit voltage of 6 volts, what is the internal resistance of the battery if the resulting current is 0.8 A?

Answer 1

$I=E/(R+r)=E/(V^2/P+r)\to$

$r=(E-IV^2/P)/I=(6-0.8\cdot 6^2/5)/0.8=0.3\ (\Omega)$