Answer
a)$r=0.65\Omega$
b)$P= 3.44W$
Work Step by Step
We have $I =2.3A$ and $\epsilon = 1.5V$
a) internal resistance $ r= \frac{\epsilon}{I} = \frac{1.5}{2.3}\Omega $
$r=0.65\Omega$
power dissipated inside battery(by internal resistance) i
$P = I^2r = (2.3)^2*0.65 W $
$P= 3.44W$
