Answer
a. Internal resistance of the cell is $0.65Ω$
b. Power dissipation across the cell is $3.5 W$
Work Step by Step
a. Here the current is generated by only the internal resistance as the ammeter is ideal
$V = IR$
$1.5V = 2.3A\times R$
$R = 0.65Ω$
b. Power dissipation of a cell can be taken by the equation
$P=VI$ Where V is the emf of the cell and I is the current generated
$P=1.5V\times 2.3A = 3.45W \approx 3.5W$
