A certain x-ray tube operates at a current of 7.00 mA and a potential difference of 80.0 kV. What is its power in watts?
The power dissipated is given by the product of the current and the potential difference:
P=i V=\left(7.0 \times 10^{-3} \,A \right)\left(80 \times 10^3\, V \right)=560 \,W .