As we know that resistance of the resistor will remain same while connecting across different batteries
So here we will say by ohm's law
![R = (V)/(i)](https://img.qammunity.org/2020/formulas/physics/middle-school/dkl7hqsnci3wcpivs1ih1av9jh50n4uu0f.png)
here we know that
![i = 0.60 A](https://img.qammunity.org/2020/formulas/physics/middle-school/bsp19z9refdx7vk1km2xulx0k35gwcyysi.png)
![V = 373 Volts](https://img.qammunity.org/2020/formulas/physics/middle-school/99h28i9ng4p7hoikjhp56sz68thjhv28nr.png)
now we have
![R = (373)/(0.60) = 621.67 ohm](https://img.qammunity.org/2020/formulas/physics/middle-school/taye7lv7fjzxitbxt06pxw8kqz7vsikp7s.png)
now we have
another voltage as V = 139 volts
now we have current
![i = (V)/(R) = (139)/(621.67)](https://img.qammunity.org/2020/formulas/physics/middle-school/6i4ihdrvtoepbzz7r37239scrijxpvt7ee.png)
![i = 0.22 A](https://img.qammunity.org/2020/formulas/physics/middle-school/62s5569twdn1lipfpmdluffgnr3qnn15s6.png)
so final current is 0.22 A