The power consumed by the resistor can be expressed as,
[tex]P=\frac{V^2}{R}[/tex]Substitute the given values,
[tex]\begin{gathered} P=\frac{(9.0V)^2}{50\text{ }\Omega}(\frac{1\text{ W}}{1V^2\Omega^{-1}}) \\ =1.62\text{ W} \end{gathered}[/tex]The thermal energy produced in the given time can be expressed as,
[tex]E=Pt[/tex]Plug in the known values,
[tex]\begin{gathered} E=(1.62\text{ W)(7.5 min)(}\frac{60\text{ s}}{1\text{ min}}) \\ =729\text{ J} \\ \approx7.3\times10^2\text{ J} \end{gathered}[/tex]Thus, the thermal energy produced is 7.3*10^2 J which means fourth option is correct.
