Question

A 50-ω resistor is connected to a 9.0 V battery. How much thermal energy is produced in 7.5 minutes?1.2 102 J1.3 103J3.0 102 J7.3 102 J

Answer 1

The power consumed by the resistor can be expressed as,

`P=(V^2)/(R)`

Substitute the given values,

`\begin{gathered} P=\frac{(9.0V)^2}{50\text{ }\Omega}(\frac{1\text{ W}}{1V^2\Omega^(-1)}) \\ =1.62\text{ W} \end{gathered}`

The thermal energy produced in the given time can be expressed as,

`E=Pt`

Plug in the known values,

`\begin{gathered} E=(1.62\text{ W)(7.5 min)(}\frac{60\text{ s}}{1\text{ min}}) \\ =729\text{ J} \\ \approx7.3*10^2\text{ J} \end{gathered}`

Thus, the thermal energy produced is 7.3*10^2 J which means fourth option is correct.