The current through the battery and resistors 1 and 2 in Figure-ais 2.00 A. Energy is transferred from the current to thermal energy E_thin both resistors. Curves 1 and 2 in Figure-bgive that thermal energy E_th for resistors 1 and 2, respectively, as a function of time t. The vertical scale is set by ${\mathit{E}}_{\mathbf{t}\mathbf{h}\mathbf{,}\mathbf{s}}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mJ}$, and the horizontal scale is set by ${\mathit{t}}_{\mathbf{s}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathit{s}$. What is the power of the battery?

• The thermal energy is$E_{th}$.
• The time is t.
• The vertical scale is set by$E_{th,s}=40.0\mathrm{mJ}$.
• The horizontal scale is set by $t_s=5.00\mathrm{s}$.

In this problem, using the information from the given graph in the formula for the thermal energy, we can find the corresponding powers for resistance 1 and 2. By applying energy conservation, we can find the power of the battery.

Formulae:

The thermal energy transferred in the system is,

$E_{th}=Pt$ … (i)

According to the conservation of energy, the power supplied by the battery is,

$P_{battery}=P_1+P_2$ … (ii)

From equation (i), we can get the value of power as follows:

$P=\frac{E_{th}}{t}$ … (iii)

Thus, from the given graph, we get

$$\begin{gathered} P=\frac{E_{th}}{t} \\ =\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{graph} \end{gathered}$$

From the given graph, for resistance 1, we get:

$$\begin{gathered} E_{th,s}=40.0\mathrm{mJ} \\ {t}_s=5.00\mathrm{s} \end{gathered}$$

Now, for resistance 1, the power value using the given data in equation (iii) can be calculated as follows:

$$\begin{gathered} P_1=\frac{40.0\mathrm{mJ}}{5.00\mathrm{s}}\left(\frac{1\mathrm{mW}}{1\mathrm{mJ}/\mathrm{s}}\right) \\ =8.00\mathrm{mW} \end{gathered}$$

From the given graph, for resistance 2, we get:

$$\begin{gathered} E_{th,s}=20.0\mathrm{mJ} \\ {t}_s=5.00\mathrm{s} \end{gathered}$$

Similarly, for resistance 2, the power value using the given data in equation (iii) can be calculated as follows:

$$\begin{gathered} P_2=\frac{20.0\mathrm{mJ}}{5.00\mathrm{s}}\left(\frac{1\mathrm{mW}}{1\mathrm{mJ}/\mathrm{s}}\right) \\ =4.00\mathrm{mW} \end{gathered}$$

According to the conservation of energy, the power supplied by the battery can be calculated using equation (ii) as follows:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{battery}}=8.00\mathrm{mW}+4.00\mathrm{mW} \\ =12\mathrm{mW} \end{gathered}$$

Hence, the value of the power of the battery is 12 mW.