Question

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?

Short answer

The power of the battery is 12mW.

Step-by-step solution

Identification of the given data

• 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 $t_s=5.00\mathrm{s}$.

Understanding the concept of the power

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)

Calculation of the power of the battery

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.