Q38P - Fundamentals Of Physics

Question

In Figure-a, a $\mathbf{20}\mathit{\Omega }$ resistor is connected to a battery. Figure-b shows the increase of thermal energy ${\mathit{E}}_{\mathbf{t}\mathbf{h}}$ in the resistor as a function of time t. The vertical scale is set by ${\mathit{E}}_{\mathbf{th}\mathbf{,}\mathbf{s}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathit{m}\mathit{J}$, and the horizontal scale is set by  ${\mathit{t}}_{\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}$. What is the electric potential across the battery?

Step-by-Step Solution

Answer

The electric potential across the battery is 0.10 V.

Step 1: The given data

a) The resistance, $R=20\Omega$

b) The vertical scale of graph is $E_{th,s}=2.5\mathrm{mJ}$

c) The horizontal scale of graph is $t_s=4.0s$

Step 2: Significance of the electric potential

The electric potential is the amount of work done that is needed to move a unit electric charge from one reference point to another reference point in an electric field.

We have given the graph of thermal energy versus time, so by calculating the slope of the graph, we can get the power applied by the battery. Then, using the equation of power relating to electric potential and resistance, we can calculate the electric potential across the battery.

 

Formulae:

The power subjected across the load,$P=\frac{V^2}{R}$                                                          …(i)

Here, P is the power,V is the electric potential and R is the resistance. 

The slope of the graph, $\mathrm{slope}=\frac{\mathrm{rise}}{\mathrm{run}}$                                                                            …(ii)

Step 3: Determining the electric potential across the battery

The slope of the graph can be calculated using equation (ii) as follows:

$$\begin{gathered} \mathrm{slope}=\frac{2.0\mathrm{mJ}-0.0\mathrm{mJ}}{4.0\mathrm{s}-0.0\mathrm{s}} \\ =5\times {10}^{-4}\mathrm{W} \end{gathered}$$

Now, the value of the electric potential across the battery can be calculated by using equation (i) as follows:

$$\begin{gathered} V=\sqrt{P\times R} \\ =\sqrt{5\times {10}^{-4} W\times 20 \Omega } \\ =0.10 V \end{gathered}$$

Hence, the value of the potential is 0.10V.