The following table gives the electric potential difference${\mathbf{V}}_T$across the terminals of a battery as a function of currentbeing drawn from the battery.

(a) Write an equation that represents the relationship between${\mathbf{V}}_T$and$\mathit{i}$. Enter the data into your graphing calculator and perform a linear regression fit of${\mathbf{V}}_T$versus.$\mathit{i}$From the parameters of the fit, find

(b) the battery’s emf and

(c) its internal resistance.

Terminal voltage is equal to the difference between the battery voltage and voltage drop across the internal resistance. Determine the current through the load is$i$, internal resistance of the battery is rand emf is$V$, then using Ohm’s law,determine the relationship between$V_T$and i. Plot the given data would give us the value of emf of the battery as well as internal resistance using the graphing calculator and linear regression fit.

Formula:

$V=ir$

All the current through the load would pass through the internal resistance of the battery. if we know the terminal voltage $V_T$,the current through the load as i and internal resistance of the battery with emf V . The required equation is:

$V_T=V-ir$ (1)

the least square fit method for the $V_T$against i values given to us, we can write

$V_T=13.6107-0.05985i$ (2)

Compare equation (1) and (2), determine the emf of the battery is$13.6107\text{\hspace{0.17em}V}$.

Therefore,$V\approx 13.6V$

Compare equation (1) and (2), we can conclude that internal resistance as follows:

$\begin{array}{c}r=0.05985\Omega \\ \approx 0.060\Omega \end{array}$

Determine the plot for the given condition as: