Please resolve the questions again using current scale as 0.00,0.19,0.23,0.27,0.33 And potential different with 3.90,3.69,3.65,3.61,3.55

The provided data for current ($I$) and potential difference ($V$) are:

• Current ($I$): $0.00 \text{ A}, 0.19 \text{ A}, 0.23 \text{ A}, 0.27 \text{ A}, 0.33 \text{ A}$
• Potential Difference ($V$): $3.90 \text{ V}, 3.69 \text{ V}, 3.65 \text{ V}, 3.61 \text{ V}, 3.55 \text{ V}$

The terminal potential difference ($V$) of a battery is related to its electromotive force ($E$), the current ($I$) flowing through the circuit, and its internal resistance ($r$) by the equation:

$$V = E - Ir$$

This equation is in the form of a straight line $y = c + mx$, where $V$ is on the y-axis, $I$ is on the x-axis, $E$ is the y-intercept, and $-r$ is the gradient of the line.

From the equation $V = E - Ir$, when the current $I = 0 \text{ A}$ (i.e., the circuit is open), the terminal potential difference $V$ is equal to the electromotive force $E$.
Looking at the given data, the first data point is $(I=0.00 \text{ A}, V=3.90 \text{ V})$.
Therefore, the electromotive force ($E$) of the battery is:

$$E = 3.90 \text{ V}$$ $$\boxed{E = 3.90 \text{ V}}$$

The internal resistance $r$ is the negative of the gradient of the $V$ versus $I$ graph. We can calculate the gradient ($m$) using two points from the data. Let's use the first point $(I_1, V_1) = (0.00 \text{ A}, 3.90 \text{ V})$ and the last point $(I_2, V_2) = (0.33 \text{ A}, 3.55 \text{ V})$ to determine the gradient of the best-fit line.

The gradient $m$ is calculated as:

$$m = \frac{\Delta V}{\Delta I} = \frac{V_2 - V_1}{I_2 - I_1}$$

Substitute the values:

$$m = \frac{3.55 \text{ V} - 3.90 \text{ V}}{0.33 \text{ A} - 0.00 \text{ A}}$$ $$m = \frac{-0.35 \text{ V}}{0.33 \text{ A}}$$ $$m \approx -1.0606 \text{ } \Omega$$

Since $m = -r$, the internal resistance $r$ is:

$$r = -m$$ $$r = -(-1.0606 \text{ } \Omega)$$ $$r \approx 1.06 \text{ } \Omega$$

Rounding to two decimal places, the internal resistance ($r$) is:

$$\boxed{r = 1.06 \text{ } \Omega}$$