Determine the Emf of the battery and its internal resistance using the provided circuit data.

Question 3 asks to measure and record the electromotive force (Emf) of the battery with the circuit open. When the circuit is open, no current flows ($I = 0 \text{ A}$). In this condition, the voltmeter measures the Emf ($E$) of the battery, as the voltage drop due to internal resistance ($Ir$) is zero.
From the given data:

• When Current ($I$) $= 0.00 \text{ A}$, Potential Difference ($V$) $= 3.90 \text{ V}$. Therefore, the electromotive force ($E$) of the battery is:

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

Question 4 instructs to record ammeter and voltmeter readings for different external resistances. The provided data is:

Reading	Current $I$ (A)	Potential Difference $V$ (V)
1	0.00	3.90
2	0.19	3.69
3	0.23	3.65
4	0.27	3.61
5	0.33	3.55

The relationship between terminal potential difference ($V$), Emf ($E$), current ($I$), and internal resistance ($r$) is given by:

$$V = E - Ir$$

To find the internal resistance ($r$), we can rearrange the formula or use the gradient of a $V$ vs $I$ graph. Since we already found $E$, we can use any data point $(I, V)$ to calculate $r$. Let's use the last data point $(I=0.33 \text{ A}, V=3.55 \text{ V})$ and the calculated $E=3.90 \text{ V}$.

Substitute the values into the equation:

$$3.55 \text{ V} = 3.90 \text{ V} - (0.33 \text{ A}) \times r$$

Rearrange the equation to solve for $r$:

$$(0.33 \text{ A}) \times r = 3.90 \text{ V} - 3.55 \text{ V}$$ $$(0.33 \text{ A}) \times r = 0.35 \text{ V}$$ $$r = \frac{0.35 \text{ V}}{0.33 \text{ A}}$$ $$r \approx 1.0606 \text{ } \Omega$$

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

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