Ohm's law states that the current flowing through a conductor between two points is directly proportional to the potential difference across the two points, and inversely proportional to the resistanc

Here are the solutions to the questions:

8.1.1

Ohm's law states that the current flowing through a conductor between two points is directly proportional to the potential difference across the two points, and inversely proportional to the resistance between them, provided the temperature and other physical conditions remain constant.

8.1.2

You need to draw a straight line of best fit through the plotted points on the graph. Ensure this line extends to cut both the potential difference (y-axis) and current (x-axis).

8.1.3

The terminal potential difference ($V$) is related to the EMF ($\mathcal{E}$), current ($I$), and internal resistance ($r$) by the equation $V = \mathcal{E} - Ir$. When the current ($I$) is zero, the terminal potential difference ($V$) is equal to the EMF ($\mathcal{E}$). From the graph, the y-intercept (where $I = 0 \, \text{A}$) is approximately $5.5 \, \text{V}$. Therefore, the EMF of the battery is: $\mathcal{E} = \text{5.5 V}$

8.1.4

The internal resistance ($r$) is the negative of the gradient of the potential difference versus current graph. We will use two points from the line of best fit: Point 1 (y-intercept): $(I_1, V_1) = (0 \, A, 5.5 \, V)$ Point 2 (x-intercept, estimated): $(I_2, V_2) = (4.5 \, A, 0 \, V)$ Step 1: Calculate the gradient ($m$) of the line. $m = \frac{\Delta V}{\Delta I} = \frac{V_2 - V_1}{I_2 - I_1}$ $m = \frac{0 \, V - 5.5 \, V}{4.5 \, A - 0 \, A}$ $m = \frac{-5.5 \, V}{4.5 \, A} \approx -1.222 \, \Omega$ Step 2: Determine the internal resistance ($r$). $r = -m$ $r = -(-1.222 \, \Omega) = 1.222 \, \Omega$ The internal resistance of the battery is $\boxed{1.22 \Omega}$.

8.2.1

To calculate the current in the $8 \, \Omega$ resistor, we use Ohm's law. Step 1: Apply Ohm's law ($I = \frac{V}{R}$) to the $8 \, \Omega$ resistor. $I_{8\Omega} = \frac{V_{8\Omega}}{R_{8\Omega}}$ $I_{8\Omega} = \frac{21.84 \, V}{8 \, \Omega}$ $I_{8\Omega} = 2.73 \, \text{A}$ The current in the $8 \, \Omega$ resistor is $\boxed{2.73 A}$.

8.2.2

To calculate the equivalent resistance of the resistors in parallel, we use the formula for two parallel resistors. Step 1: Calculate the equivalent resistance for the $30 \, \Omega$ and $20 \, \Omega$ resistors in parallel. $R_{parallel} = \frac{R_1 \times R_2}{R_1 + R_2}$ $R_{parallel} = \frac{30 \, \Omega \times 20 \, \Omega}{30 \, \Omega + 20 \, \Omega}$ $R_{parallel} = \frac{600 \, \Omega^2}{50 \, \Omega} = 12 \, \Omega$ The equivalent resistance of the resistors in parallel is $\boxed{12 \Omega}$.

8.2.3

To determine the internal resistance $r$ of the battery, we use the total current and the terminal potential difference formula. Step 1: Determine the total current ($I_{total}$) in the circuit. Since the $8 \, \Omega$ resistor is in series with the parallel combination, the current through it is the total current. $I_{total} = I_{8\Omega} = 2.73 \, \text{A}$ Step 2: Calculate the total external resistance ($R_{ext}$). This is the sum of the $8 \, \Omega$ resistor and