Let's address each part of the question systematically.
2.1.a) Complete the table:
Given the table:
\begin{tabular}{|l|l|l|l|l|l|}
\hline
Voltage (V) & 4 & 8 & 12 & 16 & 20 \\
\hline
Current (I) & 0.4 & 0.8 & 1.2 & [tex]$\ldots \ldots \ldots \ldots$[/tex] & [tex]$\ldots \ldots \ldots \ldots$[/tex] \\
\hline
\end{tabular}
We know that [tex]\( \frac{I}{V} \)[/tex] should be constant if the relationship is linear. To find the missing currents:
1. Determine the proportional constant (slope, [tex]\( m \)[/tex]) using the given points:
[tex]\[ m = \frac{I}{V} = \frac{0.4}{4} = \frac{0.8}{8} = \frac{1.2}{12} = 0.1 \][/tex]
2. Use this constant to find the missing values for the Current:
[tex]\[ I = m \times V = 0.1 \times V \][/tex]
For [tex]\( V = 16 \)[/tex]:
[tex]\[ I = 0.1 \times 16 = 1.6 \][/tex]
For [tex]\( V = 20 \)[/tex]:
[tex]\[ I = 0.1 \times 20 = 2.0 \][/tex]
The completed table now is:
\begin{tabular}{|l|l|l|l|l|l|}
\hline
Voltage (V) & 4 & 8 & 12 & 16 & 20 \\
\hline
Current (I) & 0.4 & 0.8 & 1.2 & 1.6 & 2.0 \\
\hline
\end{tabular}
2.1.b) Show that the relationship between [tex]\( V \)[/tex] and [tex]\( I \)[/tex] is linear:
By plotting [tex]\( I \)[/tex] against [tex]\( V \)[/tex], it will show a straight-line relationship if linear.
The points we have are:
- [tex]\((4, 0.4)\)[/tex]
- [tex]\((8, 0.8)\)[/tex]
- [tex]\((12, 1.2)\)[/tex]
- [tex]\((16, 1.6)\)[/tex]
- [tex]\((20, 2.0)\)[/tex]
This shows that:
[tex]\[ I = 0.1V \][/tex]
This is a linear equation of the form [tex]\( I = mV \)[/tex] where [tex]\( m = 0.1 \)[/tex].
2.2.a) Plot the graph:
On the grid, you plot the points given above and connect them. The [tex]\( x \)[/tex]-axis represents Voltage [tex]\( V \)[/tex] and the [tex]\( y \)[/tex]-axis represents Current [tex]\( I \)[/tex].
2.2.b) Is the graph of the relationship between [tex]\( V \)[/tex] and [tex]\( I \)[/tex] an example of direct or indirect proportion? Justify your answer:
The graph shows a straight line through the origin (0,0). This is an example of direct proportion because as Voltage [tex]\( V \)[/tex] increases, Current [tex]\( I \)[/tex] increases proportionally. The relationship can be expressed as [tex]\( I \propto V \)[/tex] (I is directly proportional to V).
2.2.c) Use any two points on the graph to determine the gradient of the graph, hence the equation in the form [tex]\( y = mx + c \)[/tex]:
To determine the gradient [tex]\( m \)[/tex], let's use two points from our graph. We can use the points [tex]\((4, 0.4)\)[/tex] and [tex]\((20, 2.0)\)[/tex]:
[tex]\[ m = \frac{\Delta I}{\Delta V} = \frac{2.0 - 0.4}{20 - 4} = \frac{1.6}{16} = 0.1 \][/tex]
The y-intercept [tex]\( c \)[/tex] in this case is 0 because the line passes through the origin.
So the equation of the straight line is:
[tex]\[ I = 0.1V + 0 \][/tex]
or simply:
[tex]\[ I = 0.1V \][/tex]
