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Worksheet 8b - MATHS

Activity 2: Groupwork

2.1 Study the table below if the values for the Voltage (V) and the Current (I) for your circuit were as indicated below and answer the questions that follow:

\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}

a) Complete the table.

b) Show that the relationship between V and I is given by:

2.2 Use the grid provided to answer the questions that follow:

a) Plot the graph showing the relationship above where [tex]$V$[/tex] is the [tex]$x$[/tex]-axis and [tex]$I$[/tex] is the [tex]$y$[/tex]-axis.

b) Is the graph of the relationship between V and I an example of direct or indirect proportion? Justify your answer.

c) Use any two points on the graph to determine the gradient of the graph, hence the equation in the form of [tex]$y = mx + c$[/tex].



Answer :

GinnyAnswer
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]

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