Activity 2: Group Work

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}{|c|c|c|c|c|c|}
\hline Voltage (V) & 4 & 8 & 12 & 16 & 20 \\
\hline Current (I) & 0.4 & 0.8 & 1.2 & 1.6 & 2.0 \\
\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:



Answer :

Certainly! To answer the questions given the provided data, let's go through a detailed, step-by-step solution without referring to any external code. Let's focus on finding the missing current, verifying the relationship between Voltage (V) and Current (I), and completing the table.

## Given Data
Voltage (V): 4, 8, 12, 16, 20
Current (I): 0.4, 0.8, 1.2, (missing), (missing)

### a) Complete the table

#### Step 1: Define the known linear relationship
In circuit analysis, Ohm's Law states that Voltage (V) is the product of Current (I) and Resistance (R):

[tex]\[ V = I \times R \][/tex]

We can rearrange this to solve for Resistance (R):

[tex]\[ R = \frac{V}{I} \][/tex]

#### Step 2: Calculate the Resistance for each known data point
Using the known values of Voltage (V) and Current (I):

1. For V = 4 and I = 0.4:
[tex]\[ R_1 = \frac{4}{0.4} = 10 \, \Omega \][/tex]

2. For V = 8 and I = 0.8:
[tex]\[ R_2 = \frac{8}{0.8} = 10 \, \Omega \][/tex]

3. For V = 12 and I = 1.2:
[tex]\[ R_3 = \frac{12}{1.2} = 10 \, \Omega \][/tex]

4. For V = 16 and I = (missing), we assume Resistance is constant at 10 Ω.

#### Step 3: Calculate the missing Current for V = 16
Given [tex]\( R = 10 \, \Omega \)[/tex]:

[tex]\[ I = \frac{V}{R} = \frac{16}{10} = 1.6 \, A \][/tex]

So, the missing Current for V = 16 is [tex]\( I = 1.6 \, A \)[/tex].

#### Step 4: Calculate the missing Current for V = 20
Using the same Resistance [tex]\( R = 10 \, \Omega \)[/tex]:

[tex]\[ I = \frac{V}{R} = \frac{20}{10} = 2.0 \, A \][/tex]

So, the missing Current for V = 20 is [tex]\( I = 2.0 \, A \)[/tex].

#### Completed table:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline Voltage (V) & 4 & 8 & 12 & 16 & 20 \\ \hline Current (I) & 0.4 & 0.8 & 1.2 & 1.6 & 2.0 \\ \hline \end{tabular} \][/tex]

### b) Show the relationship between V and I

Observing the completed table, we can calculate Resistance (R) for all data points:

1. For V = 4 and I = 0.4:
[tex]\[ R = \frac{4}{0.4} = 10 \, \Omega \][/tex]

2. For V = 8 and I = 0.8:
[tex]\[ R = \frac{8}{0.8} = 10 \, \Omega \][/tex]

3. For V = 12 and I = 1.2:
[tex]\[ R = \frac{12}{1.2} = 10 \, \Omega \][/tex]

4. For V = 16 and I = 1.6:
[tex]\[ R = \frac{16}{1.6} = 10 \, \Omega \][/tex]

5. For V = 20 and I = 2.0:
[tex]\[ R = \frac{20}{2.0} = 10 \, \Omega \][/tex]

The relationship between Voltage (V) and Current (I) is linear, and the Resistance (R) is constant for all data points at [tex]\( 10 \, \Omega \)[/tex]. This confirms that:

[tex]\[ V = I \times R \][/tex]

With [tex]\( R \)[/tex] consistently at [tex]\( 10 \, \Omega \)[/tex], Voltage (V) is directly proportional to Current (I).

### Conclusion
The relationship between Voltage (V) and Current (I) in this circuit is given by Ohm's Law, where [tex]\( V = I \times 10 \, \Omega \)[/tex]. The completed table of values is as follows:

[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline Voltage (V) & 4 & 8 & 12 & 16 & 20 \\ \hline Current (I) & 0.4 & 0.8 & 1.2 & 1.6 & 2.0 \\ \hline \end{tabular} \][/tex]