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Sagot :
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]
## 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]
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