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Sagot :
Sure, let's go through the solution step by step.
### Given Data:
The table provides the following values for Voltage (V) and Current (I):
| Voltage (V) | 4 | 8 | 12 | 16 | 20 |
|-------------|----|----|----|----|----|
| Current (I) | 0.4| 0.8| 1.2| | |
You are required to determine the missing values in the table.
### Step-by-Step Solution:
1. Calculate the Resistance Using Ohm's Law:
Ohm's Law states that [tex]\( V = IR \)[/tex] or rearranged [tex]\( R = \frac{V}{I} \)[/tex].
We can use the given pairs of voltage and current to calculate the resistance at each point where both values are available.
#### For [tex]\( V = 4V \)[/tex] and [tex]\( I = 0.4A \)[/tex]:
[tex]\[ R_1 = \frac{V}{I} = \frac{4V}{0.4A} = 10 \, \Omega \][/tex]
#### For [tex]\( V = 8V \)[/tex] and [tex]\( I = 0.8A \)[/tex]:
[tex]\[ R_2 = \frac{V}{I} = \frac{8V}{0.8A} = 10 \, \Omega \][/tex]
#### For [tex]\( V = 12V \)[/tex] and [tex]\( I = 1.2A \)[/tex]:
[tex]\[ R_3 = \frac{V}{I} = \frac{12V}{1.2A} = 10 \, \Omega \][/tex]
2. Calculate the Average Resistance:
Since resistance in a circuit with the same components should be constant, we can take an average of the calculated resistances for a more accurate prediction, though in our case, it's already consistent:
[tex]\[ R_{\text{average}} = \frac{R_1 + R_2 + R_3}{3} = \frac{10 + 10 + 10}{3} = 10 \, \Omega \][/tex]
3. Predict the Missing Currents:
Using the average resistance ([tex]\( R_{\text{average}} \)[/tex]), we can predict the missing current values for [tex]\( V = 16V \)[/tex] and [tex]\( V = 20V \)[/tex].
#### For [tex]\( V = 16V \)[/tex]:
[tex]\[ I = \frac{V}{R} = \frac{16V}{10 \, \Omega} = 1.6A \][/tex]
#### For [tex]\( V = 20V \)[/tex]:
[tex]\[ I = \frac{V}{R} = \frac{20V}{10 \, \Omega} = 2.0A \][/tex]
### Final Values:
Thus, the completed table with all the predicted currents is:
| Voltage (V) | 4 | 8 | 12 | 16 | 20 |
|-------------|----|----|----|----|----|
| Current (I) | 0.4| 0.8| 1.2| 1.6| 2.0|
In conclusion, the resistances at known points calculate to 10 Ω each, the average resistance is 10 Ω, and using this, the missing current values for 16V and 20V are predicted as 1.6A and 2.0A respectively.
### Given Data:
The table provides the following values for Voltage (V) and Current (I):
| Voltage (V) | 4 | 8 | 12 | 16 | 20 |
|-------------|----|----|----|----|----|
| Current (I) | 0.4| 0.8| 1.2| | |
You are required to determine the missing values in the table.
### Step-by-Step Solution:
1. Calculate the Resistance Using Ohm's Law:
Ohm's Law states that [tex]\( V = IR \)[/tex] or rearranged [tex]\( R = \frac{V}{I} \)[/tex].
We can use the given pairs of voltage and current to calculate the resistance at each point where both values are available.
#### For [tex]\( V = 4V \)[/tex] and [tex]\( I = 0.4A \)[/tex]:
[tex]\[ R_1 = \frac{V}{I} = \frac{4V}{0.4A} = 10 \, \Omega \][/tex]
#### For [tex]\( V = 8V \)[/tex] and [tex]\( I = 0.8A \)[/tex]:
[tex]\[ R_2 = \frac{V}{I} = \frac{8V}{0.8A} = 10 \, \Omega \][/tex]
#### For [tex]\( V = 12V \)[/tex] and [tex]\( I = 1.2A \)[/tex]:
[tex]\[ R_3 = \frac{V}{I} = \frac{12V}{1.2A} = 10 \, \Omega \][/tex]
2. Calculate the Average Resistance:
Since resistance in a circuit with the same components should be constant, we can take an average of the calculated resistances for a more accurate prediction, though in our case, it's already consistent:
[tex]\[ R_{\text{average}} = \frac{R_1 + R_2 + R_3}{3} = \frac{10 + 10 + 10}{3} = 10 \, \Omega \][/tex]
3. Predict the Missing Currents:
Using the average resistance ([tex]\( R_{\text{average}} \)[/tex]), we can predict the missing current values for [tex]\( V = 16V \)[/tex] and [tex]\( V = 20V \)[/tex].
#### For [tex]\( V = 16V \)[/tex]:
[tex]\[ I = \frac{V}{R} = \frac{16V}{10 \, \Omega} = 1.6A \][/tex]
#### For [tex]\( V = 20V \)[/tex]:
[tex]\[ I = \frac{V}{R} = \frac{20V}{10 \, \Omega} = 2.0A \][/tex]
### Final Values:
Thus, the completed table with all the predicted currents is:
| Voltage (V) | 4 | 8 | 12 | 16 | 20 |
|-------------|----|----|----|----|----|
| Current (I) | 0.4| 0.8| 1.2| 1.6| 2.0|
In conclusion, the resistances at known points calculate to 10 Ω each, the average resistance is 10 Ω, and using this, the missing current values for 16V and 20V are predicted as 1.6A and 2.0A respectively.
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