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Sagot :
Certainly! Let's go through the steps to prove that the unit of potential difference (voltage, [tex]\(V\)[/tex]) is [tex]\( \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1} \)[/tex].
1. Mass:
- The SI unit for mass is the kilogram ([tex]\(\text{kg}\)[/tex]).
2. Length:
- The SI unit for length is the meter ([tex]\(\text{m}\)[/tex]).
3. Time:
- The SI unit for time is the second ([tex]\(\text{s}\)[/tex]).
4. Electric Current:
- The SI unit for electric current is the ampere ([tex]\(\text{A}\)[/tex]).
Now, let's consider the definition of voltage (potential difference):
Voltage ([tex]\(V\)[/tex]) is defined as the energy per unit charge.
Mathematically:
[tex]\[ V = \frac{\text{Energy}}{\text{Charge}} \][/tex]
First, let's break down the units for energy and charge:
### Energy:
- Energy is measured in joules (J).
- One joule (J) is defined in terms of basic SI units as:
[tex]\[ 1 \, \text{J} = 1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} \][/tex]
### Electric Charge:
- Electric charge is measured in coulombs (C).
- One coulomb (C) can be expressed in terms of basic SI units, specifically:
[tex]\[ 1 \, \text{C} = 1 \, \text{A} \cdot \text{s} \][/tex]
### Combining Units to Find Voltage:
We know:
[tex]\[ V = \frac{\text{Energy}}{\text{Charge}} \][/tex]
Substitute the units of energy and charge into the equation:
[tex]\[ V = \frac{1 \, \text{J}}{1 \, \text{C}} \][/tex]
Now, replace joules and coulombs with their respective SI units:
[tex]\[ V = \frac{1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}}{1 \, \text{A} \cdot \text{s}} \][/tex]
Simplify the equation:
[tex]\[ V = 1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} \cdot \left( \text{A} \cdot \text{s} \right)^{-1} \][/tex]
[tex]\[ V = 1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} \cdot \text{A}^{-1} \cdot \text{s}^{-1} \][/tex]
Combine the exponents of seconds ([tex]\(\text{s}\)[/tex]):
[tex]\[ V = 1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1} \][/tex]
### Final Result:
Thus, the unit of potential difference (voltage) is:
[tex]\[ V = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1} \][/tex]
This completes the proof that the unit of potential difference (voltage, [tex]\(V\)[/tex]) is [tex]\( \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1} \)[/tex].
1. Mass:
- The SI unit for mass is the kilogram ([tex]\(\text{kg}\)[/tex]).
2. Length:
- The SI unit for length is the meter ([tex]\(\text{m}\)[/tex]).
3. Time:
- The SI unit for time is the second ([tex]\(\text{s}\)[/tex]).
4. Electric Current:
- The SI unit for electric current is the ampere ([tex]\(\text{A}\)[/tex]).
Now, let's consider the definition of voltage (potential difference):
Voltage ([tex]\(V\)[/tex]) is defined as the energy per unit charge.
Mathematically:
[tex]\[ V = \frac{\text{Energy}}{\text{Charge}} \][/tex]
First, let's break down the units for energy and charge:
### Energy:
- Energy is measured in joules (J).
- One joule (J) is defined in terms of basic SI units as:
[tex]\[ 1 \, \text{J} = 1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} \][/tex]
### Electric Charge:
- Electric charge is measured in coulombs (C).
- One coulomb (C) can be expressed in terms of basic SI units, specifically:
[tex]\[ 1 \, \text{C} = 1 \, \text{A} \cdot \text{s} \][/tex]
### Combining Units to Find Voltage:
We know:
[tex]\[ V = \frac{\text{Energy}}{\text{Charge}} \][/tex]
Substitute the units of energy and charge into the equation:
[tex]\[ V = \frac{1 \, \text{J}}{1 \, \text{C}} \][/tex]
Now, replace joules and coulombs with their respective SI units:
[tex]\[ V = \frac{1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}}{1 \, \text{A} \cdot \text{s}} \][/tex]
Simplify the equation:
[tex]\[ V = 1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} \cdot \left( \text{A} \cdot \text{s} \right)^{-1} \][/tex]
[tex]\[ V = 1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} \cdot \text{A}^{-1} \cdot \text{s}^{-1} \][/tex]
Combine the exponents of seconds ([tex]\(\text{s}\)[/tex]):
[tex]\[ V = 1 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1} \][/tex]
### Final Result:
Thus, the unit of potential difference (voltage) is:
[tex]\[ V = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1} \][/tex]
This completes the proof that the unit of potential difference (voltage, [tex]\(V\)[/tex]) is [tex]\( \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1} \)[/tex].
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