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To determine the minimum cross-sectional area of the wire needed to prevent the voltage at the motor from dropping below 230 V, we can approach the problem as follows:
1. Identify the given values:
- Current ([tex]\( I \)[/tex]) = 20 A
- Voltage of the power source ([tex]\( V_{\text{source}} \)[/tex]) = 240 V
- Voltage at the motor ([tex]\( V_{\text{motor}} \)[/tex]) = 230 V
- Distance from the power source to the motor ([tex]\( d \)[/tex]) = 15 m
- Resistivity of copper ([tex]\( \rho \)[/tex]) = [tex]\( 1.68 \times 10^{-8} \)[/tex] ohm-meter
2. Calculate the allowable voltage drop:
The voltage drop ([tex]\( V_{\text{drop}} \)[/tex]) can be calculated using the difference between the voltage of the power source and the voltage at the motor.
[tex]\[ V_{\text{drop}} = V_{\text{source}} - V_{\text{motor}} \][/tex]
Substituting the given values:
[tex]\[ V_{\text{drop}} = 240 \, \text{V} - 230 \, \text{V} = 10 \, \text{V} \][/tex]
3. Determine the resistance of the wire:
The voltage drop ([tex]\( V_{\text{drop}} \)[/tex]) across a wire can be expressed using Ohm's Law:
[tex]\[ V_{\text{drop}} = I \times R \][/tex]
Here, [tex]\( R \)[/tex] is the resistance of the wire. Rearranging the formula to solve for [tex]\( R \)[/tex]:
[tex]\[ R = \frac{V_{\text{drop}}}{I} \][/tex]
Substituting the values:
[tex]\[ R = \frac{10 \, \text{V}}{20 \, \text{A}} = 0.5 \, \Omega \][/tex]
4. Relate the resistance to the physical properties of the wire:
The resistance [tex]\( R \)[/tex] of a wire is also related to the resistivity ([tex]\( \rho \)[/tex]), length ([tex]\( L \)[/tex]), and cross-sectional area ([tex]\( A \)[/tex]) by the formula:
[tex]\[ R = \rho \frac{L}{A} \][/tex]
Since the current flows to the motor and back, the total length of the wire is [tex]\( 2d \)[/tex]. Therefore, substituting [tex]\( L = 2d \)[/tex]:
[tex]\[ R = \rho \frac{2d}{A} \][/tex]
Solving for [tex]\( A \)[/tex]:
[tex]\[ A = \rho \frac{2d}{R} \][/tex]
5. Calculate the minimum cross-sectional area:
Now substitute the known values into the formula:
[tex]\[ A = \left( 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \right) \frac{2 \times 15 \, \text{m}}{0.5 \, \Omega} \][/tex]
Simplifying this:
[tex]\[ A = \frac{1.68 \times 10^{-8} \, \Omega \cdot \text{m} \times 30 \, \text{m}}{0.5 \, \Omega} \][/tex]
[tex]\[ A = \frac{50.4 \times 10^{-8} \, \Omega \cdot \text{m}^2}{0.5 \, \Omega} \][/tex]
[tex]\[ A = 1.008 \times 10^{-6} \, \text{m}^2 \][/tex]
Thus, the minimum cross-sectional area of the wire that can be used, ensuring the voltage at the motor is not lower than 230 V, is [tex]\(1.008 \times 10^{-6}\)[/tex] square meters.
1. Identify the given values:
- Current ([tex]\( I \)[/tex]) = 20 A
- Voltage of the power source ([tex]\( V_{\text{source}} \)[/tex]) = 240 V
- Voltage at the motor ([tex]\( V_{\text{motor}} \)[/tex]) = 230 V
- Distance from the power source to the motor ([tex]\( d \)[/tex]) = 15 m
- Resistivity of copper ([tex]\( \rho \)[/tex]) = [tex]\( 1.68 \times 10^{-8} \)[/tex] ohm-meter
2. Calculate the allowable voltage drop:
The voltage drop ([tex]\( V_{\text{drop}} \)[/tex]) can be calculated using the difference between the voltage of the power source and the voltage at the motor.
[tex]\[ V_{\text{drop}} = V_{\text{source}} - V_{\text{motor}} \][/tex]
Substituting the given values:
[tex]\[ V_{\text{drop}} = 240 \, \text{V} - 230 \, \text{V} = 10 \, \text{V} \][/tex]
3. Determine the resistance of the wire:
The voltage drop ([tex]\( V_{\text{drop}} \)[/tex]) across a wire can be expressed using Ohm's Law:
[tex]\[ V_{\text{drop}} = I \times R \][/tex]
Here, [tex]\( R \)[/tex] is the resistance of the wire. Rearranging the formula to solve for [tex]\( R \)[/tex]:
[tex]\[ R = \frac{V_{\text{drop}}}{I} \][/tex]
Substituting the values:
[tex]\[ R = \frac{10 \, \text{V}}{20 \, \text{A}} = 0.5 \, \Omega \][/tex]
4. Relate the resistance to the physical properties of the wire:
The resistance [tex]\( R \)[/tex] of a wire is also related to the resistivity ([tex]\( \rho \)[/tex]), length ([tex]\( L \)[/tex]), and cross-sectional area ([tex]\( A \)[/tex]) by the formula:
[tex]\[ R = \rho \frac{L}{A} \][/tex]
Since the current flows to the motor and back, the total length of the wire is [tex]\( 2d \)[/tex]. Therefore, substituting [tex]\( L = 2d \)[/tex]:
[tex]\[ R = \rho \frac{2d}{A} \][/tex]
Solving for [tex]\( A \)[/tex]:
[tex]\[ A = \rho \frac{2d}{R} \][/tex]
5. Calculate the minimum cross-sectional area:
Now substitute the known values into the formula:
[tex]\[ A = \left( 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \right) \frac{2 \times 15 \, \text{m}}{0.5 \, \Omega} \][/tex]
Simplifying this:
[tex]\[ A = \frac{1.68 \times 10^{-8} \, \Omega \cdot \text{m} \times 30 \, \text{m}}{0.5 \, \Omega} \][/tex]
[tex]\[ A = \frac{50.4 \times 10^{-8} \, \Omega \cdot \text{m}^2}{0.5 \, \Omega} \][/tex]
[tex]\[ A = 1.008 \times 10^{-6} \, \text{m}^2 \][/tex]
Thus, the minimum cross-sectional area of the wire that can be used, ensuring the voltage at the motor is not lower than 230 V, is [tex]\(1.008 \times 10^{-6}\)[/tex] square meters.
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