Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
