Respuesta :
Explanation:
To calculate the greatest length of cable that can be used without exceeding a voltage drop of 11.5 volts, we can use the formula for voltage drop:
\[ V = I \times R \times L \]
Where:
- \( V \) is the voltage drop (11.5 volts in this case)
- \( I \) is the current (13A)
- \( R \) is the resistance per unit length of the cable
- \( L \) is the length of the cable
First, let's calculate the resistance per unit length (\( R \)) of the cable using the formula:
\[ R = \frac{{\rho \times L}}{{A}} \]
Where:
- \( \rho \) is the resistivity of copper (17.2 × 10^-9 Ω/m)
- \( L \) is the length of the cable (to be determined)
- \( A \) is the cross-sectional area of the cable (2.5mm² = \( 2.5 \times 10^{-6} \) m²)
\[ R = \frac{{17.2 \times 10^{-9} \, \Omega/m \times L}}{{2.5 \times 10^{-6} \, m²}} \]
\[ R = \frac{{17.2 \times 10^{-9} \, \Omega/m \times L}}{{2.5 \times 10^{-6} \, m²}} \]
\[ R = \frac{{6.88 \times L}}{{10^{-6}}} \]
\[ R = 6.88 \times 10^3 \times L \]
Now, plug this expression for \( R \) into the voltage drop formula:
\[ 11.5 = 13 \times (6.88 \times 10^3 \times L) \]
Now, solve for \( L \):
\[ L = \frac{{11.5}}{{13 \times 6.88 \times 10^3}} \]
\[ L ≈ \frac{{11.5}}{{89.44 \times 10^3}} \]
\[ L ≈ \frac{{11.5}}{{8.944 \times 10^4}} \]
\[ L ≈ 0.1286 \]
So, the greatest length of cable that may be used without exceeding a voltage drop of 11.5 volts is approximately 0.1286 meters.
