To solve the problem, we will follow these steps:
### Step 1: Calculate the Resistance of the Copper Wire
The resistance \( R \) of the wire can be calculated using the formula:
\[
R = \frac{\rho L}{A}
\]
where:
- \( \rho \) is the resistivity of copper (\( 1.7 \times 10^{-8} \, \Omega \cdot m \)),
- \( L \) is the length of the wire (20,000 m for a round trip of 10 km),
- \( A \) is the cross-sectional area of the wire.
The area \( A \) of the wire can be calculated as:
\[
A = \pi r^2
\]
where \( r \) is the radius of the wire. Given that the radius is 0.5 cm, we convert it to meters:
\[
r = 0.5 \, \text{cm} = 0.005 \, \text{m}
\]
Now, substituting the values:
\[
A = \pi (0.005)^2 = \pi (2.5 \times 10^{-5}) \approx 7.85 \times 10^{-5} \, m^2
\]
Now substituting \( \rho \), \( L \), and \( A \) into the resistance formula:
\[
R = \frac{1.7 \times 10^{-8} \times 20000}{7.85 \times 10^{-5}} \approx 4.32 \, \Omega
\]
### Step 2: Calculate the Current at 220 V
Using the power formula:
\[
P = V \cdot I
\]
we can rearrange it to find the current \( I \):
\[
I = \frac{P}{V}
\]
Given \( P = 1 \, \text{MW} = 10^6 \, \text{W} \) and \( V = 220 \, \text{V} \):
\[
I = \frac{10^6}{220} \approx 4545.45 \, \text{A}
\]
### Step 3: Calculate the Ohmic Losses
The power loss due to resistance (ohmic losses) can be calculated using:
\[
P_{\text{loss}} = I^2 R
\]
Substituting the values of \( I \) and \( R \):
\[
P_{\text{loss}} = (4545.45)^2 \times 4.32 \approx 8.86 \times 10^6 \, \text{W}
\]
### Step 4: Calculate the Fraction of Ohmic Losses to Power Transmitted
The fraction of power loss to the total power transmitted can be calculated as:
\[
\text{Fraction} = \frac{P_{\text{loss}}}{P} = \frac{8.86 \times 10^6}{10^6} = 8.86
\]
This indicates that a significant amount of power is lost as heat in the wires.
### Step 5: Calculate the New Current at 11000 V
Now, if we boost the voltage to 11000 V, we can calculate the new current:
\[
I' = \frac{10^6}{11000} \approx 90.91 \, \text{A}
\]
### Step 6: Calculate the New Ohmic Losses
Now, we can calculate the new power loss with the new current:
\[
P'_{\text{loss}} = (I')^2 R = (90.91)^2 \times 4.32 \approx 3560.2 \, \text{W}
\]
### Step 7: Calculate the New Fraction of Ohmic Losses
Finally, we can calculate the new fraction of power loss:
\[
\text{New Fraction} = \frac{P'_{\text{loss}}}{P} = \frac{3560.2}{10^6} \approx 0.00356 \text{ or } 0.356\%
\]
### Summary of Results
1. Resistance of the wire: \( R \approx 4.32 \, \Omega \)
2. Current at 220 V: \( I \approx 4545.45 \, \text{A} \)
3. Ohmic losses at 220 V: \( P_{\text{loss}} \approx 8.86 \times 10^6 \, \text{W} \)
4. Fraction of ohmic losses at 220 V: \( 8.86 \)
5. Current at 11000 V: \( I' \approx 90.91 \, \text{A} \)
6. New ohmic losses at 11000 V: \( P'_{\text{loss}} \approx 3560.2 \, \text{W} \)
7. New fraction of ohmic losses: \( 0.356\% \)
