A generator at a utility company produces 100 A of current at 4000 V. The voltage is stepped up to 240000 V by a transformer before it is sent on a high voltage transmission line. The current in transmission line is

Options

A. 3.67 A
B. 2.67 A
C. 1.67 A
D. 2.40 A

Explanation

To solve the problem, we will use the principle of an ideal transformer, which states that the power input to the primary coil is equal to the power output from the secondary coil. This can be expressed mathematically as:

\[ V_p \cdot I_p = V_s \cdot I_s \]

Where:
- \( V_p \) = primary voltage
- \( I_p \) = primary current
- \( V_s \) = secondary voltage
- \( I_s \) = secondary current

### Step-by-Step Solution:

1. **Identify the given values:**
- Primary voltage (\( V_p \)) = 4000 V
- Primary current (\( I_p \)) = 100 A
- Secondary voltage (\( V_s \)) = 240000 V
- Secondary current (\( I_s \)) = ? (This is what we need to find)

2. **Apply the transformer equation:**
Using the formula \( V_p \cdot I_p = V_s \cdot I_s \), we can rearrange it to find \( I_s \):
\[ I_s = \frac{V_p \cdot I_p}{V_s} \]

3. **Substitute the known values into the equation:**
\[ I_s = \frac{4000 \, \text{V} \cdot 100 \, \text{A}}{240000 \, \text{V}} \]

4. **Calculate the numerator:**
\[ 4000 \, \text{V} \cdot 100 \, \text{A} = 400000 \, \text{VA} \]

5. **Divide by the secondary voltage:**
\[ I_s = \frac{400000 \, \text{VA}}{240000 \, \text{V}} = \frac{400000}{240000} = \frac{4000}{2400} = \frac{10}{6} \approx 1.67 \, \text{A} \]

6. **Conclusion:**
The current in the transmission line is approximately 1.67 A.

### Final Answer:
The current in the transmission line is **1.67 A**.