To find the voltage (V) generated by the fuel cell, you can use the formula for power (P), which is given by:
\[ P = V \times I \]
where:
- \( P \) is the power in watts,
- \( V \) is the voltage in volts,
- \( I \) is the current in amperes.
In this case, the fuel cell generates 10.0 kilowatts of power, which is equal to 10,000 watts (since 1 kW = 1,000 W), and the current is 15.8 amps.
Rearranging the formula to solve for voltage gives:
\[ V = \frac{P}{I} \]
Now, substituting in the values:
\[ V = \frac{10,000 , \text{W}}{15.8 , \text{A}} \]
Calculating this:
\[ V = 632.91 , \text{V} \]
Rounding to three significant figures, the voltage is approximately:
\[ V \approx 633 , \text{V} \]
Therefore, the correct answer is:
C. 633 volts.