Answer :
Final answer:
The ampacity for a single-phase resistive load with a nameplate rating of 38.4 kW and 240 V is 160 A. The correct answer is option a.
Explanation:
The ampacity of a single-phase resistive load can be calculated using the formula P = VI, where P is the power (in watts), V is the voltage (in volts), and I is the current (in amperes).
Given a nameplate rating of 38.4 kW and 240 V, we can rearrange the formula to solve for I (current): I = P / V.
Using the provided information, we calculate the current as follows:
- Power (P) = 38,400 W (since 38.4 kW is equal to 38,400 W)
- Voltage (V) = 240 V
- Current (I) = P / V = 38,400 W / 240 V = 160 A
Therefore, the ampacity for the given load is 160 A.
