Final answer:
The minimum conductor ampacity required for the four baseboard heaters with a total power of 4,000W at 208V is 19.23A, which aligns with option a. Option A is correct.
Step-by-step explanation:
To determine the minimum conductor ampacity for four 1,000W, 208V electric baseboard heaters, we first calculate the total power required for all the heaters combined, and then use the power formula to solve for the current.
The total power (P) needed for all heaters is:
P = number of heaters × power per heater = 4 × 1,000W = 4,000W
To calculate the current (I), we use the power formula:
I = P / V where 'V' is the voltage and 'P' is the total power.
So the current required is:
I = 4,000W / 208V = 19.23A
Therefore, the minimum conductor ampacity to handle this current is 19.23A, which corresponds to option a. 19.23A.
To calculate the minimum conductor ampacity for four 1,000W, 208V, 10 electric baseboard heaters, we can use the formula:
Ampacity = Total Power / Voltage
Given that the power of each heater is 1,000W and the voltage is 208V, the total power is 4,000W and the voltage is still 208V. Plugging in these values into the formula, we get:
Ampacity = 4,000W / 208V = 19.23A
Therefore, the minimum conductor ampacity is 19.23A, so the correct answer is option a. 19.23A.