For a 3600 watt, 240-volt load supplied from a 2-pole, 30-amp breaker, how many amps does each ungrounded conductor carry?

To determine how many amps each ungrounded conductor carries for a 3600-watt, 240-volt load, you can begin by using the power formula, which relates power (watts), voltage (volts), and current (amps):

[ P = V \times I ]

Where:

• ( P ) is the power in watts (3600 W)

• ( V ) is the voltage in volts (240 V)

• ( I ) is the current in amps

We can rearrange the formula to find the current:

[ I = \frac{P}{V} ]

Now, substituting the known values:

[ I = \frac{3600 \text{ W}}{240 \text{ V}} = 15 \text{ A} ]

This calculation shows that the total current drawn by the load is 15 amps. Since the load is supplied by a 240-volt, 2-pole circuit breaker, this means that the total current is split equally between the two ungrounded conductors. In this case, each conductor is carrying the full 15 amps of current to the load.

Therefore, 15 amps represents the current carried by each ungrounded conductor