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Is it possible to operate two 1300 W hairdryers on a 120 V circuit without blowing fuse?

To operate two 1300 W hairdryers on a 120 V circuit without blowing the fuse, you need to calculate the total power consumption and compare it with the circuit's power handling capacity.

Formula for calculating power: Power (in watts) = Voltage (in volts) × Current (in amps)

Given information:

Voltage (V) = 120 V

Power consumption of each hairdryer (P1 and P2) = 1300 W

If we assume the current required for each hairdryer (I1 and I2) is the same, we can calculate the total current (I) drawn by both hairdryers:

I1 = P1 / V = 1300 W / 120 V ≈ 10.83 A

I2 = P2 / V = 1300 W / 120 V ≈ 10.83 A

I = I1 + I2 = 10.83 A + 10.83 A ≈ 21.66 A

Now, let's calculate the total power consumption of both hairdryers (P):

P = (P1 + P2) = 1300 W + 1300 W = 2600 W

The standard North American 120 V circuit is typically protected by a 15-amp fuse. To avoid blowing the fuse, the total current drawn by the devices on the circuit should not exceed the fuse's rating. Since the total current calculated earlier (21.66 A) exceeds the fuse rating, it is not safe to operate both hairdryers simultaneously on a 120 V circuit without risking a blown fuse.

Therefore, it is not possible to operate two 1300 W hairdryers simultaneously on a 120 V circuit without blowing the fuse.