What minimum number of 150 W lightbulbs must be connected in parallel to a single 180 V household circuit to trip a 49.0 A circuit breaker?

59

Explanation:

V = Voltage = 180 V

P = Power = 150 W

I = Current = 49 A

Power is given by

The total resistance

The circuit breaker will trip when total resistance drops below 3.67346938776 Ω

The number of lightbulbs is

The number of light bulbs is 59 to trip the circuit breaker.

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