The wires in a house running from the circuit breaker box (often in the basement

Question

The wires in a house running from the circuit breaker box (often in the basement) to the outlets have a total resistance on the order of 0.2 ohm. A device with a resistance of 25 ohm is connected to the outlet. Estimate the power dissipated in both the device and in the wiring in the walls. A fault develops within the device such that it acts like it has an additional resistance of 50 ohm between the hot and ground wires. If the device is connected to a circuit protected by both a 10 A circuit breaker and a GFCI outlet, will it trip either of these safety devices

Explanation / Answer

Let the House has the supply voltage of V =120 Volts.

a) Then, Power dissipated in device, Pdevice = V2 / R = 120*120/ 25 = 576 W

Similarly,Power dissipated in wirings, Pwirings = V2 / R = 120*120/ 0.2 = 72000 W

Let us calculate the current in each:

Current in device , Idevice = V/R = 120/25 = 4.8 A

Current in wirings , Iwirings = V/R = 120/0.2 = 600 A

Current in additional 50 resistance , I50 = V/R = 120/50 = 2.4 A

And power,P50 = 120*120/ 50 = 288 W

And ,Power due to 10A ,P10 = 120*10 = 1200 W

Now ,Pdevice = V2 / R = 120*120/ 75 = 192 W due to additional 50 ohms

Now,  Pdevice< P50 <P10 <Pwirings .So,yes it will trip device

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