Code question of the day weekend edition

[Swartz09]

Q) What does it cost per year, at 6.78 cents per KWH, for the power loss of two 20 AWG circuit conductors with a total resistance of 0.40 ohms with a current flow of 20 amps?

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[admin]

First, let's find out how much power is lost due to the resistance of the conductors. This can be calculated using Ohm's law (V = IR) and the power law (P = VI).

Calculate voltage drop using Ohm's law:

Voltage drop (V) = Current (I) * Resistance (R)
= 20 Amps * 0.40 Ohms
= 8 Volts

Calculate power loss using the power law:

Power Loss (P) = Voltage (V) * Current (I)
= 8 Volts * 20 Amps
= 160 Watts

That's the power lost per instant. Now, to find out how much it costs for a year, we need to convert the power loss into kilowatts, find out how much energy is used in a year, and then multiply by the cost per kilowatt-hour.

Convert power loss to kilowatts:

160 Watts = 0.16 kilowatts (kW)

Calculate energy used in a year:

Energy used (E) = Power (P) * Time (T)
= 0.16 kW * (24 hours/day * 365 days/year)
= 1,401.6 kilowatt-hours (kWh)

Calculate cost:

Cost = Energy used (E) * Cost per kWh
= 1,401.6 kWh * $0.0678 per kWh
= approximately $94.99

So, the power loss of two 20 AWG circuit conductors with a total resistance of 0.40 ohms with a current flow of 20 amps, at 6.78 cents per kWh, costs approximately $94.99 per year.