## Wire Size and Electric Current – Joule Heating

 Ever take a peek inside the toaster while you’re waiting for the toast to pop up?  If so, you would have noticed a bright orange glow.  That glow is produced when the toasting wires heat up, which in turn creates a nice crusty surface on your bread or waffle.  It’s the same phenomenon as when the filament inside an incandescent bulb glows.  The light and heat produced in both these cases are the result of the Joule, pronounced “jewel,” effect at work.      To understand Joule heating, let’s first refresh our memories as to electrical current resistance.  We learned previously that wire is not a perfect conductor, and as such resistance to flow is encountered.  This resistance causes power to be lost along the length of wire, in accordance with this equation: Power Loss = I2 × R Where I is the electric current flowing through a wire, and R is the total electrical resistance of the wire.  The power loss is measured in units of Joules per second, otherwise known as watts, “watt” denoting a metric unit of power.  It is named after the famed Scottish mechanical engineer, James Watt, who is responsible for inventing the modern steam engine.  A Joule is a metric unit of heat energy, named after the English scientist James Prescott Joule.  He was a pioneer in the field of thermodynamics, a branch of physics concerned with the relationships between different forms of energy.      Anyway, to see how the equation works, let’s look at an example.  Suppose we have 12 feet of 12 AWG copper wire.  We are using it to feed power to an appliance that draws 10 amperes of electric current.  Going to our handy engineering reference book, we find that the 12 AWG wire has an electrical resistance of 0.001588 ohms per foot, “ohm” being a unit of electrical resistance.  Plugging in the numbers, our equation for total electrical resistance becomes: R = (0.001588 ohms per foot) × 12 feet = 0.01905 ohms And we can now calculate power loss as follows: Power = I2 × R = (10 amperes)2 × (0.01905 ohms) = 1.905 watts      Instead of using a 12 AWG wire, let’s use a smaller diameter wire, say, 26 AWG.  Our engineering reference book says that 26 AWG wire has an electrical resistance of 0.0418 ohms per foot.  So let’s see how this changes the power loss: R = (0.0418 ohms per foot) × 12 feet = 0.5016 ohms Power = I2 × R = (10 amperes)2 × (0.5016 ohms) = 50.16 watts      This explains why appliances like space heaters and window unit air conditioners have short, thick power cords.  They draw a lot of current when they operate, and a short power cord, precisely because it is short, poses less electrical resistance than a long cord.  A thicker cord also helps reduce resistance to power flow.  The result is a large amount of current flowing through a superhighway of wire, the wide berth reducing both the amount of power loss and the probability of dangerous Joule heating effect from taking place.       Our example shows that the electric current flowing through the 12 AWG wire loses 1.905 watts of power due to the inconsistencies within the wire, and this in turn causes the wire to heat up.  This is Joule heating at work.  Joule heating of 50.16 watts in the thinner 26 AWG wire can lead to serious trouble.      When using a power cord, heat moves from the copper wire within it, whose job it is to conduct electricity, and beyond, on to the electrical insulation that surrounds it.  There the heat is not trapped, but escapes into the environment surrounding the cord.  If the wire has low internal resistance and the amount of current flowing through it is within limits which are deemed to be acceptable, then Joule heating can be safely dissipated and the wire remains cool.  But if the current goes beyond the safe limit, as specified in the American Wire Gauge (AWG) table for that type of wire, then overheating can be the result.  The electrical insulation may start to melt and burn, and the local fire department may then become involved.          That’s it for wire sizing and electric current.  Next time we’ll slip back into the mechanical world and explore a new topic: the principles of ventilation. _____________________________________________

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