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Physics · Ch 2 — Current Electricity

Joule's Law

2.6.1

Joule's Law

If a current I flows through a conductor across a potential difference V for a time t, the total electrical work done -- equivalently, the electric potential energy spent -- is

W=VIt(2.66)W = VIt \qquad (2.66)

When there is no other useful effect (such as doing mechanical work), the whole of this energy is spent heating the conductor, so the amount of heat H produced is

H=VIt(2.67)H = VIt \qquad (2.67)

Using Ohm's law, V=IRV=IR, this converts into the more commonly used resistor form:

H=I2Rt(2.68)H = I^2Rt \qquad (2.68) …

Misc Example 2.27Heat produced in a resistor over a given time

Worked out. The heat energy produced in a 10 Ω10\ \Omega resistor carrying a 5 A current for 5 minutes is required. Converting time, t=5×60=300t=5\times60=300 s. Using H=I2RtH=I^2Rt: H=52×10×300=25×10×300=75000H=5^2\times10\times300=25\times10\times300=75000 J, i.e. 75 kJ. This is a direct, single-step application of Joule's law -- it shows how, over a modest five-minute span, even an ordinary 10 ohm heating element carrying a household-scale 5 A current already dissipates 75 kJ of heat energy, illustrating just how quickly resis …

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