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4.6: Dissipation of Energy

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    5436

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    When current flows through a resistor, electricity is falling through a potential difference. When a coulomb drops through a volt, it loses potential energy 1 joule. This energy is dissipated as heat. When a current of \(I\) coulombs per second falls through a potential difference of \(V\) volts, the rate of dissipation of energy is \(IV\), which can also be written (by making use of Ohm’s law) \(I^2R \text{ or }V^2/R\).

    If two resistors are connected in series, the current is the same in each, and we see from the formula \(I^2R\) that more heat is generated in the larger resistance.

    If two resistors are connected in parallel, the potential difference is the same across each, and we see from the formula \(V^2/R\) that more heat is generated in the smaller resistance.

    This page titled 4.6: Dissipation of Energy is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.