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2.4: Calculating Electric Power

  • Page ID
    687
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    Learn the Power Formula

    We’ve seen the formula for determining the power in an electric circuit — by multiplying the voltage in “volts” by the current in “amps” we arrive at an answer in “watts.” Let’s apply this to a circuit example:

    00040.webp

    How to Use Ohm’s Law to Determine Current

    In the above circuit, we know we have a battery voltage of 18 volts and a lamp resistance of 3 Ω. Using Ohm’s Law to determine current, we get:

    \[I=\frac{E}{R}=\frac{18 \mathrm{V}}{3 \Omega}=6 \mathrm{A}\]

    Now that we know the current, we can take that value and multiply it by the voltage to determine power:

    \[P=I E=(6 \mathrm{A})(18 \mathrm{V})=108 \mathrm{W}\]

    This tells us that the lamp is dissipating (releasing) 108 watts of power, most likely in the form of both light and heat.

    Increasing the Battery Voltage

    Let’s try taking that same circuit and increasing the battery voltage to see what happens. Intuition should tell us that the circuit current will increase as the voltage increases and the lamp resistance stays the same. Likewise, the power will increase as well:

    00041.webp

    Now, the battery voltage is 36 volts instead of 18 volts. The lamp is still providing 3 Ω of electrical resistance to the flow of electrons. The current is now:

    \[I=\frac{E}{R}=\frac{36 \mathrm{V} }{3 \Omega}=12 \mathrm{A}\]

    This stands to reason: if I = E/R, and we double E while R stays the same, the current should double. Indeed, it has: we now have 12 amps of current instead of 6. Now, what about power?

    \[P=I E=(12 \mathrm{A} )(36 \mathrm{V} )=432 W\]

    What does Increasing Battery do to Power?

    Notice that the power has increased just as we might have suspected, but it increased quite a bit more than the current. Why is this? Because power is a function of voltage multiplied by current, and both voltage and current doubled from their previous values, the power will increase by a factor of 2 x 2, or 4. You can check this by dividing 432 watts by 108 watts and seeing that the ratio between them is indeed 4.

    Using algebra again to manipulate the formulae, we can take our original power formula and modify it for applications where we don’t know both voltage and current:

    If we only know voltage (\(E\)) and resistance (\(R\)):

    10019.webp

    If we only know current (\(I\)) and resistance (\(R\)):

    10020.webp

    A historical note: it was James Prescott Joule, not Georg Simon Ohm, who first discovered the mathematical relationship between power dissipation and current through a resistance. This discovery, published in 1841, followed the form of the last equation (P = I2R), and is properly known as Joule’s Law. However, these power equations are so commonly associated with the Ohm’s Law equations relating voltage, current, and resistance (E=IR ; I=E/R ; and R=E/I) that they are frequently credited to Ohm.

    10021.webp

    Review

    • Power measured in watts, symbolized by the letter “W”.
    • Joule’s Law: P = I2R ; P = IE ; P = E2/R

    This page titled 2.4: Calculating Electric Power is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Tony R. Kuphaldt (All About Circuits) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.