Video: Finding the Energy Dissipated by an Electrical Component

A bulb is connected to a cell in series. The potential difference across the bulb is 4 V and the current through it is 0.1 A. How much charge passes through the bulb in 60 seconds? How much energy does the bulb dissipate as light and heat in 60 seconds?

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Video Transcript

A bulb is connected to a cell in series. The potential difference across the bulb is four volts. And the current through it is 0.1 amps. How much charge passes through the bulb in 60 seconds? How much energy does the bulb dissipate as light and heat in 60 seconds?

We have these two questions about this scenario. And we’ll start off with the first one, “How much charge passes through this bulb in 60 seconds?” Regarding this light bulb, we’re told that it’s connected in series with a cell and that the potential difference across the bulb is four volts and that the current running through the circuit is 0.1 amperes. Based on that, in this first question, we want to answer how much charge passes through the bulb in 60 seconds.

To figure this out, let’s recall the definition for current. Current 𝐼 is equal to the amount of charge 𝑄 that passes a point in a circuit every time 𝑡. Equivalently then, we can write that charge 𝑄 is equal to current times time. In our scenario, we’re told the current, 0.1 amperes. And we’re also told the time, 60 seconds. If we substitute in those two values and then multiply them together, we find a result of six coulombs. This is how much charge passes through this bulb in 60 seconds.

Knowing that, let’s move on to part two of our question, which asks “How much energy does the bulb dissipate as light and heat in 60 seconds?” Here, it will be helpful to recall an equation for energy, charge, and potential in electrical circuits. In this scenario, the energy involved is equal to the charge multiplied by the potential it experiences.

One way to better understand this equation is to consider an analogous equation from the world of mechanics. In that world, if we take an object with weight 𝑊 and raise it a height ℎ above a reference level, then we’ve given that object a potential energy, which is equal to the product of its weight times the height. In this case, we don’t have weight times height. But we do have charge times potential. And the product of those two terms gives us an electrical energy.

In any case, it’s the energy 𝐸 we want to solve for. And that’s equal to the charge 𝑄, which we solved for previously as six coulombs, multiplied by the voltage across the bulb, four volts. When we find the product of these values, we see that it’s equal to 24 joules. That’s how much energy the bulb gives off as both light and heat over 60 seconds.

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