# Question Video: Understanding Energy Dissipation in Alternating Current Circuit Physics • 9th Grade

An alternating current has a peak value of 1.75 A through a 148 Ω resistor. What is the energy dissipated by the current in a time of 365 s? Give your answer in kilojoules to one decimal place.

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### Video Transcript

An alternating current has a peak value of 1.75 amperes through a 148-ohm resistor. What is the energy dissipated by the current in a time of 365 seconds? Give your answer in kilojoules to one decimal place.

Let’s begin by drawing a diagram of a simple circuit that includes an alternating current source and a 148-ohm resistor. Recall that when there is current present, a resistor will dissipate or give off some energy, usually in the form of heat. And in this question, our job is to figure out how much energy is dissipated over a certain amount of time. To do this, it will be helpful to remember that the rate of energy dissipation by a circuit corresponds to the circuit’s power. Power, 𝑃, can be calculated using the formula 𝑃 equals 𝑊 divided by Δ𝑡, where the amount of energy dissipated is represented by work, 𝑊, and Δ𝑡 represents the period of time over which the energy is dissipated.

Since we want to know the amount of energy dissipated, we want to solve for work. So let’s copy the formula down here and multiply both sides by Δ𝑡 so that term cancels out of the right-hand side, leaving work by itself. Now, flipping the formula and writing it a bit more neatly, we have work equals power times time. We will use the amount of time given, 365 seconds, for Δ𝑡. But we don’t yet know the power of the circuit. So we’ll need to calculate it.

Recall that for an alternating current circuit, power equals the root-mean-square value of the current squared times resistance. Now, we know that the resistance equals 148 ohms. But we don’t know the root mean-square-value of current. The value that we were given, 1.75 amps, is the peak current. This is not a problem though because the root-mean-square value of an alternating current can be easily found using one over the square root of two times the peak current.

Now that we have an expression for power written in terms of values that we know, we should substitute it into our formula for work so that we have work equals one over the square root of two times the peak current squared times resistance times Δ𝑡. Now, substituting in the values for Δ𝑡, resistance, and peak current, we have one over the square root of two times 1.75 amperes squared times 148 ohms times 365 seconds. Since everything is already written in base SI units, we’re all set to calculate, and we have 82718.125 joules.

Now, since we want to give our answer in kilojoules, we should recall that one kilojoule equals 1000 joules. And to convert, we should move the decimal point of our value one, two, three places to the left. And finally, rounding our answer to one decimal place, we have found that over a time period of 365 seconds, the amount of energy dissipated by the alternating current circuit is 82.7 kilojoules.