Power in Circuits - Online Tutor, Practice Problems & Exam Prep

In electrical circuits, energy loss due to resistance is quantified by the equation Δu = q × Δv, where Δu is the change in energy, q is charge, and Δv is potential difference. Power, defined as the rate of energy change, is expressed as P = ΔuΔt. For resistors, power can also be calculated using P = IV, P = I^2R, or P = V^2R, where I is current and R is resistance. The energy dissipated as heat in resistors is crucial for understanding circuit efficiency.

Power in Circuits

Hey, guys. So we talked about resistances, resistors in circuits, and we've also talked about the relationship between the three variables in Ohm's law. So we're going to talk about energy in circuits and how quickly that energy changes. Let's go ahead and check it out. So remember when we talked about resistance, we said that resistance was an internal friction that occurs in conductors. As these charges move through this circuit, they encounter some resistance and they lose some energy due to, like, this friction. So the energy is lost because you have these charges moving through a potential difference, Δv. We have the equation for the amount of energy that charges lose through a potential difference. It's Δu, and it's equal to q times Δv. But a lot of times, we're more interested in how quickly this power - sorry, how quickly this energy gets changed. And remember, that's the definition of power. Power is defined as how quickly energy changes. So in a circuit, it's how quickly it's gained or lost. And we have an equation for that as well. That's just Δu divided by Δt, which is the change in energy versus how long it took to change. Right? So we can actually use this equation to come up with the power output in any circuit element. So when we're talking about circuits, this equation, p = vi, is going to be valid for anything. Batteries, resistors, we'll see other kinds of circuit elements later on. So how do we get to this equation, p = vi? All we have to do is take this Δu over Δt, and just substitute our equation for the charge lost through a resistor. So we have q times Δv, and once we substitute that, divided by Δt. And if you take a look here, this q divided by Δt is actually the definition of what current is. Remember, current is an amount of charge that's flowing through in a certain amount of time. So this really is just I. Remember that this is kind of like in the numerator here. Right? So this just becomes Δv times I, and then we just drop the Δ. Right? Because that's the voltage. Cool. So that's where that equation comes from. Now, when we're talking about specifically a resistor, we can actually use Ohm's law, which remember is v = Ir. That only is applicable to resistors to come up with two alternative forms of this equation. We know p is equal to v times I, but if we substitute this v in Ohm's law for this equation, we can come up with another expression, which is I²r. And if we sort of substitute this I for this equation, manipulate it and solve for v and r, we can actually see that this equation is also equal to v² / r. So here are the three alternate forms of this equation. All three of them are equally valid. You can use any single one of them when you're talking about circuits. The only thing that's going to determine which one of these three you're going to use is just what variables you're given in a problem. That's basically it. Okay. So this is the power that's dissipated by a resistor. But where does that energy go? Remember, we said that this energy is internal friction, and friction always generates some form of heat. So what happens is that the power that's dissipated by these resistors is released in the form of heat, and in the extreme case, heat and light. So good examples of resistors are going to be light bulbs, flashlights, toasters, or hairdryers, any kind of thing that changes electrical energy into heat is a resistor. Alright? That's basically it, so let's go ahead and take a look at some examples. We've got a battery that operates at a voltage, and the battery's outputting 540 watts of energy. So how much current is the battery producing? So we've got p, which is the power output, and we've got I. And then all we have to do is figure out what the I is, and we have what the voltage is. Right? So if we're looking for the currents, we have to relate it to the power output of any circuit formula. We can't use the three equations because this is not a resistor. Remember, this is a battery, So we have to use p = vi, so that we just have to move the voltage over to the bottom. And we have p / v is equal to I. In other words, 540 / 9 is equal to 60, and that's going to be in amps. So that's basically our answer, 60 amps. Alright. Let's take a look at this bottom one here. A little bit more complicated. We've got a resistor now, so we're going to be able to use those three equations. It's attached to a battery, so it forms a simple circuit, and we're told the resistance and the currents of this circuit, and we need to figure out how much energy is released in the form of heat in 1 minute. So we're actually not solving for power in this equation, we're solving for the amount of energy, which is also equal to Δu. Both of these letters mean the same exact thing. So really, this is what we need to find. But how do we relate that back to power? Remember that this Δu, this power, is the definition of the change in energy over change in time. So we're going to relate it back to this power formula. Now we have a resistor that's attached to a battery, so we can use those three equations. And in this specific instance, we have what the resistance is and the current. So let's take a look at which one of these equations we can use. We've got p = vi through a resistor = I²r = v² / r. Now we've got I and we've got I and r, so that means that we're not going to use this one or this one. We're just going to use the moreirect power equation. Right? So you could probably relate this using using Ohm's law, but that's just creating more work for yourself. Just go ahead and figure out what you have, and then just plug that into the appropriate equation. So we've got power is equal to we've got 60 milliamps. So that's 60 times 10 to the -3, and now we're going to square that. Don't forget that. Now, we have to multiply it by the resistance, which is equal to 30 kilo-ohms. So that's 30 times 10 to the 3. So this is actually going to be 108 watts of power that's dissipated. So now, we take this number here, and if we want to figure out what the amount of energy is released in 1 minute, then we have what this Δt is equal to. This is just 60 seconds. So we have what this Δt is. And we just need to figure out right here. So we've got Δu is equal to p times Δt. So in other words, it's 108 watts times the time, which is 60 seconds. So we got the amount of energy released in 1 minute due to this battery and this resistor is equal to 6,480, and that's in joules. Alright? Let me know if you guys have any questions, and we're going to take a look at a couple more practice problems. Thanks for watching.

A hair dryer operates at 120 V (the voltage produced by a household outlet), and outputs 1200 W of energy. For this problem, treat the hair dryer as a single resistor. 

(a) At what current does the hair dryer operate?
(b) What is the resistance of the hair dryer?

• A. (a) i=0.1 A;
  (b) R=8.3×10⁻⁵ Ω
• B. (a) i=10 A;
  (b) R=0.083 Ω
• C. (a) i=10 A;
  (b) R=12 Ω
• D. (a) i=0.1 A;
  (b) R=120,000 Ω

An incandescent lightbulb produces 100 W of light. If this lightbulb operates at 25% efficiency (meaning that out of all the power it generates, only 25% is released as light), what resistance must the lightbulb have if it operates at 120 V?

• A. 36 Ω
• B. 144 Ω
• C. 576 Ω
• D. 5.76 × 10⁶ Ω