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concept

## Power in Circuits

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Hey guys. So we talked about resistance is resistors in circuits, and we've also talked about the relationship between the three variables in alms law. So we're gonna talk about energy in circuits and how quickly the energy changes. Let's go and check it out. So remember we talked about resistance. We said that that resistance was an internal friction that occurs in conductors. So as these charges moved through this circuit, they encounter some resistance and they lose some energy to do like this friction. So the energy is lost because you have these charges moving through a potential difference. Delta V. We have the equation for the amount of energy that charges lose through a potential difference. Delta you is that that's equal to Q times Delta V. But a lot of times were more interested in how quickly this power or 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 Delta you divided by Delta T. What's 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 elements. So we're talking about circuits. This equation p equals V. I is gonna be valid for anything. Batteries, resistors. We'll see other kinds of circuit elements later on. So how do we get to this equation? Peak was V I All we have to dio is take this, tell to you over Delta T and just substitute our equation for the charge lost to a resistor. So we have Q times Delta V once we substitute that divided by Delta T. And if you take a look here, this Q divided by Delta T is actually the definition of what current is Remember, current is the 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 basically just becomes delta v times I And then we just dropped the delta 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 owns law, which remember is V equals IR. That on Lee is applicable to resisters to come up with to also alternative forms of this equation. We know P is equal to v times I. But if we substitute this V and owns Law in for this equation, we can come up with another expression, which is I squared. Times are. And if we sort of substitute this I and for this equation manipulated in sulfur V and R, we can actually see that this equation is also equal to V squared, divided by our so here, the three sort of 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 three. Only thing that's going to determine which one of these three you're gonna 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 generate some sort of 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. They're gonna be light bulbs, flashlights, toasters, hair dryers. 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 the the batteries out putting 540 watts of energy. So how much current is the battery producing? So we've got P, which is the power output we've got I and then all we have to do is figure out the Sorry, we're trying to figure out what the I is and we have with 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 equals V times I so that we just have to move the voltage over to the bottom and we have p over V is equal toe I in other words, 540 over nine is equal to 60 and that's gonna be in amps. So that's basically our answer. 60 amps. Alright, let's take a look. This bottom one here a little bit more complicated. We've got a resistor now, so we're gonna be able to use those three equations attached to a battery. So it forms of some simple circuit and we're told the resistance and the currents of the circuit and we need to figure out how much energy Israel is released in form of heat in one minutes. So we're actually not solving for power in this equation, we're solving for the amount of energy which is also equal to delta. You 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 delta u This this power is the definition of the change in energy over change in time. So we're gonna 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 the resistance is and the current. So let's take a look Which one of these equations we can use we've got p equals vie through resistor equals I squared r equals V squared over R Now we've got I and we've got I and are So that means that we're not gonna use this one or this one. We're just gonna use the more direct power equation, right? So you could probably relate this using using alms law, but that's just creating more work for yourself. Just so just go ahead and figure out what you have, and then just plug that into the appropriate equation. So you've got power is equal to got 60 mila amps, so that's 60 times 10 to the minus three. And now we're gonna square that. Don't forget that. Now we have to multiply it by the resistance, which is equal to 30 Kila homes. So that's 30 times 10 to the three. So this is actually going to be 108 watts off power. That's dissipated so that we take this number here. And if we want to figure out what the amount of energy is released in one minutes. Then we have with this delta T is equal to This is just 60 seconds. So we have with this delta t is and we just need to figure out, uh, let's see, right now we have this power output right here. So we've got Delta. U is equal to p times Delta t. So, in other words, it's 108 watts times the time, which is 60 seconds. So we got the amount of energy released in one minute due to this battery in this resistor is equal to 6480 that's in jewels. All right, let me know if you guys have any questions and we're gonna take a look at a couple more practice problems. Thanks for watching.

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Problem

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

(b) R=8.3×10

^{−5}ΩB

(a) i=10 A;

(b) R=0.083 Ω

(b) R=0.083 Ω

C

(a) i=10 A;

(b) R=12 Ω

(b) R=12 Ω

D

(a) i=0.1 A;

(b) R=120,000 Ω

(b) R=120,000 Ω

3

Problem

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 × 10

^{6}ΩAdditional resources for Power in Circuits

PRACTICE PROBLEMS AND ACTIVITIES (10)

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- Consider the circuit of Fig. E25.30. (c) At what rate is electrical energy being converted to other forms in t...
- Consider the circuit of Fig. E25.30. (b) What is the power output of the 16.0-V battery?
- Consider the circuit of Fig. E25.30. (a) What is the total rate at which electrical energy is dissipated in th...
- A 1500-W electric heater is plugged into the outlet of a 120-V circuit that has a 20-A circuit breaker. You pl...
- The heating element of an electric dryer is rated at 4.1 kW when connected to a 240-V line. (b) What is the re...