Lesson Video: Energy Conservation in Circuits Physics

In this video, we will learn how to apply Kirchhoff’s laws to circuits to find the values of currents and potential differences in these circuits.


Video Transcript

In this video, our topic is energy conservation in circuits. The principle of energy conservation applies to all physical systems. And in this lesson, we’re going to learn two laws for electrical circuits that describe how energy is conserved in these cases. As we get started, let’s remind ourselves that energy conservation refers to this idea that the total energy possessed by a closed system is constant, where a closed system is some collection of objects where energy is neither added to that collection nor taken away.

As an example of this, say that we have a system that consists of ourselves, a heavy rock, a staircase, and the ground. If we pick up this rock and start to climb the staircase, then we’re converting energy within our body to gravitational potential energy, both of ourselves as well as the rock. And then, if we release the rock, it will fall back to Earth and its gravitational potential energy is converted to kinetic energy.

So we see then that within this closed system of these four objects, energy can be transferred from one type to another. But, nonetheless, the total energy possessed by this system is constant. And this principle holds true for electrical systems as well. If we think of an electrical circuit on a very small scale, we know that when there’s current in these circuits, that means there’s charge in motion through them. And the reason these charges move is because they’re exposed to a potential difference, what we can call 𝑉.

If we think of a single charge 𝑄 moving through a potential difference 𝑉, then the electrical energy involved is equal to 𝑄 times 𝑉. And then, if we let this charge 𝑄 equal the total charge that passes a point in a circuit over some amount of time, then that ratio, 𝑄 divided by 𝑡, would be equal to the current 𝐼 in the circuit. Or multiplying both sides of the equation by the time 𝑡, the total charge passing a point in some amount of time is equal to the current times that time. And so if we substitute this expression for 𝑄 into our equation for electrical energy, we see that the electrical energy in the circuit is equal to current times time times potential difference.

Two of these quantities, current and potential difference or voltage, are each the subject of an energy conservation law for circuits. These laws are known as Kirchhoff’s laws, and the first one has to do with current in a circuit. It says that whenever there is a junction in an electrical circuit, that would be where there’s a split in a circuit, so there’s more than one current path. Whenever that happens, the total current into the junction must equal the total current out.

So, considering our circuit on screen, let’s say that this voltage source produces a current that we’ll call 𝐼 one in this direction. And then, let’s say that these two voltage sources working together generate a current 𝐼 two pointed this way. We can see that these two currents will meet here at this junction point in the circuit. When they do, a third current magnitude, we can call it 𝐼 three, will exist in this branch of the circuit moving down.

So we can say that the total current moving into this junction, 𝐼 in, is equal to 𝐼 one plus 𝐼 two, while the total current coming out of it is 𝐼 three. And Kirchhoff’s first law, sometimes called Kirchhoff’s current law, says that in this case, 𝐼 one plus 𝐼 two is equal to 𝐼 three. The total current coming into the junction equals the total current coming out.

This current law is always true no matter how many branches enter a junction and no matter how many leave. And we can connect this law with energy conservation like this. If we think of an individual electrical charge, an electron, moving along as part of one of these currents, that electron has a charge and it experiences a potential difference. So it has some amount of electrical energy. Now, let’s say that as this charge approaches the junction, rather than be diverted downward to become part of current 𝐼 three, somehow the charge gets stuck in the junction.

If this happened, it would violate Kirchhoff’s current law because it would mean that the total current into this junction does not equal the total current out. But this would also mean that the total electrical energy into this junction doesn’t equal the total energy out. In other words, it would violate energy conservation. Physically, it’s not the case that charge is moving into a junction gets stuck and pile up there. Rather, any that do enter are able to exit as well. And this enables all three of these currents — 𝐼 one, 𝐼 two, and 𝐼 three — to exist.

So that’s the first of Kirchhoff’s laws covering the current in a circuit. The second of these laws, which also has to do with energy conservation, describes voltage. Unlike Kirchhoff’s current law, where we look at junctions in electrical circuits, for the voltage law, we consider loops in a circuit. One of those loops we’ll look at is here on the left, and another is here on the right. And notice that each one of these loops has a direction we’ve associated with it. In each case, this arrow points in the direction that charge flows through that loop.

So in this loop on the left, that charge flows clockwise, while in the loop on the right, that charge moves counterclockwise. And the direction of charge flow has to do with the direction of voltage sources in these loops. So let’s pick one of these two loops, say this one here. And we’ll follow an individual charge as it moves all throughout the loop. Now, let’s say that our charge starts right here at what we can call the negative terminal of our voltage source. As this charge moves across our voltage source, it experiences a rise in electrical potential.

So if we sketch out a little graph of potential difference in this loop, where movement in this direction along our horizontal axis corresponds with clockwise motion around this loop, we can say that as a charge moves across our voltage source, it experiences a gain in potential. Then as the charge continues to move clockwise like this and then turns this corner and becomes part of the current 𝐼 three, there’s no gain or loss of electrical potential. So we’ll represent that with a horizontal segment.

But then, our charge comes to encounter this resistor here. And let’s say that this resistor possesses a resistance, we’ll call 𝑅 one. As our charge moves from one end of the resistor to another, it will experience a drop in potential. And by Ohm’s law, this potential drop will be equal to the current through the resistor times the resistor value. So on our graph, we can plot that loss in potential like this. And we can say that the magnitude of that loss is 𝐼 three times 𝑅 one. So then, once our charge has crossed this resistor, it moves down to this junction point.

And since we’re specifically considering this current loop, we’ll say that when our charge reaches this junction, it ends up going left out of it. Our charge then continues moving. And we can plot that on our sketch using this horizontal line segment. Once again, we’re not gaining or losing potential. But once our charge reaches this resistor, and we’ll say that this one has a resistance 𝑅 two, as the charge crosses over, it will experience a potential drop. And in magnitude, that drop will equal 𝐼 one, the current that exists through this resistor, multiplied by the resistance value 𝑅 two.

Once our charge has made it through this last obstacle, it’s essentially home free. It can complete its circuit back at the negative terminal of our voltage supply. So if we look at our sketch of voltage across this current loop, we see that we start at zero and we also ended at zero volts. This is the essence of Kirchhoff’s voltage law. Also known as Kirchhoff’s second law, it says that the sum of emfs of voltage sources across a loop equals the sum of voltage drops across that same loop. And these drops, as we saw, are due to components in the circuit, for example, these resistors 𝑅 one and 𝑅 two.

This law also connects with energy conservation because it says that the potential difference we put into a current loop is also the potential difference we take out of it. No matter how complicated a particular current loop, no matter how many sources of voltage and how many resistors and other components, if we do make a plot of potential difference across the complete loop, that plot will always start and end at the same level. That’s the meaning of Kirchhoff’s voltage law.

Now, in a typical example exercise, it’s not necessary to graphically depict the voltage throughout the current loop. But instead, we often represent this change using symbols and algebra. Considering once again this current loop that we moved through, let’s say that our voltage supply generated a potential difference 𝑉.

Applying Kirchhoff’s second law to this loop, we would then say that 𝑉, which is the sum of emfs of voltage sources in this particular loop, is equal to the sum of voltage drops across that same loop. And we saw that those two drops equal 𝐼 three times 𝑅 one and 𝐼 one times 𝑅 two, respectively. Writing the whole thing out this way, we could then solve for one of these values if it was unknown. Knowing all this, let’s get some practice now with Kirchhoff’s laws, which we could also call energy conservation laws for circuits.

The current in three wires of the circuit shown are known. The currents 𝐼 one and 𝐼 two are unknown. Find 𝐼 one. Find 𝐼 two.

So taking a look at this circuit, we see five different currents labeled. There are three that are known — this one here, 1.5 amps; this one here, 2.0 amps; and then this current, 2.5 amps. And along with this, there are two unknown currents, 𝐼 one and 𝐼 two.

We want to solve for both of these unknown currents. And to do this, we can recall Kirchhoff’s current law, also called Kirchhoff’s first law. We can state this law in words like this: the current entering a junction equals the current leaving. This means that at any junction in an electrical circuit, if we add up all the currents moving into that junction point, that sum will equal the sum of all the currents leaving that point.

We can apply this law to certain junction points in this circuit over here to answer the question of what is 𝐼 one and what is 𝐼 two. Let’s first solve for the current 𝐼 one. To do that, we’ll pick a junction point in this circuit, where 𝐼 one is either entering or leaving the junction. The junction we’ll pick is right here. At this location, we have these two currents entering and then current 𝐼 one leaving.

If we apply Kirchhoff’s current law to this junction, we can say that the sum of currents entering it, 1.5 amps added to 2.5 amps, is equal to the sum of currents leaving this junction. But we see that only one is leaving, 𝐼 one. So then, 𝐼 one is the only current on the right side of our equation. And adding together 1.5 amps and 2.5 amps, we get 4.0 amps. This is the value of the current 𝐼 one.

Now let’s use the same law to solve for current 𝐼 two. This time, we’ll pick a different junction point in our circuit. A good one to choose will be this junction right here. We can see that there are two currents, 𝐼 two, and then the current of 2.0 amps coming into this junction point and one current of 2.5 amperes leaving it.

Once again, applying Kirchhoff’s current law, we can say that the sum of currents entering this junction is 𝐼 two added to 2.0 amps and that that equals the total current leaving, which we can see is 2.5 amps. Using this equation to solve for 𝐼 two, if we subtract 2.0 amps from both sides, we see then that 𝐼 two is equal to 0.5 amps. So 𝐼 one is 4.0 amps and 𝐼 two is 0.5 amps.

Let’s look now at a second example exercise.

The resistor in the circuit shown is powered by two batteries in parallel that have terminal voltages of 2.5 volts each. What is the potential drop across the resistor?

So here in this circuit, we have these two batteries, here’s one and here’s the other, and they’re arranged in parallel. And they supply voltage to this resistor here so that there’s a potential drop of 𝑉 volts across the resistor. Our task is to solve for this value. And to do this, we can recall Kirchhoff’s voltage law. In words, this law says that the sum of emfs of voltage sources across a current loop equals the sum of potential drops.

So this is saying that anytime we have a voltage source such as these two batteries in our circuit, then if we consider those sources in the context of a given current loop, the emfs supplied by the sources are equal to the sum of potential drops experienced throughout the rest of the loop due to circuit components such as resistors. Especially in our situation, considering only those voltage sources and circuit components that are part of the same current loop is very important.

If we forget this part about the current loop, we might look at this circuit and say we have 2.5 volts here and 2.5 volts here. And both batteries seem oriented in the same direction, so we expect these voltages to combine. And so we might say then that 𝑉 is equal to the sum of these two voltages, 5.0 volts.

But if we constrain ourselves to thinking in terms of current loops, as the voltage law tells us, then we’ll come up with a different response. In this circuit, there are two separate loops that contain both a battery as well as this resistor. Here’s one of those loops that includes the lower of the two batteries. And then, here’s the second loop. As we analyze this circuit, we’ll consider what’s going on in these two different loops separately.

So first, let’s consider the interior loop. We’ll call this loop one. Following Kirchhoff’s voltage law, this says that the sum of emfs of voltage sources, which for this particular loop we can see is 2.5 volts, is equal — Kirchhoff’s voltage law says — to the sum of potential drops across this loop. As we look at loop one, we see that there’s only one circuit component where voltage drop can occur. It’s over this resistor here. And we’re told that that voltage drop is some amount we can call capital 𝑉.

So then applying Kirchhoff’s voltage law to this first loop, we get 2.5 volts is equal to 𝑉. There are no other components in this current loop. So our answer is simply that 𝑉, the potential drop across the resistor, is 2.5 volts. But now, what if we consider the other loop, we can call that loop two, instead. If we did that and applied Kirchhoff’s voltage law, once again our sum of emfs of voltage sources around this loop would be 2.5 volts.

And also just like before, the only component where this potential can drop is the resistor. And therefore, that potential drop, capital 𝑉, must be equal to 2.5 volts. So no matter which of the two loops we consider, we end up with the same answer that the potential drop across the resistor is 2.5 volts.

Let’s summarize now what we’ve learned about energy conservation in circuits. In this lesson, we saw that energy conservation in circuits is described by Kirchhoff’s laws. The first of these, often called Kirchhoff’s current law, says that the total current entering a junction in a circuit equals the total current leaving that junction. So, for example, if we have a current junction like this with 𝐼 one and 𝐼 two entering and 𝐼 three leaving, then 𝐼 three must be equal to 𝐼 one plus 𝐼 two.

The second of these laws is called Kirchhoff’s voltage law. This says that the sum of emfs of voltage sources across a current loop equals the sum of potential drops across that same loop. So, for example, in a circuit like this, where there’s just one loop of current, the total emfs supplied by voltage sources, what we’ve called 𝑉, is equal to the potential drops across the components of this loop — in this case, resistors 𝑅 one and 𝑅 two. This is a summary of energy conservation in circuits.

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