Lesson Explainer: Energy Conservation in Circuits Physics

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

Let us start by understanding what we mean by energy in a circuit. We may be used to energy in other forms, such as mechanical energy, but what do we mean by energy in an electric circuit?

Consider two points in a circuit, point 𝐴 and point 𝐵, with potential difference 𝑉 across them caused by an electric field such that point 𝐴 is at a higher potential than point 𝐵. Now, consider a positive charge, 𝑄, placed at point 𝐴. The potential difference will cause the charge to move from point 𝐴 to point 𝐵, as illustrated in the following diagram.

The charge moves because the potential difference causes a force to act on the charge, which does work on it as it moves. The work done on the charge, 𝐸, is equal to the charge multiplied by the potential difference between the two points: 𝐸=𝑄𝑉.

As the charge moves through from higher to lower potential, energy is transferred from electrical potential energy to other categories of energy.

In an electric circuit, we can consider the amount of charge, 𝑄, that has moved past a point over some time, 𝑡. Recall that the current in an electric circuit, 𝐼, is equal to the amount of charge that has passed the point in the circuit that is being measured divided by time: 𝐼=𝑄𝑡.

This can be rearranged to give an expression for the charge that has passed this point in the circuit: 𝑄=𝐼𝑡.

We can substitute this back into our equation for electrical energy, 𝐸=𝐼𝑡𝑉, then we can divide both sides of the equation by 𝑡: 𝐸𝑡=𝐼𝑉.

This is the amount of energy per unit of time that the electric circuit consumes—also known as power, 𝑃=𝐼𝑉.

In an electric circuit, energy is conserved; the circuit cannot spontaneously create or destroy energy. Electrical energy can, however, be transformed to other categories of energy between points where the potential of the circuit changes (e.g., across a resistor).

Let us consider a point in a circuit where the circuit splits in two. We call the point at which a circuit splits or joins together a “node,” or a “junction.”

The diagram below shows this point in a circuit, with current 𝐼 into the node and 𝐼 and 𝐼 out of the node:

The amount of energy into the node per unit of time, the power in, is equal to the sum of currents into the node multiplied by the voltage of the node, 𝑉: 𝑃=𝐼𝑉.in

The amount of energy out of the node per unit of time, the power out, is equal to the sum of currents out of the node multiplied by the voltage of the node: 𝑃=(𝐼+𝐼)𝑉.out

Energy at the node is conserved; the power into the node is equal to the power out of the node: 𝑃=𝑃.inout

This can be expressed in terms of current and potential as 𝐼𝑉=(𝐼+𝐼)𝑉.

We can divide both sides of this equation by 𝑉 to give an expression for the currents in and out of the node: 𝐼=𝐼+𝐼.

Now, imagine a node with many currents in and many out:

Again, we can calculate the power into the node: 𝑃=𝐼+𝐼+⋯𝑉.ininin

Similarly, we can calculate the power out of the node: 𝑃=𝐼+𝐼+⋯𝑉.outoutout

Energy is conserved, so 𝑃=𝑃.inout

Substituting the expressions for 𝑃in and 𝑃out, 𝐼+𝐼+⋯𝑉=𝐼+𝐼+⋯𝑉,ininoutout then dividing both sides by 𝑉 gives an expression relating the currents in and out of a node: 𝐼+𝐼+⋯=𝐼+𝐼+⋯.ininoutout

This is the first of Kirchhoff’s two laws. Kirchhoff’s first law states that the sum of the currents into a node equals the sum of the currents out of a node.

Definition: Kirchhoff’s First Law

The sum of the currents into a node in a circuit, 𝐼+𝐼+⋯inin, equals the sum of the currents out of the node, 𝐼+𝐼+⋯outout: 𝐼+𝐼+⋯=𝐼+𝐼+⋯.ininoutout

This can be used in circuits to calculate currents in each branch of a circuit. For example, in the following diagram, the currents in two branches are known, and can be used to calculate the current in the third branch.

Without knowing the value of either resistor, or even the potential difference across the battery, we can use Kirchhoff’s first law to relate the three currents: 𝐼=𝐼+𝐼.

Rearranging this for 𝐼 and substituting the values 𝐼=0.75A and 𝐼=0.25A, we get 𝐼=𝐼−𝐼𝐼=0.75−0.25𝐼=0.5.AAA

We will now work through an example question using Kirchhoff’s first law.

Example 1: Using Kirchhoff’s First Law to Calculate Current in a Circuit

The currents in two wires of the circuit shown are known. The currents 𝐼 and 𝐼 are unknown.

  1. Find 𝐼.
  2. Find 𝐼.

Answer

Part 1

Kirchhoff’s first law states that the sum of currents into a node in a circuit is equal to the sum of currents out of the node. To find 𝐼, we can consider the node at the top of the circuit. The total current into the node, 𝐼in, is equal to 𝐼=𝐼,in and the total current out of the node, 𝐼out, is equal to 𝐼=2.5+1.5.outAA

Equating the two, we get 𝐼=2.5+1.5𝐼=4.0.AAA

Part 2

To find 𝐼, we can consider the node at the bottom of the circuit. The total current into this node, 𝐼in, is 𝐼=2.5+1.5,inAA and the current out of this node, 𝐼out, is 𝐼=𝐼.out

Equating the two, we get 2.5+1.5=𝐼𝐼=4.0.AAA

Note that 𝐼 could have been found by looking at the right branch of the circuit and noticing that 𝐼=𝐼 because the currents are at two points in series.

We can also look at the electrical potential of a circuit in terms of energy.

Take a look at the following circuit, comprised of a battery and two resistors. The potential difference across the battery is equal to 𝑉 and that across the first and second resistors is 𝑉 and 𝑉 respectively. We can imagine looking at a zoomed-in view of a positive charge, 𝑄, moving around the circuit. In this case, we will follow a positive charge moving in the direction of the current.

We can calculate the work done on the positive charge as it moves between points 1 to 6 of the circuit. The work done is equal to the magnitude of the charge multiplied by the potential difference across the points it travels between. This is equal to the change in the electrical potential energy of the charge and can be visualized on the following graph of electrical potential energy, 𝐸, against distance traveled around the circuit, 𝑑.

Note that the work done on the charge only tells us the change in electrical potential energy; the charge has some base level of electrical potential energy of 𝐸.

As seen, the electrical potential energy of the charge by the time it gets back to point 1 of the circuit must be the same as the electrical potential energy of the charge the previous time it was at point 1. This is because energy in the circuit is conserved; within each loop of the circuit, the particle cannot gain or lose electrical potential energy.

This means that the total work done on the charge over a single loop of the circuit must equal zero: 𝑄𝑉+𝑄𝑉+𝑄𝑉=0.

The magnitude of the charge is constant, so we can divide this equation by 𝑄: 𝑉+𝑉+𝑉=0.

Similarly to electrical potential energy, we can also plot a graph of the electrical potential of the charge, 𝑉, against distance traveled around the circuit, 𝑑.

Note that there is some base-level electrical potential of the charge, 𝑉; we can only measure potential difference between two points in the circuit.

This means that because energy is conserved, around a loop in a circuit, the sum of the potential differences across each component in the loop must equal zero: 𝑉+𝑉+⋯+𝑉=0.

This is Kirchhoff’s second law; the sum of the potential differences across all components in a loop in a circuit is equal to zero.

Definition: Kirchhoff’s Second Law

The sum of the potential difference across each component in a loop in a circuit is equal to zero: 𝑉+𝑉+⋯+𝑉=0.

We will now use Kirchhoff’s second law to calculate unknown potential differences in a circuit.

Example 2: Using Kirchhoff’s Second Law to Calculate Voltage in a Circuit

The potential drop across the resistor in the circuit shown is 15 V. The terminal voltage of one of the batteries powering the circuit is 5.5 V. Find the terminal voltage 𝑉 of the other battery powering the circuit.

Answer

Kirchhoff’s second law states that the sum of the potential difference across each component in a loop in a circuit is equal to zero.

In this circuit, there is a known 5.5 V increase in potential across the first battery, a 𝑉 gain across the second battery, and a 15 V decrease in potential across the resistor.

Writing Kirchhoff’s second law for this loop, we have 5.5+𝑉−15=0𝑉=9.5.VVVV

We can apply Kirchhoff’s second law to circuits that split into multiple branches. For example, the following circuit has three loops that a charge could take, as shown in the following diagram.

Here, we will say that 𝑉=𝑉. We can follow a charge moving in either direction around each loop, and graphs of electrical potential relative to base-level electrical potential, 𝑉−𝑉, against distance traveled around the loop, 𝑑, can be plotted. These graphs are shown in the following diagram.

As before, across a whole loop, the work done on the charge, and therefore the total potential difference across each component in the loop, must equal zero. This means that Kirchhoff’s second law can be applied to any loop in a circuit.

We will now work through an example question applying Kirchhoff’s second law to a circuit with multiple loops.

Example 3: Using Kirchhoff’s Second Law to Calculate Voltage in a Circuit with Multiple Loops

Find the decrease in potential across the resistor in the circuit shown. The batteries powering the circuit each have a terminal voltage of 2.5 V.

Answer

This circuit has three loops that can be followed in either direction, which we can label on the circuit diagram.

To find the decrease in potential across the resistor, 𝑉, we can look at either loop A or loop C.

Recall that, according to Kirchhoff’s second law, the sum of the potential differences across each component in a loop is equal to zero: 𝑉=0.

For loop A, this is 2.5+𝑉+2.5=0.VVV

Rearranging gives 𝑉=−5.0.V

So, the decrease in potential across the resistor is equal to 5 V.

When we have batteries in parallel, Kirchhoff’s second law shows us something interesting.

Let us consider the circuit shown in the following diagram.

If the magnitude of 𝑉 is not equal to that of 𝑉, then the loop appears to not follow Kirchhoff’s second law; this is not possible, as energy must always be conserved! In that case, there is an internal resistance in each battery that must be considered that causes the drop in potential required to conserve energy in the circuit.

So, when batteries in parallel do not have identical voltage, internal resistance must be known to be able to analyze them.

We will now work through a question considering two uneven batteries in parallel.

Example 4: Kirchhoff’s Second Law Applied to Batteries in Parallel

The resistor in the circuit shown is powered by two batteries in parallel that are combined in two different configurations. The batteries have terminal voltages of 3.5 V and 2.5 V respectively. In the first configuration, the positive terminals of the battery are directly connected to each other and the negative terminals are connected to each other. In the second configuration, the positive terminals of each battery are directly connected to the negative terminal of the other battery. Which of the following is a correct statement about how the decrease in potential across the resistor compares in the two configurations?

  1. The decrease in potential will be the same in both configurations.
  2. The decrease in potential in both configurations will depend on the internal resistances of the batteries.
  3. The decrease in potential will be greater in the second configuration.
  4. The decrease in potential will be greater in the first configuration.

Answer

First, we will consider the top circuit. Applying Kirchhoff’s second law to the loop containing the top battery and the resistor gives 3.5−𝑉=0.V

Applying Kirchhoff’s second law to the loop containing the bottom battery and the resistor gives 2.5−𝑉=0.V

The two expressions give different answers! Energy must be conserved in the circuit, so there must be some internal resistance in each battery that must be known to be able to analyze the circuit correctly.

We can repeat the same process on the bottom circuit. Applying Kirchhoff’s second law to the loop containing the top battery and the resistor gives 3.5−𝑉=0.V

Repeating this for the loop containing the bottom battery and the resistor gives −2.5−𝑉=0.V

Again, the two expressions produce different answers. This means that the internal resistance of the batteries must be known to correctly apply Kirchhoff’s laws to the circuit.

The only way to know the exact effect of changing the configuration of the batteries is if the internal resistances of the batteries are known. Therefore, the answer is B.

We can summarize what we have learned in this explainer in the following key points.

Key Points

  • Energy is conserved in a circuit. This means that, for a charged particle moving around the circuit, the work done on the particle is equal to the change in electrical potential energy of the particle.
  • Kirchhoff’s first law states that the sum of the currents into a junction/node in a circuit, 𝐼+𝐼+⋯inin, must be the same as the sum of the currents out of the junction/node, 𝐼+𝐼+⋯outout: 𝐼+𝐼+⋯=𝐼+𝐼+⋯.ininoutout
  • Kirchhoff’s second law states that the sum of the potential difference across each component in a loop, 𝑉,𝑉,…,𝑉, is equal to zero: 𝑉+𝑉+⋯+𝑉=0.

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