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Introduction

Kirchhoff’s second law deals with the conservation of energy around a closed loop series path. We call this law Kirchhoff’s Voltage Law, or KVL, for short.

What’s a Closed Loop Series Path?

A closed loop series path is where there is a direct path through circuit elements from one point of the circuit around through other points of the circuit, and back to the initial point — each joined by a conductor in a sequence. Here’s a quick and easy to understand representation of what loops are and how to identify them.

A voltage source, such as a battery in a closed loop circuit provides a certain amount of voltage. As current flows through elements within the circuit, a potential difference occurs across the elements or a drop or gain in voltage is detected across each element within the circuit.

There could be other elements within a circuit that provide energy besides the battery source, such as other voltage or current sources, which could provide voltage gains, but many elements in a circuit are passive, which do not require an external source to function – they just influence the flow of power (P = IV), like resistors, capacitors and inductors.

The algebraic sum of these voltage potentials around any closed loop in a circuit is equal to zero because a circuit loop is a closed conducting path and no energy should be gained or lost within the circuit – meaning there’s no external influence on the circuit for energy to be gained or lost – it’s “contained” within the circuit.

Let’s Make a Rule!

If we were to set up a rule for ourselves and say when there is a voltage potential across components in a closed loop path, we just need to add up all those voltages – giving them the proper sign convention based on passive or active components – and set our summation of values equal to zero.

If you’re unfamiliar with the concept of passive and active elements in a circuit and their sign conventions, I encourage you to check out our post on active and passive elements here.

\begin{equation}
\Sigma \; Voltage \; Potentials \; in \; Closed \; Loop \; Path \; = \; 0
\end{equation}

Where Σ is the capital Greek letter Sigma, which in mathematics is used to express the summation of values – literally meaning “the sum of”. You can think of the letter Σ telling you, “Just add this stuff together, okay?!”

Said another way, we can say that we add all the voltage potentials of the passive elements and add all the voltage potentials of the active elements (which is the summation of negative values), then set our summation of values equal to zero.

\begin{equation}
\Sigma \; (VP \; across \; each \; PE) \; + \; \Sigma \; (VP \; across \; each \; AE) \; = \; 0
\end{equation}

Where the abbreviations above stand for the following:

• VP: Voltage Potential
• PE: Passive Element
• AE: Active Element

Using the image above as an example, we’d do our summation of the closed loop as follows:

\begin{equation}
\Sigma \; (VP \; across \; each \; PE) \; + \; \Sigma \; (VP \; across \; each \; AE) \; = \; 0
\end{equation}

\begin{equation}
V_{R1} \; + \; V_{R2} \; + \; V_{S} \; = \; 0
\end{equation}

Quick Review of The Conservation of Energy

The conservation of energy is written as:

\begin{equation}
\Delta U \; = \; W \; + \; Q
\end{equation}

Where the abbreviations above stand for the following:

• ΔU: The change in energy of the system (the total energy minus the initial energy)
• W: The work done by or on the system
• Q: The heat added or removed from the system

You can think of ΔU as being equivalent to (Σ (VP across each PE) + Σ (VP across each AE)), the change of energy in the form of voltage potentials throughout the loop of a circuit, and (W + Q) as being zero — because our circuit is a closed loop circuit, therefore any work done on the system is negated by the heat removed by the system, or any work done by the system is being negated by the heat added to the system.

But, this is not a physics class — we’re talking about voltage here. Just know that a closed loop path circuit has some initial energy given to the system (the circuit) via a battery, for example. That energy is dispersed across the elements of the circuit via voltage potentials across passive elements, which are given a positive sign convention. If elements in the circuit are active elements, then they provide energy to the system or circuit and they have a negative sign convention.

---

A Voltage Potential Across a Single Element

Let’s look at a basic case of KVL using a single element in a circuit. As you can see in the image below, the circuit contains only a battery (voltage source) and a resistor.

The battery provides 1.5 volts to the system. The system is the circuit containing only the battery and the resistor. So, if the battery provides 1.5V, then the voltage potential across the resistor must be 1.5V, no matter its value of resistance.

\begin{equation}
\boxed{V_{R} \; = \; V_{BC} \; = \; 1.5V}
\end{equation}

Why? Well, looking at the image above, we see between points A and B in the circuit that there’s no element there. Therefore, according to Ohm’s Law, there’s no voltage potential there because there’s no resistance there. Similarly, between points C and D there’s no elements there, so no voltage potential there either. The only other place — besides between points A and D, where the battery is — for a voltage potential to show up in the circuit is across the resistor (R), between points B and C.

---

Voltage Potentials Around a Series Circuit Loop

Let’s say we have a single closed loop path for a circuit that contains a battery as its voltage source and two resistors, as shown in the image below. Imagine we’re told that the voltage source V_S is 9 volts and the resistor values for R₁ and R₂ are both 1 kilo-ohms.

Now, we’re asked to see if energy is conserved within the circuit. Are we loosing any energy out of the circuit? Are we gaining any energy into the circuit? According to Kirchhoff’s voltage law we shouldn’t be if our circuit is a closed loop path.

Kirchhoff’s voltage law (KVL) says that as we go around the circuit, the sum of the voltages around the circuit should equal to zero – meaning the energy should be conserved. Looking at our circuit below, now with the values shown, let’s determine if it follows KVL.

I’ve added the letters A, B, C, and D to the circuit as a guide for us as we traverse across the circuit. These letters are of no significance to the circuit itself, nor do they contain any values or affect our calculations. They’re strictly for use as a guide for us.

Going From Point A to Point B

If we start at point A going to point B in the circuit, we see that current travels from the negative (-) side of the 9V battery to the positive (+) side. We are going to be using passive convention here for our calculations. If you’re unfamiliar with the concept of passive convention, check out our post on passive convention here real quick, then come back.

A battery is an active element, not a passive element, so this means we will right down our first value of summation as minus 9 volts (-9V), since current is traveling from the minus-side to the plus-side of the battery:

\begin{equation}
V_{AB} \; = \; -9V
\end{equation}

Going From Point B to Point C

Next, we go from point B to point C in the circuit and in doing so we see that current travels through a resistor. We label the B-side of the resistor as positive (+) since this side is connected to the positive-side of the battery – we’re following passive convention here, so since the resistor is a passive element, current passes through it from positive (+) to negative (-).

We’re adding voltages here and resistor R₁ is given in ohms, so we need to use the help of Ohm’s Law to use both the resistor and current values to calculate the voltage potential across the resistor.

Ohm’s Law \begin{equation}
V_{BC} \; = \; IR_{1}
\end{equation}

\begin{equation}
V_{BC} \; = \; (4.5mA)(1kΩ)
\end{equation}

\begin{equation}
V_{BC} \; = \; (0.0045A)(1000Ω)
\end{equation}

\begin{equation}
V_{BC} \; = \; 4.5V
\end{equation}

Note that we must convert from milliamps to amps, and from kilo-ohms to ohms for our calculation. We calculated 4.5V across resistor R₁. Notice that it’s a positive value, because the resistor is a passive element and we used passive convention. So, this voltage value is our second value of summation:

\begin{equation}
V_{BC} \; = \; 4.5V
\end{equation}

Going From Point C to Point D

Going from point C to point D in the circuit, we see that current travels through another resistor, R₂, with a value of 1kΩ. Again, we’re using passive convention here, and we need to use Ohm’s Law to calculate the voltage potential across R₂ to be able to add this value to our summation of voltage values for this circuit.

Ohm’s Law \begin{equation}
V_{CD} \; = \; IR_{2}
\end{equation}

\begin{equation}
V_{CD} \; = \; (4.5mA)(1kΩ)
\end{equation}

\begin{equation}
V_{CD} \; = \; (0.0045A)(1000Ω)
\end{equation}

\begin{equation}
V_{CD} \; = \; 4.5V
\end{equation}

Note that we must convert from milliamps to amps, and from kilo-ohms to ohms for our calculation. We calculated 4.5V across resistor R₂. Notice that it’s a positive value, because the resistor is a passive element and we used passive convention. So, this voltage value is our second value of summation:

\begin{equation}
V_{CD} \; = \; 4.5V
\end{equation}

Going From Point D to Point A

Going from point D to point A in the circuit, we see that current does not travel through any elements, just through the wire or conductor of this section of the circuit. So, there’s no voltage potential across this part of the circuit, since there’s no resistance from an element here:

\begin{equation}
V_{DA} \; = \; IR
\end{equation}

\begin{equation}
V_{DA} \; = \; (4.5mA)(0Ω)
\end{equation}

\begin{equation}
V_{DA} \; = \; 0V
\end{equation}

Summation of Values

Well, we’ve made it around the closed loop path of our circuit and found all the voltages across the elements in the circuit. Now, we’re able to use KVL to check and make sure that energy is conserved in our circuit. All we need to do is add up all the voltages!

\begin{equation}
V_{AB} \; + \; V_{BC} \; + \; V_{CD} + \; V_{DA} \; = \; 0
\end{equation}

\begin{equation}
-9V \; + \; 4.5V \; + \; 4.5V + \; 0V \; = \; 0
\end{equation}

\begin{equation}
-9V \; + \; 9V \; = \; 0
\end{equation}

\begin{equation}
\boxed{9V \; = \; 9V}
\end{equation}

Energy is conserved and Kirchhoff’s Voltage Law proved it!

---

Using KVL in a Parallel Circuit

Let’s say we have a parallel circuit containing a battery as its voltage source and three resistors in parallel, as shown in the image below.

Imagine we’re told that the voltage source V_S is 12 volts and the resistor values for R₁, R₂, and R₃ are 1 kilo-ohm, 170 ohms, and 10 mega-ohms. Again, we’re asked to see if energy is conserved within the circuit.

You may be asking yourself, “Wait a minute. I thought that for KVL we deal with closed loop series paths. This is a parallel circuit!”

Ah, you are correct! But, you forget that a loop is any closed path in a circuit – not just for a series circuit. If you’re confused at this point, have a look at this excellent explanation of what a loop is and come right back – it’ll only take you a minute. We can see that our circuit has 6 loops in total.

As you can see in the image above, I’ve added some points labeled with capital letters in black and loops that are numbered within the circuit in different colors to help explain the use of KVL in a parallel circuit. Let’s start with loop 1.

KVL Around Loop 1

We’ll step around the loop, starting at point A, work our way toward point B, then down to point G, over to point H, and back to point A to complete the loop.

Point A to Point B

Going from point A to point B, we see that there’s no element in this part of the loop of the circuit – meaning there’s no resistance here of any kind to provide a voltage potential. So, the voltage here, V_AB, is zero volts:

\begin{equation}
V_{AB} \; = \; 0V
\end{equation}

Point B to Point G

Now, working our way down from point B to point G of the circuit, we have a resistor (R₁) with a value of 1kΩ. If we knew the value of the current flowing through this part of the circuit, we could calculate the voltage across this resistor using Ohm’s Law, but since we don’t, we’ll just leave it as V_BG, for the voltage across the resistor, R₁.

\begin{equation}
V_{BG}
\end{equation}

Point G to Point H

Going from point G to point H, we see that there’s no element in this part of the loop of the circuit – meaning there’s no resistance here of any kind to provide a voltage potential. So, the voltage here, V_GH, is zero volts:

\begin{equation}
V_{GH} \; = \; 0V
\end{equation}

Point H to Point A

Now, we complete the loop and end back at point A from point H. The voltage source, our 12V battery, lies between these two points of loop 1. So, the voltage here is -12V.

Why negative 12 volts? Recall, that we are using passive convention – meaning that when we go from point H to point A, we are going from the negative terminal to the positive terminal, which means we write the voltage with a negative value based on the passive convention we’re using.

\begin{equation}
V_{HA} \; = \; -12V
\end{equation}

Summation of Voltage Potentials Around Loop 1

As Kirchhoff’s voltage law states, all we need to do is add all the voltages around loop 1 to see if energy is conserved:

\begin{equation}
V_{AB} \; + \; V_{BG} \; + \; V_{GH} + \; V_{HA} \; = \; 0
\end{equation}

\begin{equation}
0V \; + \; V_{BG} \; + \; 0V \; + \; (-12V) \; = \; 0
\end{equation}

\begin{equation}
0V \; + \; V_{BG} \; + \; 0V \; – \; 12V \; = \; 0
\end{equation}

\begin{equation}
V_{BG} \; – \; 12V \; = \; 0
\end{equation}

\begin{equation}
\boxed{V_{BG} \; = \; 12V}
\end{equation}

So, the voltage across resistor R₁ is 12V. This makes sense, since loop 1 is like the voltage potential across a single element we looked at earlier on in our discussion above.

KVL Around Loop 2

We’ll step around the loop, starting at point A, work our way toward point B, then to point C, down to point F, over to point G, moving to point H, and back to point A to complete the loop.

Point A to Point B

Going from point A to point B, we see that there’s no element in this part of the loop of the circuit – meaning there’s no resistance here of any kind to provide a voltage potential. So, the voltage here, V_AB, is zero volts:

\begin{equation}
V_{AB} \; = \; 0V
\end{equation}

Point B to Point C

Just as it was from going to point B from point A, going to point C from point B provides a path in the circuit that has no circuit element. Therefore, the voltage here, V_BC, is zero volts:

\begin{equation}
V_{BC} \; = \; 0V
\end{equation}

Point C to Point F

Now, working our way down from point C to point F of the circuit, we have a resistor (R₂) with a value of 170Ω. If we knew the value of the current flowing through this part of the circuit, we could calculate the voltage across this resistor using Ohm’s Law, but since we don’t, we’ll just leave it as V_CF, for the voltage across the resistor, R₂.

\begin{equation}
V_{CF}
\end{equation}

Point F to Point G

Going from point F to point G, we see that there’s no element in this part of the loop of the circuit – meaning there’s no resistance here of any kind to provide a voltage potential. So, the voltage here, V_FG, is zero volts:

\begin{equation}
V_{FG} \; = \; 0V
\end{equation}

Point G to Point H

Just as it was from going to point G from point F, going to point H from point G provides a path in the circuit that has no circuit element. Therefore, the voltage here, V_GH, is zero volts:

\begin{equation}
V_{GH} \; = \; 0V
\end{equation}

Point H to Point A

Now, we complete the loop and end back at point A from point H. The voltage source or our 12V battery lies between these two points of loop 2, so the voltage here is -12V.

Recall, that we are using passive convention – meaning that when we go from point H to point A, we are going from the negative terminal to the positive terminal, which means we write the voltage with a negative value based on the passive convention we’re using.

\begin{equation}
V_{HA} \; = \; -12V
\end{equation}

Summation of Voltage Potentials Around Loop 2

As Kirchhoff’s voltage law states, all we need to do is add all the voltages around loop 2 to see if energy is conserved:

\begin{equation}
V_{AB} \; + \; V_{BC} \; + \; V_{CF} + \; V_{FG} \; + \; V_{GH} \; + \; V_{HA} \; = \; 0
\end{equation}

\begin{equation}
0V \; + \; 0V \; + \; V_{CF} + \; 0V \; + \; 0V \; + \; (-12V) \; = \; 0
\end{equation}

\begin{equation}
0V \; + \; 0V \; + \; V_{CF} + \; 0V \; + \; 0V \; – \; 12V \; = \; 0
\end{equation}

\begin{equation}
V_{CF} \; – \; 12V \; = \; 0
\end{equation}

\begin{equation}
\boxed{V_{CF} \; = \; 12V}
\end{equation}

So, the voltage across resistor R₂ is 12V. This makes sense, since loop 2 is like the voltage potential across a single element we looked at earlier on in our discussion above (ignoring resistor R₁).

KVL Around Loop 3

In loop 3, we’d step around the loop, starting at point B, working our way toward point C, then down to point F, over to point G, and back to point B to complete the loop. But, notice that we’ve already found the values of the voltages across R₁ (V_BG = 12V) and R₂ (V_CF = 12V). So, our work here is already done! Sweet!

KVL Around Loop 4

We’ll step around the loop, starting at point B, work our way toward point C, then to point D, down to point E, over to point F, moving to point G, and back to point B to complete the loop.

Point B to Point C

Going from point B to point C, we see that there’s no element in this part of the loop of the circuit – meaning there’s no resistance here of any kind to provide a voltage potential. So, the voltage here, V_BC, is zero volts:

\begin{equation}
V_{BC} \; = \; 0V
\end{equation}

Point C to Point D

Going from point C to point D, we see that there’s no element in this part of the loop of the circuit – meaning there’s no resistance here of any kind to provide a voltage potential. So, the voltage here, V_CD, is zero volts:

\begin{equation}
V_{CD} \; = \; 0V
\end{equation}

Point D to Point E

Now, working our way down from point D to point E of the circuit, we have a resistor (R₃) with a value of 10MΩ. If we knew the value of the current flowing through this part of the circuit, we could calculate the voltage across this resistor using Ohm’s Law, but since we don’t, we’ll just leave it as V_DE, for the voltage across the resistor, R₃.

\begin{equation}
V_{DE}
\end{equation}

Point E to Point F

Going from point E to point F, we see that there’s no element in this part of the loop of the circuit – meaning there’s no resistance here of any kind to provide a voltage potential. So, the voltage here, V_EF, is zero volts:

\begin{equation}
V_{EF} \; = \; 0V
\end{equation}

Point F to Point G

Going from point F to point G, we see that there’s no element in this part of the loop of the circuit – meaning there’s no resistance here of any kind to provide a voltage potential. So, the voltage here, V_FG, is zero volts:

\begin{equation}
V_{FG} \; = \; 0V
\end{equation}

Point G to Point B

We see that going from point G to point B we’ve come back where we started for loop 4. Going from point G to point B we go through resistor R₁. Notice though that we’ve already found the value of the voltage potential across R₁ as being V_BG for when we solved it for loop 1 earlier.

V_BG is the voltage potential from point B to point G though. We need the voltage potential of V_GB, the voltage from point G to point B. All we need to do is flip the sign!

Notice that for loop 4 we go from the negative potential of the resistor R₁, to its positive potential – going from point G to point B – whereas we went from positive to negative potential of R₁ in loop 1, from point B to point G, which gave us a +12V. Going the opposite way would give us -12V:

\begin{equation}
V_{GB} \; = \; -12V
\end{equation}

Summation of Voltage Potentials Around Loop 4

As Kirchhoff’s voltage law states, all we need to do is add all the voltages around loop 4 to see if energy is conserved:

\begin{equation}
V_{BC} \; + \; V_{CD} \; + \; V_{DE} + \; V_{EF} \; + \; V_{FG} \; + \; V_{GB} \; = \; 0
\end{equation}

\begin{equation}
0V \; + \; 0V \; + \; V_{DE} + \; 0V \; + \; 0V \; + \; (-12V) \; = \; 0
\end{equation}

\begin{equation}
0V \; + \; 0V \; + \; V_{DE} + \; 0V \; + \; 0V \; – \; 12V \; = \; 0
\end{equation}

\begin{equation}
V_{DE} \; – \; 12V \; = \; 0
\end{equation}

\begin{equation}
\boxed{V_{DE} \; = \; 12V}
\end{equation}

So, the voltage across resistor R₃ is 12V. This makes sense, since loop 4 is like the voltage potential across a single element we looked at earlier on in our discussion above (ignoring resistor R₂ and knowing R₁ has a potential of 12V across it). Are you noticing a pattern here?

Resistors in Parallel

Notice that for the parallel resistors in our circuit, each has a voltage potential of 12V across their terminals. This is exactly what the study of resistors in parallel tells us – resistors in parallel have the same voltage potential across their terminals.

KVL Around Loop 5

The situation for loop 5 is similar to how loop 3 was for us – the work was already done for us!

In loop 5, we’d step around the loop, starting at point C, working our way toward point D, then down to point E, over to point F, and back to point C to complete the loop. But, notice that we’ve already found the values of the voltages across R₂ (V_CF = 12V) and R₃ (V_DE = 12V). So, our work here is already done. Nice!

KVL Around Loop 6

For our final loop, we’ll step around the loop, starting at point A, work our way toward point B, then toward point C, over to point D, then down to point E, back over to point F, continuing to point G, then to point H, and back to point A to complete the loop.

Point A to Point B

Going from point A to point B, we see that there’s no element in this part of the loop of the circuit – meaning there’s no resistance here of any kind to provide a voltage potential. So, the voltage here, V_AB, is zero volts:

\begin{equation}
V_{AB} \; = \; 0V
\end{equation}

Point B to Point C

Just as it was from going to point B from point A, going to point C from point B provides a path in the circuit that has no circuit element. Therefore, the voltage here, V_BC, is zero volts:

\begin{equation}
V_{BC} \; = \; 0V
\end{equation}

Point C to Point D

Going from point C to point D, we see that there’s no element in this part of the loop of the circuit – meaning there’s no resistance here of any kind to provide a voltage potential. So, the voltage here, V_CD, is zero volts:

\begin{equation}
V_{CD} \; = \; 0V
\end{equation}

Point D to Point E

Now, working our way down from point D to point E of the circuit, we have a resistor (R₃) with a value of 10 mega-ohms. We’ve previously found the voltage potential across resistor R₃, that being the voltage V_DE, with a value of 12V:

\begin{equation}
V_{DE} \; = \; 12V
\end{equation}

Point E to Point F

Going from point E to point F, we see that there’s no element in this part of the loop of the circuit – meaning there’s no resistance here of any kind to provide a voltage potential. So, the voltage here, V_EF, is zero volts:

\begin{equation}
V_{EF} \; = \; 0V
\end{equation}

Point F to Point G

Going from point F to point G, we see that there’s no element in this part of the loop of the circuit – meaning there’s no resistance here of any kind to provide a voltage potential. So, the voltage here, V_FG, is zero volts:

\begin{equation}
V_{FG} \; = \; 0V
\end{equation}

Point G to Point H

Just as it was from going to point G from point F, going to point H from point G provides a path in the circuit that has no circuit element. Therefore, the voltage here, V_GH, is zero volts:

\begin{equation}
V_{GH} \; = \; 0V
\end{equation}

Point H to Point A

Now, we complete the loop and end back at point A from point H. The voltage source or our 12V battery lies between these two points of loop 1, so the voltage here is -12V.

Recall, that we are using passive convention – meaning that when we go from point H to point A, we are going from the negative terminal to the positive terminal, which means we write the voltage with a negative value based on the passive convention we’re using.

\begin{equation}
V_{HA} \; = \; -12V
\end{equation}

Summation of Voltage Potentials Around Loop 6

As Kirchhoff’s voltage law states, all we need to do is add all the voltages around loop 6 to see if energy is conserved:

\begin{equation}
V_{AB} \; + \; V_{BC} \; + \; V_{CD} + \; V_{DE} \; + \; V_{EF} \; + \; V_{FG} \; + \; V_{GH} \; + \; V_{HA} \; = \; 0
\end{equation}

\begin{equation}
0V \; + \; 0V \; + \; 0V + \; V_{DE} \; + \; 0V \; + \; 0V \; + \; 0V \; + \; (-12V) \; = \; 0
\end{equation}

\begin{equation}
0V \; + \; 0V \; + \; 0V + \; V_{DE} \; + \; 0V \; + \; 0V \; + \; 0V \; – \; 12V \; = \; 0
\end{equation}

\begin{equation}
V_{DE} \; – \; 12V \; = \; 0
\end{equation}

\begin{equation}
\boxed{V_{DE} \; = \; 12V}
\end{equation}

So, the voltage across resistor R₃ is 12V, but we knew that already because we had calculated it for loop 4. This is also what we should have expected being that R₃ is a resistor in parallel with the other resistors before it, each having a voltage potential across them of 12V. This also makes sense, because loop 6 is like the voltage potential across a single element we looked at earlier on in our discussion above (ignoring resistors R₁ and R₂).

---

Conclusion

As you can see from our example circuits above that having the knowledge of Kirchhoff’s Voltage Law is a very handy tool to have in your mental toolbox. KVL tells us that whatever voltage potentials there are around a closed loop circuit, when you add them up, they should equal to zero.

KVL works with any type of circuit as long as you are following the rules of the law and are summing voltage potentials around a closed loop series path. If you’re doing it right, the algebraic sum of the voltages will always be equal to zero.

Now that you’re enlightened with the concept of KVL, next it’s time to learn about KCL or Kirchhoff’s Current Law. If you haven’t already, check out our post on Kirchhoff’s Current Law (KCL) here.

We hope that you’ve enjoyed this lesson on KVL and that it was beneficial to you. Let us know in the comments if our explanations are helpful. Remember, stay motivated and keep at it!