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# Kirchhoff's laws

Kirchhoff's Laws describe current in a node and voltage around a loop. These two laws are the foundation of advanced circuit analysis. Written by Willy McAllister.
Kirchhoff's Laws for current and voltage lie at the heart of circuit analysis. With these two laws, plus the equations for individual component (resistor, capacitor, inductor), we have the basic tool set we need to start analyzing circuits.
This article assumes you are familiar with the definitions of node, distributed node, branch, and loop.
You may want to have a pencil and paper nearby to work the example problems.

## Currents into a node

Try to reason through this example by yourself, before we talk about the theory. The schematic below shows four branch currents flowing in and out of a distributed node. The various currents are in milliamps, $\text{mA}$. One of the currents, $i$, is not known.
Problem 1: What is $i$?

Here's another example, this time with variable names instead of numerical values. This node happens to have $5$ branches. Each branch might (or might not) carry a current, labeled ${i}_{1}\phantom{\rule{0.167em}{0ex}}\text{to}\phantom{\rule{0.167em}{0ex}}{i}_{5}$.
All the arrows are drawn pointing in. This choice of direction is arbitrary. Arrows pointing inward is as good a choice as any at this point. The arrows establish a reference direction for what we choose to call a positive current.
Look at branch current ${i}_{1}$.
Where does it go?
The first thing ${i}_{1}$ does is flow into the node (represented by the black dot).
Then what?
Here's two things ${i}_{1}$ can't do: The flowing charge in ${i}_{1}$ can't stay inside the node. (The node does not have a place to store charge). And ${i}_{1}$'s charge can't jump off the wires into thin air. Charge just doesn't do that under normal circumstances.
What's left?: The current has to flow out of the node through one or more of the other branches.
For our example node, we would write this as,
${i}_{1}+{i}_{2}+{i}_{3}+{i}_{4}+{i}_{5}=0$
If ${i}_{1}$ is a positive current flowing into the node, then one or more of the other currents must be flowing out. Those outgoing currents will have a $-$negative sign.
This observation about currents flowing in a node is nicely captured in general form as Kirchhoff's Current Law.

## Kirchhoff's Current Law

Kirchhoff's Current Law says that the sum of all currents flowing into a node equals the sum of currents flowing out of the node. It can be written as,
$\sum {i}_{in}=\sum {i}_{out}$

### Kirchhoff's Current Law - concept checks

Currents are in milliamps, $\text{mA}$.
Problem 2: What is ${i}_{5}$?

Problem 3: What is ${i}_{3}$ in this distributed node?

## Voltage around a loop

Below is a circuit with four resistors and a voltage source. We will solve this from scratch using Ohm's Law. Then we will look at the result and make some observations. The first step in solving the circuit is to compute the current. Then we will compute the voltage across the individual resistors.
We recognize this as a series circuit, so there is only one current flowing, $i$, through all five elements. To find $i$, the four series resistors can be reduced to a single equivalent resistor:
${R}_{series}=100+200+300+400=1000\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }$
Using Ohm's Law, the current is:
$i=\frac{V}{{R}_{series}}=\frac{20\phantom{\rule{0.167em}{0ex}}\text{V}}{1000\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }}=0.020\phantom{\rule{0.167em}{0ex}}\text{A}=20\phantom{\rule{0.167em}{0ex}}\text{mA}$
Now we know the current. Next we find the voltages across the four resistors. Go back to the original schematic and add voltage labels to all five elements:
Apply Ohm's Law four more times to find the voltage across each resistor:
$v\phantom{{}_{\text{R1}}}=i\phantom{\rule{0.167em}{0ex}}\text{R}$
${v}_{\text{R1}}=20\phantom{\rule{0.167em}{0ex}}\text{mA}\cdot 100\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }=+2\phantom{\rule{0.167em}{0ex}}\text{V}$
${v}_{\text{R2}}=20\phantom{\rule{0.167em}{0ex}}\text{mA}\cdot 200\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }=+4\phantom{\rule{0.167em}{0ex}}\text{V}$
${v}_{\text{R3}}=20\phantom{\rule{0.167em}{0ex}}\text{mA}\cdot 300\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }=+6\phantom{\rule{0.167em}{0ex}}\text{V}$
${v}_{\text{R4}}=20\phantom{\rule{0.167em}{0ex}}\text{mA}\cdot 400\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }=+8\phantom{\rule{0.167em}{0ex}}\text{V}$
We know the current and all voltages. The circuit is now solved.
We can write the voltages for the resistors and the source on the schematic. These five voltages are referred to as element voltages. (The circuit nodes get names, $\text{a}$ to $\text{e}$, so we can talk about them.)
Let's do a quick check. Add up the voltages across the resistors,
$2\phantom{\rule{0.167em}{0ex}}\text{V}+4\phantom{\rule{0.167em}{0ex}}\text{V}+6\phantom{\rule{0.167em}{0ex}}\text{V}+8\phantom{\rule{0.167em}{0ex}}\text{V}=20\phantom{\rule{0.167em}{0ex}}\text{V}$
The individual resistor voltages add up to the source voltage. This makes sense, and confirms our calculations.
Now we add up the voltages again, using a slightly different procedure: by "going around the loop." There's no new science here, we are just rearranging the same computation.

### Procedure: Add element voltages around a loop

Step 1: Pick a starting node.
Step 2: Pick a direction to travel around the loop (clockwise or counterclockwise).
Step 3: Walk around the loop.
Include element voltages in a growing sum according to these rules:
• When you encounter a new element, look at the voltage sign as you enter the element.
• If the sign is $+$, then there will be a voltage drop going through the element. Subtract the element voltage.
• If the sign is $-$, then there will be a voltage rise going through the element. Add the element voltage.
Step 4: Continue around the loop until you reach the starting point, including element voltages all the way around.

### Apply the loop procedure

1. Start at the lower left at node $\text{a}$.
2. Walk clockwise.
1. The first element we come to is the voltage source. The first voltage sign we encounter is a $-$ minus sign, so there is going to be a voltage rise going through this element. Consulting the procedure step 3., we initialize the loop sum by adding the source voltage.

${v}_{loop}=+20\phantom{\rule{0.167em}{0ex}}\text{V}$ going through the voltage source, to node $\text{b}$.
The next element we encounter is the $100\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }$ resistor. Its nearest voltage sign is $+$. Consult the procedure again, and this time we subtract the element voltage from the growing sum.
${v}_{loop}=+20\phantom{\rule{0.167em}{0ex}}\text{V}-2\phantom{\rule{0.167em}{0ex}}\text{V}$ going through the $100\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }$ resistor, to node $\text{c}$.
Keep going. Next we visit the $200\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }$ resistor, and again we first encounter a $+$ sign, so we subtract this voltage.
${v}_{loop}=+20\phantom{\rule{0.167em}{0ex}}\text{V}-2\phantom{\rule{0.167em}{0ex}}\text{V}-4\phantom{\rule{0.167em}{0ex}}\text{V}$ going through the $200\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }$ resistor, to node $\text{d}$.
We complete the loop with the addition of two more elements,
${v}_{loop}=+20\phantom{\rule{0.167em}{0ex}}\text{V}-2\phantom{\rule{0.167em}{0ex}}\text{V}-4\phantom{\rule{0.167em}{0ex}}\text{V}-6\phantom{\rule{0.167em}{0ex}}\text{V}\phantom{\rule{0.167em}{0ex}}$ through the $300\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }$ resistor, to node $\text{e}$.
${v}_{loop}=+20\phantom{\rule{0.167em}{0ex}}\text{V}-2\phantom{\rule{0.167em}{0ex}}\text{V}-4\phantom{\rule{0.167em}{0ex}}\text{V}-6\phantom{\rule{0.167em}{0ex}}\text{V}-8\phantom{\rule{0.167em}{0ex}}\text{V}\phantom{\rule{0.167em}{0ex}}$ after the $400\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }$ resistor.
(Check the circuit diagram, make sure I got the last two $-$ signs correct.)
1. Done. We made it back home to node $\text{a}$. What does this expression for ${v}_{loop}$ add up to?

${v}_{loop}=+20\phantom{\rule{0.167em}{0ex}}\text{V}-2\phantom{\rule{0.167em}{0ex}}\text{V}-4\phantom{\rule{0.167em}{0ex}}\text{V}-6\phantom{\rule{0.167em}{0ex}}\text{V}-8\phantom{\rule{0.167em}{0ex}}\text{V}=0$
The sum of voltages going around the loop is $0$. The starting and ending node is the same, so the starting and ending voltage is the same. On your "walk" you went up voltage rises and down voltage drops, and they all cancel out when you get back to where you started. This happens because electric force is conservative. There isn't a net gain or loss of energy if you return to the same place you started.
We'll do another example, this time with variable names instead of numerical values. The following familiar schematic is labeled with voltages and node names. The voltage polarity on the resistors is arranged in a way you might not expect, with all the arrows pointing in the same direction around the loop. This reveals a cool property of loops.
Let's take a walk around the loop, adding up voltages as we go. Our starting point is node $\text{a}$ in the lower left corner. Our walk goes clockwise around the loop (an arbitrary choice, either way works).
Starting at node $\text{a}$, going up, we first encounter a minus sign on the voltage source, which says there is going to be a voltage rise of ${v}_{ab}$ volts going through the voltage source. Because it is a voltage rise, this element voltage gets a $+$ sign when we include it in the loop sum.
Continue around the loop from node $\text{b}$ to $\text{c}$ to $\text{d}$ to $\text{e}$, and finish back home at node $\text{a}$. Append resistor voltages to the loop sum along the way. The polarity labels on all the resistors are arranged so we encounter a $-$ sign as we approach each resistor. So the resistor voltages all go into the loop sum with a $+$ sign. The final loop sum looks like this:
$+{v}_{\text{ab}}+{v}_{\text{R1}}+{v}_{\text{R2}}+{v}_{\text{R3}}+{v}_{\text{R4}}$
What does this add up to? Let's reason it out.
The loop starts and ends at the same node, so the starting and ending voltages are identical. We went around the loop, adding voltages, and we end up back at the same voltage. That means the voltages have to add to zero. For our example loop, we would write this as,
${v}_{\text{ab}}+{v}_{\text{R1}}+{v}_{\text{R2}}+{v}_{\text{R3}}+{v}_{\text{R4}}=0$
This observation about voltages around a loop is nicely captured in general form as Kirchhoff's Voltage Law.

## Kirchhoff's Voltage Law

Kirchhoff's Voltage Law: The sum of voltages around a loop is zero.
Kirchhoff's Voltage Law can be written as,
$\sum _{n}{v}_{n}=0$
where $n$ counts the element voltages around the loop.
You can also state Kirchhoff's Voltage Law another way: The sum of voltage rises equals the sum of voltage drops around a loop.
$\sum {v}_{rise}=\sum {v}_{drop}$
Kirchhoff's Voltage Law has some nice properties:
• You can trace a loop starting from any node. Walk around the loop and end up back at the starting node, the sum of voltages around the loop adds up to zero.
• You can go around the loop in either direction, clockwise or counterclockwise. Kirchhoff's Voltage Law still holds.
• If a circuit has multiple loops, Kirchhoff's Voltage Law is true for every loop.

#### Voltages all positive?

If you are wondering: how can the element voltages all be positive if they have to add up to zero? It's okay. The voltage arrows and polarity signs are just reference directions for voltage. When the circuit analysis is complete, one or more of the element voltages around the loop will be negative with respect to its voltage arrow. The signs of the actual voltages always sort themselves out during calculations.

### Kirchhoff's Voltage Law - concept check

Problem 4: What is ${v}_{R3}$?
Reminder: Check the first sign of each element voltage as you walk around the loop.

## Summary

We were introduced to two new friends.
Kirchhoff's Current Law for branch currents at a node,
$\sum _{n}{i}_{n}=0$
Kirchhoff's Voltage Law for element voltages around a loop,
$\sum _{n}{v}_{n}=0$
Our new friends sometimes go by their initials, KCL and KVL.
And we learned it's important to pay close attention to voltage and current signs if we want correct answers. This is a tedious process that requires attention to detail. It is a core skill of a good electrical engineer.

## Want to join the conversation?

• what is the algebraic sum?
• It's a fancy way of saying "total summation".
Example: the algebraic sum of 8, -1, -2 is 5.
• How is this affected if there are multiple batteries, and therefore multiple currents? What if the resistors are in parallel?
• Kirchhoff's Laws work for every circuit, no matter the number of batteries or resistor configuration. KCL tells you about the sum of currents at each specific node in the circuit. KVL tells you about the sum of voltage rises/drops around every loop of a circuit. KVL and KCL aren't fooled by multiple voltage or current sources, or parallel resistors. They always work.
• When I see the words "voltage on each resistor" - what does that mean? that voltage is getting eaten up or used by the resistor? Is it wasting energy?
• The phrases "Voltage on" and "Voltage across" mean the same thing. One end of the resistor is at a higher potential than the other end. The difference in potential is called the "voltage across" the resistor. It is the nature of resistors that when they have a voltage across them, a current flows. When current flows, there is energy being dissipated. If the circuit is doing something useful, then we say "Nice job using the energy! I like this song." If the circuit is not useful, then we might say "What a waste of energy, how sad".
• How do we know that in kirchoff's voltage law we have to add or subtract the voltage
• If the first sign voltage meets in a loop is negative then it's a positive voltage and v.v
(1 vote)
• what is voltage?
• It is the electric potential between two points. You probably know potential energy, like when a bowling ball is sitting on a skyscraper it has the potential energy stored from gravity because it could fall. Voltage is the same thing but with electrons (negatively charged) that have potential because they will try and move to a postive (or "less" negative) area.
• Where does the current flow when it enters the node.? What if all the current (arrows) are pointing inward, how is the sum of current zero in that case?
• good question.

If all the arrows are drawn as going into the node, then one of the currents going 'in' will have a negative value in your calculations.

Of course, there must be just as much going in as out
• is kirchhoff's law applicable for ac circuits?
• Hello Drakshaspodia,

Yes, both KVL and KCL apply to AC circuits.

You will find AC circuit are more interesting than DC circuits as capacitors and inductors change impedance as frequency changes. Think of impedance as the resistance to an AC signal.

Regards,

APD
• what do we understand from the arrows in the voltage diagrams
what is voltage polarity
• Hello

Here is how you make the arrows.

1) Assume a direction of current flow.

2) Walk around the loop in the chosen direction of current flow. For sake of argument lets start at the 200 Ω resistor.

2) As you walk along the circuit you will encounter a component. Ask yourself if the voltage as you enter this component is greater than or less than the exit terminal.

3) If the voltage on the terminal you entered is greater than the exit terminal then the entrance terminal gets a “+”. And the other terminal gets a “-.”

For a resister you don’t need to think very hard. If you are traveling in the direction of the current then the voltage on the terminal you enter will be higher than the exit.

The battery is more interesting. As we travel we will enter the battery on the negative terminal. When we look across we see the positive terminal. Consequently the entrance terminal gets a ‘-’ and the exit terminal gets a “+” symbol.

One last point - if you were tr travel all the way around this loop the sum of the voltages is equal to zero. This property is essential to you study of electronics. We give it the special name KCL (Kirchhoff's Voltage Law).

Regards,

APD

P.S. You could have assumed the opposite direction for the current and everything would have mathematically worked out...