Kirchhoff's laws (article) | Circuits | Khan Academy (2023)

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.

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, $\text{mA}$mAstart text, m, A, end text. One of the currents, $\blueD i$istart color #11accd, i, end color #11accd, is not known.

Problem 1: What is $i$ii?

Here's another example, this time with variable names instead of numerical values. This node happens to have $5$55 branches. Each branch might (or might not) carry a current, labeled $i_1 \,\text{to} \, i_5$i1​toi5​i, start subscript, 1, end subscript, start text, t, o, end text, i, start subscript, 5, end subscript.

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 ${i_1}$i1​i, start subscript, 1, end subscript.
Where does it go?

The first thing ${i_1}$i1​i, start subscript, 1, end subscript does is flow into the node (represented by the black dot).

Then what?

Here's two things ${i_1}$i1​i, start subscript, 1, end subscript can't do: The flowing charge in ${i_1}$i1​i, start subscript, 1, end subscript can't stay inside the node. (The node does not have a place to store charge). And ${i_1}$i1​i, start subscript, 1, end subscript'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,

If ${i_1}$i1​i, start subscript, 1, end subscript 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 $-$−minusnegative sign.

This observation about currents flowing in a node is nicely captured in general form as 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,

Currents are in milliamps, $\text{mA}$mAstart text, m, A, end text.

Problem 2: What is $i_5$i5​i, start subscript, 5, end subscript?

Problem 3: What is $i_3$i3​i, start subscript, 3, end subscript in this distributed node?

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, $\blueD i$istart color #11accd, i, end color #11accd, through all five elements. To find $\blueD i$istart color #11accd, i, end color #11accd, the four series resistors can be reduced to a single equivalent resistor:

Using Ohm's Law, the current is:

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:

$v\phantom{_{\text{R1}}} = \blueD i\,\text R$vR1​=iRv, empty space, equals, start color #11accd, i, end color #11accd, start text, R, end text
$v_{\text{R1}} = 20\,\text{mA} \cdot 100\,\Omega = +2\,\text{V}$vR1​=20mA⋅100Ω=+2Vv, start subscript, start text, R, 1, end text, end subscript, equals, 20, start text, m, A, end text, dot, 100, \Omega, equals, plus, 2, start text, V, end text
$v_{\text{R2}} = 20\,\text{mA} \cdot 200\,\Omega = +4\,\text{V}$vR2​=20mA⋅200Ω=+4Vv, start subscript, start text, R, 2, end text, end subscript, equals, 20, start text, m, A, end text, dot, 200, \Omega, equals, plus, 4, start text, V, end text
$v_{\text{R3}} = 20\,\text{mA} \cdot 300\,\Omega = +6\,\text{V}$vR3​=20mA⋅300Ω=+6Vv, start subscript, start text, R, 3, end text, end subscript, equals, 20, start text, m, A, end text, dot, 300, \Omega, equals, plus, 6, start text, V, end text
$v_{\text{R4}} = 20\,\text{mA} \cdot 400\,\Omega = +8\,\text{V}$vR4​=20mA⋅400Ω=+8Vv, start subscript, start text, R, 4, end text, end subscript, equals, 20, start text, m, A, end text, dot, 400, \Omega, equals, plus, 8, start text, V, end text

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, $\greenE{\text a}$astart color #0d923f, start text, a, end text, end color #0d923f to $\greenE{\text e}$estart color #0d923f, start text, e, end text, end color #0d923f, so we can talk about them.)

Let's do a quick check. Add up the voltages across the resistors,

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.

Step 1: Pick a starting node.

Step 2: Pick a direction to travel around the loop (clockwise or counterclockwise).

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 $+$+plus, then there will be a voltage drop going through the element. Subtract the element voltage.
• If the sign is $-$−minus, then there will be a voltage rise going through the element. Add the element voltage.

• [Alternative rule set]

Step 4: Continue around the loop until you reach the starting point, including element voltages all the way around.

Let's follow the procedure step-by-step.

1. Start at the lower left at node $\greenE{\text a}$astart color #0d923f, start text, a, end text, end color #0d923f.

2. Walk clockwise.

3. The first element we come to is the voltage source. The first voltage sign we encounter is a $-$−minus 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.
   $$

$v_{loop} = +20\,\text V$vloop​=+20Vv, start subscript, l, o, o, p, end subscript, equals, plus, 20, start text, V, end text going through the voltage source, to node $\greenE{\text b}$bstart color #0d923f, start text, b, end text, end color #0d923f.

The next element we encounter is the $100\,\Omega$100Ω100, \Omega resistor. Its nearest voltage sign is $+$+plus. Consult the procedure again, and this time we subtract the element voltage from the growing sum.

$v_{loop} = + 20\,\text V - 2\,\text V$vloop​=+20V−2Vv, start subscript, l, o, o, p, end subscript, equals, plus, 20, start text, V, end text, minus, 2, start text, V, end text going through the $100\,\Omega$100Ω100, \Omega resistor, to node $\greenE{\text c}$cstart color #0d923f, start text, c, end text, end color #0d923f.

Keep going. Next we visit the $200\,\Omega$200Ω200, \Omega resistor, and again we first encounter a $+$+plus sign, so we subtract this voltage.

$v_{loop} = + 20\,\text V - 2\,\text V - 4\,\text V$vloop​=+20V−2V−4Vv, start subscript, l, o, o, p, end subscript, equals, plus, 20, start text, V, end text, minus, 2, start text, V, end text, minus, 4, start text, V, end text going through the $200\,\Omega$200Ω200, \Omega resistor, to node $\greenE{\text d}$dstart color #0d923f, start text, d, end text, end color #0d923f.

We complete the loop with the addition of two more elements,

$v_{loop} = + 20\,\text V - 2\,\text V - 4\,\text V - 6\,\text V \,$vloop​=+20V−2V−4V−6Vv, start subscript, l, o, o, p, end subscript, equals, plus, 20, start text, V, end text, minus, 2, start text, V, end text, minus, 4, start text, V, end text, minus, 6, start text, V, end text through the $300\,\Omega$300Ω300, \Omega resistor, to node $\greenE{\text e}$estart color #0d923f, start text, e, end text, end color #0d923f.

$v_{loop} = + 20\,\text V - 2\,\text V - 4\,\text V - 6\,\text V - 8\,\text V\,$vloop​=+20V−2V−4V−6V−8Vv, start subscript, l, o, o, p, end subscript, equals, plus, 20, start text, V, end text, minus, 2, start text, V, end text, minus, 4, start text, V, end text, minus, 6, start text, V, end text, minus, 8, start text, V, end text after the $400\,\Omega$400Ω400, \Omega resistor.

(Check the circuit diagram, make sure I got the last two $-$−minus signs correct.)

4. Done. We made it back home to node $\greenE{\text a}$astart color #0d923f, start text, a, end text, end color #0d923f. What does this expression for $v_{loop}$vloop​v, start subscript, l, o, o, p, end subscript add up to?
   $$

The sum of voltages going around the loop is $0$00. 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 $\greenE{\text a}$astart color #0d923f, start text, a, end text, end color #0d923f in the lower left corner. Our walk goes clockwise around the loop (an arbitrary choice, either way works).

Starting at node $\greenE{\text a}$astart color #0d923f, start text, a, end text, end color #0d923f, going up, we first encounter a minus sign on the voltage source, which says there is going to be a voltage rise of $v_{ab}$vab​v, start subscript, a, b, end subscript volts going through the voltage source. Because it is a voltage rise, this element voltage gets a $+$+plus sign when we include it in the loop sum.

Continue around the loop from node $\greenE{\text b}$bstart color #0d923f, start text, b, end text, end color #0d923f to $\greenE{\text c}$cstart color #0d923f, start text, c, end text, end color #0d923f to $\greenE{\text d}$dstart color #0d923f, start text, d, end text, end color #0d923f to $\greenE{\text e}$estart color #0d923f, start text, e, end text, end color #0d923f, and finish back home at node $\greenE{\text a}$astart color #0d923f, start text, a, end text, end color #0d923f. Append resistor voltages to the loop sum along the way. The polarity labels on all the resistors are arranged so we encounter a $-$−minus sign as we approach each resistor. So the resistor voltages all go into the loop sum with a $+$+plus sign. The final loop sum looks like this:

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,

This observation about voltages around a loop is nicely captured in general form as 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,

where $n$nn 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.

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.

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.

Problem 4: What is $v_{R3}$vR3​v, start subscript, R, 3, end subscript?

Reminder: Check the first sign of each element voltage as you walk around the loop.

We were introduced to two new friends.

Kirchhoff's Current Law for branch currents at a node,

Kirchhoff's Voltage Law for element voltages around a loop,

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.