## Kirchhoff's Law

Ohm's Law is sufficient to solve series and parallel circuits, but more complicated circuits like bridges and T‐networks can't be solved by Ohm's Law alone. In 1845, physicist Gustav Kirchhoff formulated a pair of laws dealing with the conservation of current and energy within an electrical circuit. These two laws together are called Kirchhoff's Law or Kirchhoff's Circuit Law. Kirchhoff's Law applies to DC circuits and, in some limited circumstances, to AC circuits.

Before delving into Kirchhoff's Law, we need to define a few things:

- Node
- A junction where two or more components meet.
- Principle Node
- A big honkin' node where three or more components meet.
- Branch
- Any path in a circuit having a node at each end and containing at least one component (such as a resistor or battery) but no other nodes.
- Loop
- A closed path within a circuit where no component is encountered more than once.
- Mesh
- A simple path within a circuit with no branches.

## 1. Kirchhoff's Current Law

Kirchhoff's Current Law is a statement of the principle of conservation of electric charge. The algebraic sum of the currents entering and leaving any point in a circuit must equal zero. This means if you add up all the currents flowing into a point and compare that value to the sum of all the currents flowing out of that point, the two values must be equal.

In figure 6 below, the values of I_{1} and I_{2} entering point *N* must equal the value of I_{3} leaving *N*. If I_{1} equals 5 amps and I_{2} equals 3 amps, then the value of I_{3} must equal 8 amps. I_{1} + I_{2} - I_{3} = 0.

## 2. Kirchhoff's Voltage Law

Kirchhoff's Voltage Law is also called the Conservation of Energy Law. In a closed loop within a circuit, the algebraic sum of all voltages within the loop must equal zero.

In figure 7 below, all the voltages must equal zero. To verify this, pick an arbitrary point on the circuit (point A for example) and an arbitrary direction (counter-clockwise). From point A, move counter-clockwise around the loop. We first encounter V_{1}. Since we encountered the positive terminal of V_{1} before the negative terminal, we record the voltage as a positive number. As we continue counter-clockwise, we encounter resister R_{2}. We reached the negative terminal of R_{2} first, so we record the voltage drop across R_{2} as a negative number. Next is resistor R_{1} (another negative voltage drop). Finally, we arrive back at point A where we started. The equation of Kirchhoff's Voltage Law would be: V_{1} - V_{R2} - V_{R1} = 0.

Had we moved clockwise from point A instead of counter-clockwise, the final equation would have been: V_{R1} + V_{R2} - V_{1} = 0. As you can see, the absolute magnitude of each voltage is the same regardless of the direction around the loop. Only the polarity (positive or negative) differs based on the direction you choose.

## Methods of Circuit Analysis Using Kirchhoff's Law

There are three main methods of solving circuits using Kirchhoff's Law:

## Example: Solve a Circuit by the Branch Currents Method

The Branch Currents Method can be a bit long and tedious, but it serves to illustrate Kirchhoff's Law quite well. Once you understand the Branch Currents Method, the other two methods are easy.

For the circuit in figure 8 below, we wish to determine the current and voltages across each resistor. Since there are voltage sources in two different branches, Ohm's Law alone is insufficient to solve this. We must use Kirchhoff's Law.

Since we don't know the cumulative effect of the two voltage sources, we can't be sure of the direction of current in each branch. We will need to make an educated guess. If it turns out to be incorrect, that's all right ‐ the equation results will point out our error.

We will assume that current I_{1} flows from the negative terminal of voltage source V_{1} to resistor R_{1} then to node *A*. Similarly, current I_{2} flows from the negative terminal of voltage source V_{2} towards resistor R_{2} then to node *A*. From node *A*, a current I_{3} flows through resistor R_{3} until it reaches node *B*. Here the current divides and returns to the two voltage sources.

## Using Kirchhoff's Current Law

There are two principle nodes in the figure 8 circuit, labeled *A* and *B*. At node *A*, two currents, I_{1} and I_{2}, enter while one current, I_{3}, leaves the node. From Kirchhoff's Current Law we derive the equation

.`I _{1}` +

`I`-

_{2}`I`= 0

_{3}Similarly, at node *B*, one current, I_{3}, enters the node while two currents, I_{1} and I_{2}, leave. This gives us:

. We can use either equation to express the relationship between I`I _{3}` -

`I`-

_{1}`I`= 0

_{2}_{1}, I

_{2}, and I

_{3}as

`I`_{3} = `I`_{1} + `I`_{2}

. We will refer to this equation as the current equation for discussion purposes.## Using Kirchhoff's Voltage Law

To derive equations for the voltage drops across each resistor, we need to first define some loops. There are two inside loops in figure 8 plus an outside loop that follows the circumference of the circuit. It turns out that we only need two loop equations, so we will ignore that outside loop.

The first inside loop leaves V_{1}, moves through R_{1} to node *A*, then through R_{3} to node *B* before returning to V_{1}. We will call this Loop 1. The second inside loop leaves V_{2} traveling through R_{2} to node *A*. From here it moves through R_{3} to node *B* before returning to V_{2}. This will be Loop 2. With these loops, we can now use Kirchhoff's Voltage Law to formulate the equations for the voltages.

To formulate the loop equation for Loop 1, we begin at Node *B* and move around the loop in the assumed direction of the I_{1} current flow (clockwise). Our equation becomes:

.`V _{1}` -

`V`-

_{R1}`V`= 0

_{R3}To formulate the loop 2 equation, we begin at node *B* and move counter-clockwise (the assumed direction of current), giving us:

.`V _{2}` -

`V`-

_{R2}`V`= 0

_{R3}The next step is to formulate the voltage drop equations:

`V _{R1}` =

`I`

_{1}`R`

_{1}`V`_{R2} = `I`_{2}`R`_{2}

`V`_{R3} = `I`_{3}`R`_{3}

We can now rewrite the loop equations using the voltage drop equations and known component values and simplifying. Note: we can also use the current equation (`I`

) formulated above to substitute _{3} = I_{1} + I_{2}

for `I _{1}` +

`I`

_{2}`I`_{3}

, eliminating a variable. We will also eliminate the volt and ohm symbols for clarity.### Loop 1:

`V _{1}` -

`V`-

_{R1}`V`= 0

_{R3}`V`_{1} - `I`_{1}`R`_{1} - `I`_{3}`R`_{3} = 0

`130 - 20``I`_{1} - 10`I`_{3} = 0

`13 - 2``I`_{1} - `I`_{3} = 0

`13 - 2``I`_{1} - (`I`_{1} + `I`_{2}) = 0

`3``I`_{1} + `I`_{2} = 13

### Loop 2:

`V _{2}` -

`V`-

_{R2}`V`= 0

_{R3}`V`_{2} - `I`_{2}`R`_{2} - `I`_{3}`R`_{3} = 0

`20 - 5``I`_{2} - 10`I`_{3} = 0

`4 - ``I`_{2} - 2`I`_{3} = 0

`4 - ``I`_{2} - 2(`I`_{1} + `I`_{2}) = 0

`2``I`_{1} + 3`I`_{2} = 4

We now have a solvable problem consisting of two simplified loop equations with two variables. That's why we didn't need the outside loop equation mentioned earlier. If you remember your algebra, you can manipulate one or both of the loop equations so that when you add them together, one variable is eliminated. We will do it the longer way, rewriting the Loop 1 equation to solve for `I _{2}` as follows:

`I`_{2} = 13 - 3`I`_{1}

.Now, substitute this value for `I _{2}` in the Loop 2 equation to get:

`2``I`_{1} + 3(13 - 3`I`_{1}) = 4

`2``I`_{1} + 39 - 9`I`_{1} = 4

`7``I`_{1} = 35

`I`_{1} = 5 amps

Knowing `I _{1}`, we can now solve the Loop 1 equation for

`I`.

_{2}`I`_{2} = 13 - 3`I`_{1} = 13 - 3(5)

= -2 ampsNow, with values for `I _{1}` and

`I`, we can use the current equation derived above to solve for

_{2}`I`.

_{3}`I`_{3} = `I`_{1} + `I`_{2} = 5 amps + (-2) amps = 3 amps

We know all the branch currents now, so how about those voltage drops?

`V _{R1}` =

`I`

_{1}`R`= 5 × 20 = 100 volts

_{1}`V`_{R2} = `I`_{2}`R`_{2} = -2 × 5 = -10 volts

`V`_{R3} = `I`_{3}`R`_{3} = 3 × 10 = 30 volts

Finally, what about that negative current for `I _{2}` and the negative voltage for

`V`? Well, the negative signs means that our original assumption about the direction of

_{R2}`I`was incorrect. The current is actually going in the opposite direction. Our bad assumption also led to a reverse polarity for the

_{2}`V`voltage drop. The magnitudes are correct, just reverse the direction of

_{R2}`I`and you have a final solution!

_{2}