Kirchhoff’s Laws

Kirchhoff's two circuit laws are commonly used in the analysis of voltages and currents in electrical circuits. Kirchhoff's Current Law (KCL) equates the currents acting at any point in a circuit, while Kirchhoff's Voltage Law (KVL) considers the potential differences around any closed circuit loop.

These theorems were first described in 1845 by (1824–1887).

Kirchhoff's Current Law (KCL)

Kirchhoff's current law may be used to analyse the currents flowing into and out of a circuit junction or node. It states that, at any instant, the sum of the currents entering a junction is equal to the sum of the currents leaving that junction.

If we assume that currents flowing into a node are positive and currents flowing away are negative, then, for the above example, we can say that I1 + I2 = I3 + I4 + I5, or I1 + I2I3I4I5 = 0.

In general, for a system with k currents acting at a node, sequentially numbered I1, I2, ... Ik then

A principal node is a point in a circuit where three or more conductors meet, while a secondary node is a connection of only two wires.

Kirchhoff's Voltage Law (KVL)

Kirchhoff's voltage law states that the algebraic sum of the EMFs and potential differences acting around a closed circuit loop is zero.

In this example, the EMF produced by the battery equal the sum of the potential differences across the four resistors. Thus E1 = V1 + V2 + V3 + V4, or E1V1V2V3V4 = 0. The potential differences may also be expressed as the product of current and resistance, so the original expression can be written as
E1 = I R1 + I R2 + I R3 + I R4, or E1 = I (R1 + R2 + R3 + R4).

An electromotive force (EMF), such as that produced by a battery or power supply will tend to produce a conventional current which flows from the positive terminal to the negative (in the absence of an opposing EMF). Conversely, a resistor will produce a potential difference which 'opposes' the current flowing through the resistor.

In general, for a closed loop with k voltages (either EMFs or potential differences), numbered V1, V2, ... Vk, the algebraic sum of these voltages may be expressed as:

Worked Example 1

Use Kirchhoff's laws to find all currents in the above circuit. Hence find the voltage at node B.

The first step is to 'choose' the assumed directions for currents I1 and I2 flowing in loops ABEF and BCDE respectively.

Currents I1 and I2 both flow into node B, hence from KCL, the current flowing downwards through R3 is given by (I1 + I2).

If an initial assumed current direction turns out to be incorrect, the corresponding calculated value will be negative.

The basic method is to apply KVL to each loop, obtaining a pair of simultaneous equations, which must then be solved to find the two unknown currents I1 and I2.

Applying KVL clockwise around loop ABEF:

E1 - I1 R1R3 (I1 + I2) = 0

Grouping currents and rearranging to make E1 the subject gives

E1 = I1 (R1 + R3) + I2 R3

Substituting values gives the first simultaneous equation.

12 = 20 I1 + 16 I2 . . . (1)

Applying KVL anticlockwise around loop BCDE:

E2I2 R2R3 (I1 + I2) = 0

Grouping currents and making E2 the subject gives

E2 = I1 R3 + I2 (R2 + R3)

Now substitute values to give the second simultaneous equation.

6 = 16 I1 + 20 I2 . . . (2)

Having obtained two simultaneous equations, the next step is to eliminate one of the unknown currents in order to find the other.

Multiplying (2) by 20/16 gives

7.5 = 20 I1 + 25 I2 . . . (3)

Now calculate (1) - (3) to eliminate I1.

4.5 = − 9 I2, hence I2 = − 0.5 A . . . (4)

(so the initial assumed direction of I2 was incorrect)

Now substitute (4) into (1) and solve to find I1.

12 = 20 I1 − 16 × 0.5

20 = 20 I1, hence I1 = 1 A . . . (5)

It is always a good idea to check any working by performing an alternative calculation. A simple check is to substitute calculated values (4) and (5) back into simultaneous equation (2), Hence

6 = 16 × 1 + 20 × (− 0.5), so 6 = 16 − 10

The final step is to calculate the voltage at node B, relative to node DEF (which is assumed to be at 0 V).

VB = E1I1 R1 = 12 − (1 × 4) = 8 V

Worked Example 2

Use Kirchhoff's laws to find all currents I1 and I2 in the above circuit. Hence calculate the current flowing through resistor R2 and the voltage at node B.

Consider loop ABEF:

E1E2 = I1 R1 + I1 R2I2 R2 = I1 (R1 + R2) − I2 R2

3 = 10 I1 - 4 I2 . . . (1)

Now consider loop BCDE:

E2 = − I1 R2 + I2 ( R2 + R3)

6 = − 4 I1 + 7 I2 . . . (2)

Solving these two simultaneous equations gives:

I1 = 0.833 A, I2 = 1.333 A (to 3 d.p.)

From KCL, a current of 0.5 A flows upwards through R2.

The voltage on node B, relative to node DEF (assumed to be 0 V) may be found using several different methods, but the simplest is:

VB = I2 R3 = 1.333 × 3 = 4 V