# 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 Gustav Kirchhoff (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 I_{1} + I_{2} = I_{3} + I_{4} + I_{5}, or I_{1} + I_{2} − I_{3} − I_{4} − I_{5} = 0.

In general, for a system with k currents acting at a node, sequentially numbered I_{1}, I_{2}, ... I_{k} then

Aprincipal nodeis a point in a circuit where three or more conductors meet, while asecondary nodeis 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 E_{1} = V_{1} + V_{2} + V_{3} + V_{4}, or E_{1} − V_{1} − V_{2} − V_{3} − V_{4} = 0. The potential differences may also be expressed as the product of current and resistance, so the original expression can be written as

E_{1} = I R_{1} + I R_{2} + I R_{3} + I R_{4}, or E_{1} = I (R_{1} + R_{2} + R_{3} + R_{4}).

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 V_{1}, V_{2}, ... V_{k}, 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*.

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The first step is to 'choose' the assumed directions for currents I_{1} and I_{2} flowing in loops *ABEF* and *BCDE* respectively.

Currents I_{1} and I_{2} both flow into node *B*, hence from KCL, the current flowing downwards through R_{3} is given by (I_{1} + I_{2}).

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 I_{1} and I_{2}.

**Applying KVL clockwise around loop ABEF:**

E_{1} - I_{1} R_{1} − R_{3} (I_{1} + I_{2}) = 0

Grouping currents and rearranging to make E_{1} the subject gives

E_{1} = I_{1} (R_{1} + R_{3}) + I_{2} R_{3}

Substituting values gives the first simultaneous equation.

12 = 20 I_{1} + 16 I_{2} . . . (1)

Applying KVL anticlockwise around loop *BCDE*:

E_{2} − I_{2} R_{2} − R_{3} (I_{1} + I_{2}) = 0

Grouping currents and making E_{2} the subject gives

E_{2} = I_{1} R_{3} + I_{2} (R_{2} + R_{3})

Now substitute values to give the second simultaneous equation.

6 = 16 I_{1} + 20 I_{2} . . . (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 I_{1} + 25 I_{2} . . . (3)

Now calculate (1) - (3) to eliminate I_{1}.

4.5 = − 9 I_{2}, hence I_{2} = − 0.5 A . . . (4)

(so the initial assumed direction of I_{2} was incorrect)

Now substitute (4) into (1) and solve to find I_{1}.

12 = 20 I_{1} − 16 × 0.5

20 = 20 I_{1}, hence I_{1} = 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).

V_{B} = E_{1} − I_{1} R_{1} = 12 − (1 × 4) = 8 V

#### Worked Example 2

Use Kirchhoff's laws to find all currents I_{1} and I_{2} in the above circuit. Hence calculate the current flowing through resistor R_{2} and the voltage at node *B*.

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Consider loop *ABEF*:

E_{1} − E_{2} = I_{1} R_{1} + I_{1} R_{2} − I_{2} R_{2} = I_{1} (R_{1} + R_{2}) − I_{2} R_{2}

3 = 10 I_{1} - 4 I_{2} . . . (1)

Now consider loop *BCDE*:

E_{2} = − I_{1} R_{2} + I_{2} ( R_{2} + R_{3})

6 = − 4 I_{1} + 7 I_{2} . . . (2)

Solving these two simultaneous equations gives:

I_{1} = 0.833 A, I_{2} = 1.333 A (to 3 d.p.)

From KCL, a current of 0.5 A flows upwards through R_{2}.

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:

V_{B} = I_{2} R_{3} = 1.333 × 3 = 4 V