**Kirchhoff’s Current (KCL)**

Kirchhoff’s Current Law (KCL) which is commonly referred to as Kirchhoff’s first law states that the total current flowing into a node (or a junction) is equal to the total current flowing out of it. In other words the algebraic sum of ALL the currents entering and leaving a node (junction) must be equal to zero as: Σ I_{IN} = Σ I_{OUT}. This concept is known as the conservation of charge.

Based on Kirchhoff’s Current Law, the sum of the current entering node (n) above is equal to the sum of the current leaving node (n). This can be represented algebraically as:

At node n,

Recalling that current is a signed (positive or negative) quantity reflecting direction towards or away from a node, this principle can be succinctly stated as:

The term node in an electrical circuit generally refers to a connection or junction of two or more current carrying paths or elements such as cables and components. Also for current to flow either in or out of a node a closed circuit path must exist.

**Kirchhoff’s Voltage Law (KVL)**

Kirchhoff’s Voltage Law (which is also referred to as Kirchhoff’s second law or Kirchhoff’s loop (or mesh) rule) states that the sum of all potential differences (voltages) round a close loop is equal to zero. This is because a circuit loop is a closed conducting path so no energy is lost.

In other words the algebraic sum of ALL the potential differences around the loop must be equal to zero as: ΣV = 0. Note here that the term “algebraic sum” means to take into account the polarities and signs of the sources and voltage drops around the loop.

To avoid wrong calculations, it is important that when applying Kirchhoff’s voltage law to a specific circuit element, we must pay special attention to the algebraic signs, (+ and -) of the voltage drops across elements and the emf’s of sources otherwise.

**Applying Kirchhoff’s Voltage Law to a single loop**

Analysing the above circuit using Kirchhoff’s voltage law which states that the algebraic sum of the potential differences in any loop must be equal to zero as: ΣV = 0. Since the three resistors, R_{1}, R_{2} and R_{2 }are wired together in a series connection, they are both part of the same loop so the same current must flow through each resistor.

Thus the voltage drop across resistor R_{1} = I*R_{1} and the voltage drop across resistor, R_{2} = I*R_{2} and the voltage drop across resistor, R_{3 } = I* R3

Therefore applying KVL we have:

From the analysis above, we can see that applying Kirchhoff’s Voltage Law to this single closed loop produces the formula for the equivalent or total resistance in the series circuit and we can expand on this to find the values of the voltage drops around the loop.

**Worked example using Kirchhoff’s Voltage Law in a single loop**

**Example 1**

Three resistors of values: 20 ohms, 30 ohms and 50 ohms, respectively are connected in series across a 24V battery supply.

Calculate:

a) The total resistance,

b) The circuit current,

c) The current through each resistor,

d) The voltage drop across each resistor,

e) Verify that Kirchhoff’s voltage law, KVL holds true.

**Solution**

**c.) Current through each resistor**

Since the three resistors are connected in series, they are all part of the same loop and therefore each experience the same amount of current. Thus:

I = I_{R1} + I_{R2} + I_{R3} = 240mA

**d.) Voltage drop across each resistor**

V_{R1} = I x R_{1} = 0.24 x 20 = 4.8V

V_{R2} = I x R_{2} = 0.24 x 30 = 7.2V

V_{R3} = I x R_{3} = 0.24 x 50 = 12V

**e.) Verify Kirchhoff’s Voltage Law**

V_{s} + (- IR_{1}) + (- IR_{2}) + (- IR_{3}) = 0

24 + (- 4.8) + (- 7.2) + (- 12) = 0

24 – 4.8 – 7.2 – 12 = 0

From the above, Kirchhoff’s voltage law holds true since the total voltage drop in the close loop is equal to zero.