Kirchhoff laws

Kirchhoff's Laws

There are some simple relationships between currents and voltages of different branches of an electrical circuit. These relationships are determined by some basic laws that are known as Kirchhoff laws or more specifically Kirchhoff Current and Voltage laws. These laws are very helpful in determining the equivalent electrical resistance or impedance (in case of AC) of a complex network and the currents flowing in the various branches of the network. These laws are first derived by Guatov Robert Kirchhoff and hence these laws are also referred as Kirchhoff Laws.

Kirchhoff's First Law – The Current Law, (KCL)

Kirchhoff's Current Law or KCL, states that the “total current or charge entering a junction or node is exactly equal to the charge leaving the node“. In other words the algebraic sum of ALL the currents entering and leaving a node must be equal to zero, I_(entering) + I_(leaving) = 0. This idea by Kirchhoff is commonly known as the Conservation of Charge.

Kirchhoff's Second Law – The Voltage Law, (KVL)

Kirchhoff's Voltage Law or KVL, states that “In any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop”. In other words the algebraic sum of all voltages within the loop must be equal to zero. This idea by Kirchhoff is known as the Conservation of Energy.

Kirchhoff's Circuit Law Example:

Find the current flowing in the 40Ω Resistor, R₃

The circuit has 3 branches, 2 nodes (A and B) and 2 independent loops.

Using Kirchhoff's Current Law, KCL the equations are given as;
At node A :    I₁ + I₂ = I₃
At node B :    I₃ = I₁ + I₂

Using Kirchhoff's Voltage Law, KVL the equations are given as;
Loop 1 is given as :    10 = R₁ I₁ + R₃ I₃ = 10I₁ + 40I₃
Loop 2 is given as :    20 = R₂ I₂ + R₃ I₃ = 20I₂ + 40I₃
Loop 3 is given as :    10 – 20 = 10I₁ – 20I₂

As I₃ is the sum of I₁ + I₂ we can rewrite the equations as;
Eq. No 1 :    10 = 10I₁ + 40(I₁ + I₂)  =  50I₁ + 40I₂
Eq. No 2 :    20 = 20I₂ + 40(I₁ + I₂)  =  40I₁ + 60I₂

We now have two “Simultaneous Equations” that can be reduced to give us the
values of I₁ and I₂
Substitution of I₁ in terms of I₂ gives us the value of I₁ as -0.143 Amps
Substitution of I₂ in terms of I₁ gives us the value of I₂ as +0.429 Amps
As :    I₃ = I₁ + I₂

The current flowing in resistor R₃ is given as :  -0.143 + 0.429 = 0.286 Amps

and the voltage across the resistor R₃ is given as :   0.286 x 40 = 11.44 volts

The negative sign for I₁ means that the direction of current flow initially chosen was wrong, but never the less still valid. In fact, the 20v battery is charging the 10v battery.