In this topic, you study Kirchhoff’s Law – Explanation & Examples.
In the simple series and parallel circuits, it is possible to calculate the equivalent resistance and circuit or branch current by the direct application of Ohm’s law. But such a simple solution is not possible if the electric circuit is complex one with more number of branches and energy sources. Kirchhoffs laws are very useful for the solution of such complex electrical networks.
KIRCHHOFF’S LAWS
In 1847, Gustave Kirchhoff, a German physicist, formulated two fundamental laws which are of immense use in the analysis of electric networks. These laws are.
- Kirchhoff’s Current Law (KCL)
- Kirchhoff’s Voltage Law (KVL)
Kirchhoff’s Current Law (KCL)
This law relates to the currents at a junction or point of a circuit. According to this law, “the algebraic sum of all the Currents flowing towards a point or junction is equal to the sum of the currents going away from that point or junction”.
Or
“The algebraic sum of the currents meeting at a point or junction is equal to zero.”
Let Il, 12, 13 are the currents flowing towards a point O and 14, 15 are flowing away from O as shown in Fig. 7.13. So, according to this law algebraic sum of the current flowing towards the junction is equal to the currents flowing away from the junction.
Fig. 7.13. Kirchhoff’s Current law
The Kirchhoff’s law is true because Current is the flow of electrons, which cannot be accumulated without an accumulating source.
Kirchhoff’s Voltage Law (KVL)
This law relates to voltage fed and drops so it is called as the voltage law. According to this law, ‘in any closed circuit or mesh the algebraic sum of the voltage drops i.e. IR drops plus the algebraic sum of all the e.m.f.s. fed in that circuit is zero”.
Or
In any closed circuit, the sum of the e.m.f. fed is equal to the sum of all voltage drops.
As shown in Fig. 7.14, this circuit diagram, VI and V2 are the two e.m.f.s and drops are IRI, IR2 and IR R so according to the second law, i.e. voltage law.
Fig. 7.14. Kirchhoff’s voltage law
Determination of sign of e.m.f.s. and voltage drops.
In case of Kirchhoff’s second law, the voltage impressed and the IR drops are of great significance, the mesh or closed circuit should be very carefully considered and watched.
(i) Signs of e.m.f.’s. A rise in potential should be considered positive while fall in potential should be considered negative. Thus, if we go from the – ve terminal of a battery or source to the positive terminal, there is a rise in potential and it must be considered positive. On the other hand, if we go from the +ve terminal of a battery or source to the —ve terminal, there is a fall in potential and it should be considered negative. It should be noted that sign Of e.m.f. is independent of the direction of current through that branch.
(ii) Signs of voltage drops. There is a voltage drop in resistance due to the flow of current through it. If we go with the current, the voltage drop should be taken negative, because the Current flows from higher potential to lower potential i.e. fall in potential. On the Other hand, if we go against the current flow, the voltage drop should be taken positive because it is a rise in potential. It should be noted here that sign of voltage drop depends upon the direction of current and is independent of the polarity Of e.m.f. in the circuit under consideration.
Method for solving circuits by Kirchhoff’s laws
(i) Mark the direction of currents in various branches of the circuit in accordance with first law.
(ii) Choose any closed circuit and mark it.
(iii) Find the algebraic sum of voltage drops and e.m.f.s. in that circuit.
(iv) Put the algebraic sum of voltage drops plus algebraic sum of e.m.f.s. equal to zero.
Application of Kirchhoff’s Laws for Network Solutions (Branch -Current Method)
When properly applied, Kirchhoff’s laws are very useful in solving the complex circuits. As already mentioned previously, while applying the second law to a specific problem, it is necessary to adhere to a certain fixed sign convention to avoid errors.
In solving the network problems using Kirchhoffs laws, one should follow the following steps:
Draw the circuit diagram from the given descript•on. Letter it and insert the values of all resistances and e.m.f.s.
Mark on the diagram assumed currents and their directions using Kirchhoffs first law at all junctions. Keep the number of assumed currents to a minimum. The directions for the various currents can be assumed arbitrarily. However, once chosen, these directions must remain unchanged throughout the solution of the problem (after solving the equations, if the value for the particular current is found to be negative, it will indicate that the assumed direction for that current is wrong).
Noting down the directions of currents through various resistors, mark their terminals of higher and lower potentials by posit•ve and negative signs respectively. This gives a quick idea about the potential rise or drop across a resistor while applying Kirchhoffs second law.
Apply Kirchhoff’s second law to different closed loops in the network and obtain the corresponding equations. Each equation must contain some element which has not been considered in any previous equation. The law must be applied for sufficient number of loops to include every element in the network at least once. Total number of independent equations obtained in this manner must be equal to the number of unknowns. It is advisable to go round all the meshes in the same direction (clockwise or anticlockwise) so as to reduce the errors.
Solve the simultaneous equations for the unknowns. The method is illustrated by solving few problems given below.
Example 1: Determine the current in and the power supplied to 10 0 resistor in the network shown below in Fig. 218.
Fig. 2.18: Network for Example 2.15
Solution: Firstly, the circuit diagram is lettered suitably as shown in Fig. 2.18. Then assumed currents and their directions are marked on the diagram using Kirchhoff’s first law at junctions B and E. Noting down the directions of currents through various resistors, their terminals are marked with positive and negative signs as indicated. Since there are two unknowns (Il and 12), we will need two equations. Therefore, let us select following two closed circuits and apply Kirchhoffs voltage law to them.
Circuit ABCDEFA: -3011-5 (11-12) + 50 – 5(11-12) – 3011 + 100 O
Circuit ABEFA:
3011
– 7011 + 1012 -150
7011-1012 150
1012-3011 4 100 -o
-6011 – 1012 -100
6011 + 1012
100
Multiplying Equation (I) by 6 and Equation (Il) by 7, we have
420 Il -6012 900
42011 + 7012 700
Subtract•ng Equation (IV) from Equation (Ill), we get
-13012 200
200
– 1.54 A
130
Thus, the magnitude of the current flowing through the 10 Q resistor is 1.54 A. However, since its value is found to be negative, it ind•cates that assumed direction for this current is wrong. Actually, the current will flow from E to B and not from B to E as assumed.
Power consumed by 10 Q resistor