Gustov Robert Kirchhoff derived two basic laws governing the distribution of electric currents in a network, one commonly known as first law or current law (KCL) or point law, whereas the second is called second law or voltage law (KVL) or mesh law.

**Law # 1. Kirchhoff’s First Law or Current Law (KCL) or Point Law:**

According to this law in any network of wires carrying currents, the algebraic sum of all currents meeting at a point (or junction) is zero or the sum of incoming currents towards any point is equal to the sum of outgoing currents away from that point. If I_{1}, I_{2}, I_{3}, I_{4}, I_{5}, and I_{6} are the currents meeting at junction O, flowing in the directions of arrowheads marked on them (Fig. 2.28); taking incoming currents as positive and outgoing currents as negative, according to Kirchhoff’s first law (KCL).

**Law # 2. Kirchhoff’s Second Law or Voltage Law (KVL) or Mesh Law:**

According to this law in any closed circuit or mesh the algebraic sum of emfs acting in that circuit or mesh is equal to the algebraic sum of the products of the currents and resistances of each part of the circuit.

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If the circuit shown in Fig. 2.29 is considered, then according to Kirchhoff’s second law (or KVL).

**Application of Kirchhoff’s Laws on Circuits: **

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First of all the current distribution in various branches of the circuit is made with directions of their flow complying with first law of Kirchhoff. Then Kirchhoff’s second law is applied to each mesh (one by one) separately and algebraic equations are obtained by equating the algebraic sum of emfs acting in a mesh equal to the algebraic sum of respective drops in the same mesh. By solving the equations so obtained unknown quantities can be determined. While applying Kirchhoff’s second law, the question of algebraic signs may be troublesome and is a frequent source of error. If, however, the following rules are kept in mind, no difficulty should occur.

The resistive drops in a mesh due to current flowing in clockwise direction must be taken positive drops.

The resistive drops in a mesh due to current flowing in counter-clockwise direction must be taken as negative drops.

Similarly the battery emf causing current to flow in clockwise direction in a mesh must be taken as positive emf and the battery emf causing current to flow in counter-clockwise direction in a mesh must be taken as negative emf.

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For example, for the circuit shown in Fig. 2.29 let the current distribution be made as shown, which satisfy Kirchhoff’s first law fully.

Taking first, mesh AFCBA for the application of Kirchhoff s second law, we see that there is only one emf acting in the mesh (E_{1}) and since it tries to send current in clockwise direction so E_{1} be taken as positive, similarly all the resistive drops i.e. R_{1} (I_{1} + I_{2}), R_{2} (I_{1} + I_{2}) and R_{5} I_{1} are clockwise, so these must be taken as positive.

.** ^{.}**. According to Kirchhoff s second law in mesh AFCBA

E_{1} = R_{1 }(I_{1} + I_{2}) + R_{2 }(I_{1} + I_{2}) + R_{5}I_{1}

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In mesh FEDCF, there is only one emf acting in the mesh (E_{2}) and since it tries to send current in counter-clockwise direction through the mesh under consideration, it may be taken as negative. Since all of the resistive drops R_{3 }I_{2}, R_{4} I_{2}, R_{2 }(I_{1} + I_{2}) and R_{1} (I_{1} +1_{2}) are counter-clockwise, these may be taken as negative.

**Hence according to Kirchhoff s second law in mesh FEDCF we get:**

– E_{2} = – R_{3 }I_{2} – R_{4 }I_{2}– R_{4} I_{2} – R_{2} (I_{1} + I_{2}) – R_{1} (I_{1} + I_{2})

or E_{2} = (R_{3} + R_{4}) I_{2} + (R_{1} + R_{2}) (I_{1} + I_{2})

ADVERTISEMENTS:

In mesh AFEDCBA, emf E_{1} tries to cause current in clockwise direction so be taken as positive and emf E_{2} tries to cause current in counter-clockwise direction so be taken as negative. Similarly resistive drop R_{5}I_{1} being clockwise be taken as positive and resistive drops R_{3 }I_{2} and R_{4 }I_{2} being counter – clockwise be taken as negative.

Hence E_{1} – E_{2} = – R_{3 }I_{2} – R_{4 }I_{2} + R_{5} I_{1} = R_{5 }I_{1}, – (R_{3} + R_{4}) I_{2}.