# Calculation of current circuits with the help of the first and second Kirchhoff’s laws (KCL and KVL)

**Given**

R

_{1}=16 ohms;

R

_{2}=31 ohms;

R

_{3}=24 ohms;

R

_{4}=13 ohms;

R

_{5}=33 ohms;

R

_{6}=40 ohms;

R

_{7}=22 ohms;

R

_{8}=7 ohms;

E

_{1}=30 V;

E

_{2}=24 V;

E

_{7}=16 V;

E

_{8}=11 V.

**Find**

Currents in the circuit with the help of

**Kirchhoff’s laws**.

## Solution

We make equations according to Kirchhoff’s laws.

**The first Kirchhoff’s law**says that the sum of entering and leaving currents at any node is equal to zero.**The second Kirchhoff’s law**says the algebraic sum of the potential differences in a circuit loop is equal to the sum of the electromotive force in this loop. There are m=7 branches and n=4 nodes in the given scheme. Therefore, according to the first Kirchhoff’s law n-1 = 3 equations must be made , and according to the second Kirchhoff’s law m-(n-1)=4 equations must be made. We mark randomly chosen currents directions, bypass circuits, circuit nodes.**Answer:**

I

_{1}=0,265 А; I

_{2}=0,082 А; I

_{3}=0,347 А; I

_{5}=0,131 А; I

_{6}=0,214 А; I

_{7}=0,140 А; I

_{8}=-0,273 А.

Found currents coincide with currents, calculated using the method of mesh or loop current that confirms the correctness of the solution.