# Loop method, example of the problem solving

**Given**

R

_{1}=16 ohms;

R

_{2}=31 ohms;

R

_{3}=24 ohms;

R

_{4}=13 ohms;

R

_{5}=33 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 branch by

**the loop method**.

## Solution

Mark up randomly chosen directions of the currents, bypass circuits, circuit nodes.

Make matrix equation of the loop currents.

I

I

I

I

Having found all

I

I

I

I

I

I

I

Found currents coincide with currents calculated using Kirchhoff’s laws that confirms the correctness of the solution

**(Z)(I)=(U)**, where**(Z)**— matrix of loop resistances;**(I)**— matrix of unknown loop currents;**(U)**— matrix of the electromotive force loops.I

_{I}=0,265 А;I

_{II}=0,347 А;I

_{III}=0,133 А;I

_{IV}=0,273 А.Having found all

**loop currents**, we show through them currents in branches:I

_{1}=I_{I}=0,265 А;I

_{2}=I_{II}-I_{I}=0,347-0,265=0,082 А;I

_{3}=I_{II}=0,347 А;I

_{5}=I_{I}-I_{III}=0,265-0,133=0,132 А;I

_{6}=I_{II}-I_{III}=0,347-0,133=0,214 А;I

_{7}=I_{IV}-I_{III}=0,273-0,133=0,140 А;I

_{8}=-I_{IV}=-0,273 А.Found currents coincide with currents calculated using Kirchhoff’s laws that confirms the correctness of the solution