Electronics and schemes 1

Ministry of Information Technologies and

Communications of the Republic of Uzbekistan

Tashkent University of Information technologies

named after Muhammad al-Kharazmi

Electronics and schemes 1

Tashkent 2021

Student:Ibragimova Zuxra

Faculty: Economics and Management

Group: 124-20 Ece




Е1=40V, Е2=25V,Е3=30V, R1=5 Ohm, R2=7.5 Ohm, R3= 10 Ohm, R4=15.5 Ohm, R5=150 Ohm, R6= 17.5 Ohm
Example in Multisim

1.We can optionally define I1 – I6

2. We mark 3 units of contour I11 I22 I33 . We calculate the direction by the hour stretch.

3.We determine its resistance of the contours. To do this, we add resistance in each contour.

R11=R1+R2+R5=5+15.5+150=170.5 Ohm

R22=R2+R4+R6=7.5+15.5+17.5 =40.5Ohm

R33=R3+R5+R6=10+150+17.5=177.5 Ohm

Then we determine the common resistances, common resistances are easy to detect, they belong to several circuits at once, for example, the resistance R4 belongs to contour 1 and contour 2. Therefore, for convenience, we designate such resistances by the numbers of the contours to which they belong.

R12=R21=R4=15.5 Ohm

R23=R32=R6=17.5 Ohm

R31=R13=R5=150 Ohm

4. We proceed to the main stage - the compilation of a system of equations of contour currents. The left part of the equations includes voltage drops in the circuit, and in the right the EMF of the sources of this circuit.

Since we have three contours, therefore, the system will consist of three equations. For the first contour , the equation will look like this:

I11*R11-I22*R21-I33*R31=E1

The current of the first circuit I11, multiplied by the intrinsic resistance R11 of the same circuit, and then subtract the current I22 multiplied by the total resistance of the first and second circuits R21 and the current I33 multiplied by the total resistance of the first and third circuits R31. This expression will be equal to the EMF E1 of this circuit. The EMF value is taken with a plus sign, since the direction of the bypass (clockwise) coincides with the direction of the EMF, otherwise it would have to be taken with a minus sign.

Based on these, we get:

I11=1.86 A , I22=0.53, I33=1.79

I1=I11=1.86

-I2=-I22=-0.53

I3=I33=1.79

I4=I11-I22=1.33

I5=I11-I33=0.07

I6=I33-I22=0.53