Tashkent university of information technologies named after muhammad al-khwarizmi

TASHKENT UNIVERSITY OF INFORMATION TECHNOLOGIES NAMED AFTER MUHAMMAD AL-KHWARIZMI

Department of "Electronics and Radoi Engineering"

INDEPENDENT WORK
Topic. Methods for calculating electrical circuits

Performed : 214-20 Abdullayev Sh.Sh.

Checked by: Narkulov Kh.Yu.
Tashkent-2021

1. 
   Solve the problem
   

Е1=80V, Е2=50V,Е3=60V, R1=10 Ohm, R2=15 Ohm, R3= 20 Ohm, R4=25 Ohm, R5=30 Ohm, R6= 35 Ohm

1. Select the direction of the currents, 2. Define the contours and indicate the direction of the loop currents. 3. Determine the intrinsic and total loop resistance. 4. Determine the currents I1, I2, I3, I4, I5, I6 by the loop current method.

So, after getting acquainted with the theory, we suggest starting to practice! Let's look at an example.

Given: E1=80 V E2=50 V E3=60 V R1=10 Ohm R4=25 Ohm R2=15 Ohm R5=30 Ohm R3=20 Ohm R6=35 Ohm

FIND: I1,I2,I3,I4,I5,I6 - ?

We carry out everything in stages.

1.  Arbitrarily choose the direction of the actual currents I ₁ -I ₆ .

2. 
   Select three contours, and then indicate the direction of the loop currents I ₁₁ , I ₂₂ , I ₃₃ . We will
   

choose a clockwise direction.

3.  Determine the intrinsic resistances of the contours. To do this, add up the resistances in each circuit.

R ₁₁ = R ₁ + R ₄ + R ₅ = 10 + 25 + 30 = 65 Ohm

R ₂₂ = R ₂ + R ₄ + R ₆ = 15 + 25 + 35 = 75 Ohm

R ₃₃ = R ₃ + R ₅ + R ₆ = 20 + 30 + 35 = 85 Ohm

Then we determine the total resistances, the total resistances are easy to find, they belong to several circuits at once, for example, the resistance R ₄  belongs to circuit 1 and circuit 2. Therefore, for convenience, we will designate such resistances by the numbers of the circuits to which they belong.

R ₁₂ = R ₂₁ = R ₄ = 25 Ohm

R ₂₃ = R ₃₂ = R ₆ = 35 Ohm

R ₃₁ = R ₁₃ = R ₅ = 30 Ohm

4. We  proceed to the main stage - drawing up a system of equations for loop currents. On the left side of the equations, there are voltage drops in the circuit, and in the right EMF of the  sources of  this circuit.

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

The current of the first circuit I ₁₁ , we multiply by its own resistance R _{11 of}  the same circuit, and then we subtract the current I ₂₂ , multiplied by the total resistance of the first and second circuits R ₂₁  and the current I ₃₃ , multiplied by the total resistance of the first and third circuits R ₃₁ . This expression will be equal to the EMF E _{1 of}  this circuit. We take the EMF value with a plus sign, since the bypass direction (clockwise) coincides with the EMF direction, otherwise it would have to be taken with a minus sign.

We do the same actions with two other circuits and, as a result, we get the system:

We substitute the already known resistance values ​​into the resulting system and solve it in any known way.

5.  The last step is to find the real currents, for this you need to write down expressions for them.

The loop current is equal to the actual current that belongs only to this loop . That is, in other words, if the current flows only in one circuit, then it is equal to the circuit.

But, you need to take into account the direction of the bypass, for example, in our case, the current I2 does not coincide with the direction, so we take it with a minus sign.

The currents flowing through the common resistances are defined as the algebraic sum of the contour ones, taking into account the direction of the bypass.

For example, through a resistor R _4,  a current flows I ₄ , its direction coincides with the direction of traversal of the first circuit and the second circuit opposite direction. This means that for him the expression will look like

And for the rest