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Example of problem solution




Example 3 The electric circuit consists of two sources of EMF e1=20 V, e2=5 V and three resistors R 1, R 2=19W and R 3=10W. A current I 1=0,2 flows through the inner resistances of the sources r 1=2W, r 2=1W and through the resistor R 1, at the direction, shown in the picture.

Find: 1) resistance R 1 and a current, which flows through the resistors R 2 and R 3;

2) the potential difference between the points and .

Input data: e 1=20 V e 2=5 V r 1=2 W r 2=1 W R 2=19 W R 3=10 W I 1=0,2  
Find: I 2, I 3, R 1, DjA?

Solution:

1. Lets use the Kirchhoffs rules for the solution of the branched chain.

In order to find one magnitude of resistance and two magnitudes of a current, its necessary to make three equations. Before compiling the equations its necessary arbitrarily to choose:

a) direction of currents (if they arent set in the condition); b) direction of path-tracing.

The direction of a current I 1 is set, and lets choose directions of currents I 2 and I 3, as its shown on the scheme. Lets agree to trace-paths clockwise (dashed line on the scheme). The given scheme has two junctions: and . In order to compile the equations with the first Kirchhoffs rule its necessary to take into account, that a current, which flows in the junction, enters to the equation with plus-sign, and it is necessary to write a current, which flows out of the junction with minus-sign.

With the first Kirchhoffs rule for the junction

(1)

It is no sense to compile the equation for the junction , as it reduces to the equation (1).

Well get two more necessary equations with the second Kirchhoffs rule. It is necessary to follow the sign rules: a) the voltage drop (product IR or Ir) enters to the equation with plus-sign, if the currents direction coincides with the direction of path-tracing, in other case with minus-sign; b) EMF enters to the equation with plus-sign, if it enlarges the potential in the direction of path-tracing (pass through the source from minus to plus), in other case with minus-sign.

With the second Kirchhoffs rule for the closed loop and :

; (2)

. (3)

The set magnitudes inserted in the equation (1), (2), (3), well get the set of equations

 

Lets find I 3 from the equation (4) and insert to the equation (6)

.

Whence

Minus-sign in the sense of a current I 2 means, that the currents direction I 2 has been chosen reverse to the acting. In reality current I 2 runs from junction to the junction .

From the equation (4) we search out I 3:

From the equation (5) we search out R 1

W.

2. The difference of potential U=DjA,B=j BjA can be found, if we use Ohms law for Non-uniform site of a chain (subcircuit) in a proper way, for a example

(7)

In Ohms law its taken into consideration, that positive direction of currents strength coincides with the direction of foreign forces work of the source, which fits the enlarging of the potential. Then the required potential difference is

We make the calculations

.

Results: I2 = 0,1 A; I 3 = 0,3 ; R 1=83 W, .

 

 

Individual tasks for PROBLEM 1.5.

BRANCHED CIRCUITS

 

To compose the scheme from three adjacent branches, which are shown in the picture 4. The numbers of branches, EMF of sources e i, the inner resistance of sources r, the external resistance of branches R i (or the current , which flows along one of the branches from the point to ) set with the variants in the table. 1.5. Find: 1) magnitudes of quantities, which are indicated in the last column of the Table 1.5;

2) the potential difference from the points and .

Example of the scheme, which corresponds 25th variant is shown in the picture 4 .

TABLE OF TASK VARIANTS

Table 1.5

Variant The branch number εi, V ri , W Ri , W , Find
  1, 2, 3 ε1 =11, ε2 =4, ε3 =6 r1= r2 = r3 =0 R1 =25, R2 =50, R3 =10 1, 2, 3
  4, 5, 6 ε4 =9, ε5 =10 r4 =1, r5 =2 R4 =19, R5 =38 6 =0,1 4, 5, R6
  1, 2, 4 ε1 =16, ε2 =5, ε4 =7 r1= r2 = r4 =0 R2 =30, R4 =50 1 =0,4 2, 4, R1
  5, 4, 1 ε1 =9, ε4 =6, ε5 =2 r1= r4 = r5 =0 R4 =50, R5 =10 1 =0,2 4, 5, R1
  1, 2, 6 ε1 =10, ε2 =8 r1 =2, r2 =1 R1 =8, R2 =19, R6 =60 1, 2, 6
  3, 2, 1 ε2 =4, ε3 =5 r1= r2 = r5 =0 R1 =30, R2 =40, R3 =20 1 =0,1 2, 3, ε1
  1, 4, 6 ε1 =8, ε4 =2 r1 =2, r4 =1 R1 =18, R4 =39, R6 =80 1, 4, 6
  1, 4, 2 ε2 =11, ε4 =7 r1= r2 = r4 =0 R1 =50, R2 =20, R4 =30 1 =0,1 2, 4, ε1
  2, 1, 3 ε1 =9, ε2 =8, ε3 =1 r1= r2 = r5 =0 R1 =50, R2 =20, R3 =10 1, 2, 3
  4, 1, 5 ε4 =4, ε5 =2 r1= r4 = r5=0 R1 =25, R4 =50, R5 =10 1 =0,4 4, 5, ε1
  1, 3, 2 ε2 =16, ε3 =3 r1= r2 = r5 =0 R1 =70, R2 =20, R3 =10 1 =0,1 2, 3, ε1
  6, 4, 1 ε1 =3, ε4 =7 r1 =2, r4 =1 R1 =78, R4 =39 6 =0,1 1, 4, R6
  5, 4, 1 ε4 =4, ε5 =14 r1= r4 = r5=0 R1 =90, R4 =20, R5 =40 1 =0,1 4, 5, ε1
  4, 6, 5 ε4 =10, ε5 =5 r4 =2, r5 =1 R4 =33, R5 =19 6 =0,3 4, 5, R6
  1, 6, 4 ε1 =4, ε4 =3 r1 =2, r4 =1 R1 =18, R4 =9, R6 =60 1, 4, 6
  4, 1, 6 ε1 =2, ε4 =12 r1 =3, r4 =2 R1 =97, R4 =18 6 =0,1 2, 4, R6
  4, 1, 5 ε1 =22, ε4 =8, ε5 =4 r1= r4 = r5=0 R1 =25, R4 =50, R5 =10 1, 4, 5
  2, 1, 6 ε1 =20, ε2 =6 r2 =1 R1 =82, R2 =29, R6 =10 1 =0,2 2, 6, r1
  2, 3, 1 ε1 =19, ε2 =4, ε3 =5 r1= r2 = r3 =0 R2 =20, R3 =10 1 =0,2 2, 3, R1
  4, 1, 6 ε1 =13, ε4 =1 r4 =1 R1 =27, R4 =24, R6 =40 1 =0,3 4, 6, r1
  2, 1, 4 ε1 =12, ε2 =9, ε4 =5 r1= r2 = r4 =0 R1 =30, R2 =60, R4 =20 1, 2, 4
  2, 1, 6 ε1 =8, ε2 =6 r1 =3 R1 =27, R2 =9, R6 =25 2 =0,1 1, 6, r2
  5, 1, 4 ε1 =19, ε4 =6, ε5 =2 r1= r4 = r5=0 R4 =50, R5 =10 1 =0,2 4, 5, R1
  1, 6, 2 ε1 =18, ε2 =15 r1 =2, r2 =1 R1 =58, R2 =9, R6 =30 1, 2, 6
  4, 1, 2 ε2 =4, ε4 =2 r1= r2 = r4 =0 R1 =50, R2 =20, R4 =80 1 =0,2 2, 4, ε1
  1, 6, 5 ε1 =8, ε5 =6 r1 =2, r5 =3 R1 =8, R5 =12, R6 =10 - 1, 5, 6
  2, 4, 5 ε2 =8 r2 =2, r4 =1, r5 =5 R2 =18, R4 =14, R5 =25 4=0,2, 5=0,3 2, ε4, ε5
  3, 6, 4 ε3 =36, ε4 =9 r3 =2, r4 =1 R3 =16, R4 =8 6 =0,5 4, 3, R6
  3, 1, 5 ε3 =40, ε5 =30 r1=r5=2, r3=5 R3 =35, R1 =28, R5 =28 1 =0,7 5, 3, ε1
  2, 3, 4 ε2=20, ε4=40, ε3=10 r2=10, r4=15, r3 =5 R2 =110, R4 =105 3 =0,2 4, 2, R3




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